# AP Inter 1st Year Chemistry Important Questions Chapter 6 Thermodynamics

Students get through AP Inter 1st Year Chemistry Important Questions 6th Lesson Thermodynamics which are most likely to be asked in the exam.

## AP Inter 1st Year Chemistry Important Questions 6th Lesson Thermodynamics

Question 1.
What is the information given by the term thermodynamics?
Thermodynamics gives the infonnation about energy changes (chemical energy, electrical energy, nuclear energy) involved in a chemical reaction.

Question 2.
What is the relationship between the laws of thermodynamics and equilibrium state?
Laws of thermodynamics are applicable only when a system is in equilibrium (or) moves from one equilibrium state to another equilibrium state.

Question 3.
Define a system. Given an example.
System:
A part of the universe chosen for thermodynamic study is called system.
Ex: Ice in a beaker.

Question 4.
The wall is adiabatic and AU = Wat). What do you understand about the heat and work with respect to the system.
In adiabatic process, no transfer of heat takes place between system and surroundings.
∴ ∆q = 0.
Work done in adiabatic process is equal to change in internal energy of the system.

Question 5.
The system loses ‘q’ amount of heat though no work is done on the system. What type of wall does the system have?
Here ∆U = -q. Hence, heat transfer takes place. So the wall is thermally conducting wall. Question 6.
Work is done by the system and ‘q’ amount of heat is supplied to the system. What type of system would it be?
∆U = q-w. It is a closed system.

Question 7.
What is the work done in the free expansion of an ideal gas in reversible and irreversible processes?
Expansion of a gas into vaccum is called free expansion.

No work is done during free expansion of an ideal gas whether the process is reversible or irreversible. Because the external pressure applied (Pex = 0) is zero. Here i.e., W = 0

Question 8.
From the equation ∆U = q-pex∆V, if the volume is constant. What is the value of ∆U?
If the volume V is constant then ∆V = 0
∴ ∆U = q-pex ∆V ⇒ ∆U = q – pex(0) = q-0 = q
Hence, internal energy change is equal to the heat supplied.

Question 9.
In isothermal free expansion of an ideal gas find the value of q and ∆U?
We know ∆U = q + w.
In isothermal free expansion of a gas, we have w = 0 and q = 0
∴ ∆U = 0 + 0 = 0

Question 10.
In isothermal irreversible change of ideal gas what is the value of q?
In isothermal process, internal energy change is zero (∆U = 0).
∆U = q + w = 0 ⇒ q = -w
In the case of an isothermal irreversible change q = -w = -Pex(Vf – Vi) = -Pex ∆V Question 11.
In isothermal reversible change of an ideal gas, what is the value of q?
For isothermal reversible change
q = -w = nRT ln $$\frac{V_f}{V_i}$$ = 2.303 nRT log $$\frac{V_f}{V_i}$$

Question 12.
For an adiabatic change in an ideal gas what is the relationship between its ∆U and w (adiabatic)?
For an adiabatic change q = 0
∆U= q + w= 0 + w = w
∴ Change in internal energy = Work done in the adiabatic process.

Question 13.
State the first law of the thermodynamics. [AP 16][TS 18, 22]
First Law of thermodynamics:
The law of conservation of energy is taken as the first law of thermodynamics.

Statements:

1. “Energy can neither be created nor destroyed, but it can be converted from one form to another”.
2. “The total energy of a system and its surroundings is constant”.
3. “It is impossible to construct perpetual motion machine of first kind”.

Question 14.
What are the sign conventions of the work done on the system and work done by the system?
‘+w’, stands for work done ‘on the system’.
‘-w’, stands for work done ‘by the system’.

Question 15.
Volume (V), Pressure (P) and Temperature (T) are state functions. Is the statement true?
Volume (V), Pressure (P) and Temperature (T) are state functions. These do not depend upon path of reaction, but only depend upon the state of reaction.

Question 16.
What are the heat (q) sign conventions when heat is transferred from the surroundings to the system and that transferred from system to the surrounding?
‘+q’ is taken when heat is transferred from surroundings to the system.
‘-q’ is taken when heat is transferred from system to the surroundings. Question 17.
No heat is absorbed by the system from the surroundings, but work (W) is done on the system. What type of wall does the system have?
When no heat is absorbed, and work is done on the system, then the system consists of adiabatic wall.

Question 18.
No work is done on the system, but heat (q) is taken from the system by the surroundings. What type of wall does the system have?
When no work is done, but heat is transferred, then the system contains thermally conducting wall.

Question 19.
Work is done by the system and heat (q) is supplied to the system. What type of system would it be?
If work is done by the system and heat is supplied to the system then ∆U = q – w. It is a closed system.

Question 20.
q = w = -Pext (vf – vi) is for irreversible …….. change.
The given equation holds true for irreversible isothermal change.

Question 21.
q = – w = nRT In (vf / vi) is for isothermal ………. change.
The given equation holds true for isothermal reversible change.

Question 22.
What are the ‘∆H’ sign conventions for exothermic and endothermic reactions? [TS 16]
∆H = -ve ⇒ Exothermic reactions.
∆H = +ve ⇒ Endothermic reactions.

Question 23.
What are intensive and extensive properties? [AP 15, 19]
Intensive properties :
The properties which do not depend on the total amount of substance are called intensive properties.
Ex: Density, viscosity, specific heat, temperature, pressure, vapour pressure etc.

Extensive properties :
The properties which depend on the total amount of substance are called extensive properties.
Ex: Mass, Volume, Internal energy, enthalpy, entropy, heat capacity, refractive index etc. Question 24.
In the equation q = c.m. ∆T. If ∆T is change in temperature ‘m’ mass of the substance and ‘q’ is heat required, What is ‘c’?
In the equation q = c × m × ∆T, the term ‘c’ indicates specific heat.

It is the amount of heat required to raise the temperature of 1 gm of the substance by 1°C.

Question 25.
Give the equation that gives the relationship between ∆U and ∆H.
The required equation: ∆H = ∆U + AnRT
∆H = Change in Enthalpy
∆U = Change in internal energy
∆n = nP – nR,
R = Universal gas constant
T = Temperature.

Question 26.
What is the relationship between Cp and Cv?
Cp – Cv = R
Cp = Heat capacity at constant pressure per mole.
Cv = Heat capacity at constant volume per mole
R = Universal gas constant.

Question 27.
1 g of graphite is burnt in a bomb calorimeter in excess of O2 at 298K and 1 atm pressure, according to the equation. C(graphite) + O2(g) → CO2(g) During the reaction the temperature rises from 298K to 299K. Heat capacity of the bomb calorimeter is 20.7KJK-1. What is the enthalpy change for the above reaction at 298K and 1 atm?
Heat evolved in the reaction
∆U = qv = ms∆T = 12 × 20.7 × 1 = 248.4kJ/mol
In the given reaction: ∆ng = 0
∴ ∆H = ∆U= -2.48 × 10² kJmol-1.

Question 28.
For the above reaction what is the internal energy change, ∆U?
For the above reaction Internal energy change ∆U = -2.48 × 10² kJmol-1. Question 29.
What is ∆rH for
CH4(g) + 2O2(g) → CO2(g) + 2H2O(I)
interms of molar enthalpies of the respective reactants and products?
rHf = {∆Hf(CO2) + 2∆Hf(H2O)} – {∆Hf (CH4) + 2∆Hf(O2)}

Question 30.
Enthalpy decrease is not the criterion for spontaneity. Why?
Decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases.
Because some endothermic reactions are spontaneous at high temperatures.
C(gra) +2S → CS2(l); ∆H = +128.5 KJ/ mole

Question 31.
Is increase of entropy, the criterion for spontaneity? Why?
No. If the entropy change is positive, the process is spontaneous. For an isolated system, the change in energy remains’ constant. Therefore increase in entropy is not a necessary and sufficient condition for the spontaneous nature of a reaction.

Question 32.
Explain the relationship between Gibbs energy change and equilibrium constant.
∆G° = -2.303RTlog10 Keq.
∆G = Change in Gibbs energy
K = Equilibrium constant

Question 33.
If we measure ∆H° and ∆S° it is possible to estimate ∆G°. Is it true? Why?
Yes. ∆G° = ∆H° – T∆S°
If the values of ∆H° and ∆S° are measured then we find ∆G° from the above equation.

Question 34.
Equilibrium constant ‘K’ is measured accurately in the laboratory at given temperature. Is it possible to calculate ∆G° at any other temperature? How?
Yes.
∆G° = -2.303RTlog10 Keq
Using the above formula, we can calculate the value of ∆G° at a given temperature.

Question 35.
Comment on the thermodynamic stability of NO(g) given that
$$\frac{1}{2}$$N2(g) + $$\frac{1}{2}$$O2(g) → NO(g). ∆H° =90Kj mol-1
NO(g) + $$\frac{1}{2}$$O2(g) → NO2(g) ; ∆H° = -74KJmol-1
For NO(g): ∆H0f = +ve
The positive value of ∆H indicates that heat is absorbed during the formation of NO(g)
This means that NO(g) has higher energy than the reactants (N2 and O2)
Hence, NO(g) is unstable.

For NO2(g) : ∆H0f = -ve
The negative value of ∆H indicates that heat is evolved during the formation of NO2(g) The product NO2(g) is stabilised with minimum energy.
Hence, unstable NO(g) changes to stable NO2(g) Question 36.
Calculate the entropy change in surroundings when 1.00 mole of H2O(l) is formed under standard conditions ∆H0f = -286 KJmol-1.
qrev = -∆H0f = -(286) = 286kJmol-1
∆Ssurrounding =$$\frac{q_{rev}}{T}=\frac{286}{298}$$= 0.959kJmol-1K-1

Question 37.
The equilibrium constant for a reaction is 10. What will be value of ∆G°?
R = 8.3|4J/Kmol, T = 300K. [Imp.Q]
Given K = 10, R = 8.314J/K.mole, T = 300K
Formula: ∆G° = -RTlnK
= – 2.303RT log K
= -2.303 × 8.314 × 300 × log 10
= – 5774.14J/mole

Question 38.
State third law of thermodynamics. [AP 19][TS 16]
Third law of thermodynamics :
“The entropy of a pure and perfectly crystalline substance approaches zero when the temperature approaches absolute zero.”
Mathematically, ST = $$\int_0^T \frac{C p}{T} \cdot d T$$

Question 1.
What are open, closed and isolated systems? Give one example for each?
(a) Open system:
A system which can exchange both matter and energy with its surroundings is an open system
Ex: A cup of tea in a saucer.
Here the system can exchange both matter and heat with its surroundings.

(b) Closed system:
A system which can exchange energy but not matter with its surroundings is a closed system. [AP 19]
Ex: Chilled sealed drink bottle.
Here the system can exchange only energy heat with the surroundings.

(c) Isolated system:
A system in which neither matter nor energy is exchanged with the surroundings is an isolated system
Ex: Hot Coffee in a thermos flask.
In this system there is no exchange of either matter (or) heat with the surroundings.

Question 2.
Define the state function and state variables. Give examples,
State function(Z):
The function (Z) that describes the state of a system by means of certain measurable bulk properties (P, T,…) which depend only on initial and final states of the system, but not on the path that they are reached, is called state function. Z = f(P,T..)

State variables:
The variables involved in defining the state function are called state variables of the system.
The state variables of the gaseous system are Pressure (P), Volume (V), Temperature (T), Mass (m)
The state functions of the thermodynamic system are internal energy (U), enthalpy (H), Gibbs energy (G), entropy (S).

Question 3.
“Internal energy is a state function.” Explain.
Internal energy(U):
The total energy (like chemical, electrical, nuclear) stored in a substance at constant temperature and volume is called its Internal energy.
From the first law of thermodynamics, we have ∆U = q + w.
In case of an adiabatic process, heat transfer does not take place (∆q = 0).
Here, ∆U = U2 – U1 = wad

Thus, internal energy is a state function, whose values depend only on the initial (U1) and final (U2) states of the system.

Question 4.
“Work is not a state function.”-Explain.
The work done during a process does not depend merely on the initial and the final states of the system but it depends upon the path followed.

Suppose we wish to change the system from A to B. This may be done by two different paths as shown in the following graphs. From the graphs, it is clear that areas under the Curve are different and hence the work done is different when different paths are followed.
∴ Work is not a state function.

Question 5.
What is heat? Explain.
Heat(q) Heat is a form of energy.
Heat flows between system and surroundings due to difference in temperatures.
Units : Calorie (cal), Joule (J).

Signs of q:
‘+q’ when heat is transferred from surrounding to the system.
‘-q’ when heat is transferred from system to the surroundings.

Heat is not a state function, because the values of heat donot depend only on the initial and final states of the system.

Heat is an extensive property. Its value depends on the total mass of the substance. Question 6.
Derive the equation for ‘Wrev‘ in isothermal reversible process.
Work done(w) during the expansion of a gas, with a small increment ∆V in volume, against external pressure Pext is given by
w = Pext(-∆V) = -Pext(Vf – Vi)

In an isothermal reversible process the total work done wrev, when the gas expands from an initial volume Vi and final volume Vf is Question 7.
Two liters of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 20 litres. How much heat is absorbed and how much work is done in the expansion?
Given data:
Initial volume V1 = 2 lit, Final volume V2 = 201i
Initial pressure P = 10 atm
Since the gas expands into vacuum, external pressure Pext = 0.
Now, heat absorbed q = -w = Pex (V2 – V1)
= 0 (20 – 2) = 0(18) = 0
∴ Heat absorbed q = 0 and work done w = 0.

Question 8.
If the ideal gas given in the above problem expands against constant external pressure of 1 atm what is the q value?
Here Pext = 1 atm
∴ q = -w = Pex (V2 – V1)
= 1(20 – 2) = 18 lit atm.

Question 8.
If the ideal gas given in the above problem expands to a final volume of 10L conducted reversibly what is q value?
Here we take P = 10 atm, V = 2 lit
From ideal gas equation we have PV = nRt
⇒ nRT = PV= 10 × 2 = 20
For reversible isothermal expansion,
q = – w = 2.303nRT log $$\frac{V_2}{V_1}$$
= 2.303 × 20 log $$\frac{20}{2}$$ = 32.19594 6 2
∴ q = 32.19594 lit atm.

Question 10.
Explain the state function ‘enthalpy H’. What is the relationship between ∆U and ∆H?
Enthalpy (H):
The amount of heat exchanged by a system with its surroundings at constant pressure and temperature is called enthalpy.
H = U + PV

The energy change taking place at constant pressure and at a constant temperature is called enthalpy change (∆H).
Mathematically, ∆H = ∆U + P∆V.
Also ∆H = [Hproducts – HReactants]

Thus, Enthalpy change is a state function, because the magnitude of enthalpy change depends only upon the initial and final states of the system.
For gaseous reactions, ∆H = ∆U + ∆n(g)RT.

Question 11.
Show that ∆H = ∆U + ∆n(g)RT
For a gaseous system, let
VA = Total volume of gaseous reactants;
VB = Total volume of gaseous products.
nA = No. of moles of gaseous reactants;
nB = No. of moles of gaseous products.
From the ideal gas equation, we have
PVA = nART and PVB = nBRT
⇒ PVB – PVA = nBRT – nART
⇒ P(VB – VA) = ( nB – nA)RT
⇒ P∆V = ∆n(g)RT
We know that ∆H = ∆U + P∆V
∴ ∆H = ∆U + ∆n(g)RT Question 12.
If wafer vapour is assumed to be a perfect gas, molar enthalpy change for vapouration of 1 mole of water at 1 bar and 100°C is 41 kJmol-1 Calculate the internal energy change when
a) 1 mol of water is vapourised at 1 bar and 100°C
b) 1 mol of water is liquid is converted into ice. Question 13.
Explain extensive and intensive properties.
Measurable properties such as mass, pressure, volume, temperature, surface tension, viscosity etc., are sub divided into two categories.

1) Extensive properties :
The properties whose magnitude depends upon the total quantity of the matter present in the system are called extensive properties. These are mass dependent properties.
Ex: Weight, Volume, Internal energy, Heat, Entropy, Gibbs free energy…

2) Intensive properties :
The properties whose magnitude does not depend upon the total quantity of matter present in the system are called intensive properties.
These properties depend only on the nature of the substance.
Ex: Pressure, Temperature, Specific heat, Density, Viscosity, Surface tension, boiling point, freezing point.

Question 14.
Define heat capacity? what are Cp & Cv? Show that Cp – Cv = R. [TS 15]
Heat capacity (C) :
The amount of heat required to raise the temperature of a substance through one degree is called heat capacity.
Formula: q = C∆T ⇒ C= q/∆T
Here, q = amount of heat absorbed;
∆T = raise in temperature
For gases, heat capacity is two types.
i) Heat capacity at constant volume (Cv)
ii) Heat capacity at constant pressure (Cp)

Cv :
The heat supplied to a system to raise its temperature through 1°C by keeping its volume constant is called heat capacity at constant volume (Cv).

Cp :
The heat supplied to a system to raise its temperature through 1 °C by keeping its pressure constant is called heat capacity at constant pressure (Cv).

Relation between Cp and Cv:
At constant volume, qv = Cv ∆T = ∆U,
At constant pressure, qp = Cp∆T = ∆H
For 1 mole of an ideal gas, we have,
∆H = ∆U + R∆T
⇒ Cp∆T = Cv ∆T + R∆T
⇒ (Cp – Cv) ∆T = R∆T
⇒ Cp – Cv = R.

Question 15.
Explain the determination of ∆U of a reaction calorimetrically.

1. The experimental technique used for the calculation of heat changes in a chemical process is called Calorimetry.
2. The device used for this purpose is called Bomb Calorimeter.
3. Calorimeter consists of a steel vessel called ‘Bomb’ which is immersed in a water bath.
4. The combustile substance of known mass (m) is burnt in the steel vessel, by passing pure oxygen at constant volume.
5. Heat evolved during this process is measured from the rise in temperature (∆t). 6. The heat capacity of the calorimeter is taken as C.
7. Now the heat change at constant volume is measured using the formula qv = C × m × ∆t
8. At constant volume, no work is done. Hence qv gives the required ∆U.

Question 16.
Explain the determination of ∆H of a reaction calorimetrieuily.

1. The change of enthalpy ∆H of a reaction can be measured using a calorimeter as shown in the figure.
2. Here, the calorimeter is kept open to the atmosphere.
3. 4. The calorimeter is immersed in an insulated water bath fitted with stirrer and thermometer.
5. The temperature of the bath is recorded in the beginning and after the end of the reaction and change in temperature(∆t) is recorded.
6. Knowing the heat capacity of water bath and calorimeter (C) and also the change in temperature, the heat absorbed or evolved in the reaction is calculated using the formula: qp = C × m × ∆t
7. The heat change at constant pressure gives the required ∆H.

Question 17.
What is enthalpy of a reaction? Explain the standard enthalpy of a reaction.
Enthalpy of a reaction :
During a chemical change heat is either liberated or absorbed. The heat change(∆H) involved in a chemical reaction is called Enthalpy of the reaction.
The chemical reaction which takes place with the liberation of heat is called exothermic reaction.
Ex : N2(g) + 3H2(g) → 2NH3(g); ∆H = -92kJ

The chemical reaction which takes place with the absorption of heat is called endothermic reaction.
Ex : C(graphite) + H2O(g) → CO(g) + H2(g); ∆H = +131.4kJ

The standard enthalpy of a reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.

The enthalpy change at the standard state condition is called standard enthalpy of the reaction. It is denoted by ∆rHθ

The superscript (θ) represents standard state. Standard state of a substance is its most stable state at one bar pressure and 298K.

Question 18.
What is the standard enthalpy of formation? Explain it with example.
Standard enthalpy of formation of a compound is defined as the heat change accompanying the formation of one mole of a compound from its constituent elements, in their standard states (1 bar pressure and 298K.)

Standard enthalpy of formation of the substance is also called its standard enthalpy and denoted by Af He Standard enthalpies of free elements are taken to be zero
Ex :C(graphite) + O2(g) → CO2(g); ∆Hθ = -393.5kJ Question 19.
Define and explain enthalpy of phase transformation.
The conversion of solid into liquid is called melting or fusion; and the process of conversion of liquid into- gas is called vapourisation. These processes are collectively called phase transformations.

The enthalpy change accompanying the conversion of 1 mole of a solid substance into the liquid state at its melting point is called enthalpy of fusion.

The enthalpy change accompanying the conversion of one mole of a liquid into its vapours at its boiling point is called enthalpy of vapourisation.

Such type of enthalpies are called enthalpy of phase transformations.

Question 20.
Define and explain the standard enthalpy of fusion (Molar enthalpy of fusion).
Standard enthalpy of fusion:
The enthalpy change accompanying the conversion of 1 mole of a solid substance into the liquid at its melting point is called the standard enthalpy of fusion.

The standard enthalpy of fusion of a substance depends largely on the strength of intermolecular forces in the substance undergoing fusion.

For example, ionic solids have very strong interparticle forces. Such substances have high values of enthalpy of fusion. Molecular solids have weak interparticle forces. They have low enthalpy values of fusion.

Question 21.
Define and explain the standard enthalpy of vapourisation (Molar enthalpy of vapourisation).
Standard enthalpy of vapourisation:
The enthalpy change accompanying the conversion of one mole of a liquid into its vapours at its boiling point is called standard molar enthalpy of vapourisation.

The values of enthalpy of vapourisation give some idea about the magnitude of inter particle forces in liquids. More the enthalpy of vapourisation stronger the inter particle forces.

Question 22.
Define and explain the standard enthalpy of sublimation.
Standard enthalpy of sublimation:
It is the enthalpy change accompanying the sublimation of one mole of a solid substance into gaseous state at a constant temperature below its melting point at the standard pressure.

Sublimation is direct conversion of a solid into vapour. The enthalpy of sublimation can be calculated with the help of Hess’s law.
The enthalpy of sublimation is the sum of enthalpy of fusion and enthalpy of vapourisation.
subH = ∆fusH + ∆vapH

Question 23.
Define and explain the standard enthalpy of formation (∆rHθ)[AP 22]
Standard enthalpy of formation:
The enthalpy of formation is the heat change accompanying the formation of one mole of a compound from its constituent elements. It is generally denoted by ∆fH. For example the enthalpy of formation of carbondioxide can be represented as
C(graphite) + O2(g) → CO2(g); ∆Hθ = -393.5kJ

When all the species of the chemical reaction are in their standard states, the enthalpy of formation is called standard enthalpy of formation. It is denoted by ∆Hθ

The standard enthalpy of formation is defined as the heat change accompanying the formation of one mole of a compound from its constituent elements, all the substances being in their standard states (1 bar pressure and 298K).

Question 24.
State and explain the Hess’s law of constant Heat summation. [AP 15,16,17,18,20,22][TS 15,16,18,19,19,20,22]
Hess law :
“The total heat change in a reaction is the same, whether the chemical reaction takes place in a single step or in several steps”.
It is based on the I law of thermodynamics.

Explanation :
Consider a reaction A → D. Suppose this reaction proceeds in two paths. Path – I : A → D; ∆H
Path – II: A → B; ∆H1
B → C; ∆H2
C → D; ∆H3
Total heat change in path – II is
∆H1 + ∆H2 + ∆H3
Now, from the Hess law, we have
∆H = ∆H1 + ∆H2 + ∆H3
Ex: CO2 can be obtained from C(graphite) and O2(g) in two different ways.

Path – I : C(graphite) + O2(g) → CO2(g);
Here, ∆H = -393.5 KJ mol-1.

Path – II: C(graphite) + 1/2O2(g) → CO(g);
Here, ∆H1 =-110.5 KJ mol-1
CO(g) + 1/2O2(g) → CO2(g);
Here, ∆H2 = -283.02 KJ mol-1
Total heat change in path-II = ∆H1 + ∆H2
= (-110.5) + (-283.02) = -393.52 KJ mol-1.
∆H = ∆H1 + ∆H2
Thus Hess law is proved.

Question 25.
Define and explain the enthalpy of combustion (∆cHθ).
Enthalpy of combustion:
The heat evolved when one mole of a substance is completely burnt at constant volume in excess of oxygen is called Heat of combustion.

For example the enthalpy of combustion of carbon is represented as
C(s) + O2(g) → CO2(g); ∆Hθ = -393.5kJ

Combustion reactions are always accompanied by the evolution of heat, therefore the value of ∆cH is always negative. Question 26.
Define and explain the enthalpy of atomisation (∆aHθ).
Enthalpy of atomisation: It is the enthalpy change on breaking one mole of bonds completely to obtain neutral atoms in the gas phase.

In case of diatomic molecules, like H2, HCl etc. the enthalpy of atomisation is also the bond dissociation enthalpy. In the case of metals enthalpy of atomisation is the enthalpy of sublimation.

Question 27.
Define and explain the bond enthalpy (∆bondHθ).
Bond enthalpy:
The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase.

In the case of diatomic molecules like H2, HCl etc. the enthalpy of atomisation is also the bond dissociation enthalpy. In die case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule.

Question 28.
What is the bond enthalpy of C-H bond of CH4?
The overall thermochemical equation for atomisation reaction of CH4 is

CH4(g) → C(g) +4H(g); ∆aHθ = 1665kJmol-1

In CH4, all the four C-H bonds are similar in bond length and energy. However the energies required to break the individual C-H bonds in each successive step differ. This is because, in each step of dissociation, different fragments of CH4 are involved.

CH4(g) → CH3(g) + H(g) ; ∆bondHθ = +427 kJmol-1
CH3 (g) → CH2(g) + H(g) ; ∆bondHθ = +439 kJmol-1
CH2 (g) → CH(g) + H(g); ∆bondHθ = +452 kJmol-1
CH(g) → C(g) + H(g) ; ∆bondHθ = +347 kJmol-1
∴ for the reaction CH4(g) → C(g) + 4H(g);
aHθ = 427 + 418.4 + 460.2 + 343.1 = 665kJmol-1
Here the average bond dissociation energy is
bondHθ = $$\frac{1648.7}{4}$$ = 412.2kJ/ mol

Question 29.
Define heat of solution (∆solHθ) and heat of dilution.
Heat of solution :
The amount of heat released (or) absorbed when one mole of a solute is dissolved in excess of solvent at constant temperature is called Heat of solution.

Heat of dilution :
The change of enthalpy when a solution containing 1 mole of a solute is diluted from one concentration to another concentration.
KCl + aq → KCl(aq); ∆H = + 19.75 kJ

Question 30.
Define ionisation enthalpy and electron affinity.
Ionisation enthalpy :
The energy required to remove an electron from an isolated gaseous atom in its ground state is called Ionisation enthalpy.
X(g) → X+(g) + e
Units : kJmol-1

The enthalpy change accompanying the process of conversion of a neutral gaseous atom into negative ion by adding an electron is called electron gain enthalpy.
X(g) + e → X(g)

The electron gain enthalpy is also known as electron affinity of the atom under consideration. Electron affinity is defined at absolute zero. So at any other temperature (T) heat capacities of the reactants and the products have to be considered and thus
egH = -Ae – $$\frac{5}{2}$$RT. Question 31.
Explain the spontaneity of a process.
Spontaneous process :
A process is said to be spontaneous if it occurs on its own without the aid of any external agency of any kind.

In general, for a spontaneous reaction ∆H is -ve and ∆S is +ve.
All natural processes are spontaneous.
Ex:
a) Heat flows from hot end to cold end.
b) Water flows from higher level to lower level.
c) Gas flows from higher pressure region to lower pressure region.

Question 32.
Is decrease in enthalpy a criterion for spontaneity? Explain.
Decrease in enthalpy may be a contributory factor for spontaneity but it is not true for all cases.
(i) H2(g) + $$\frac{1}{2}$$O2(g) → H2O(l), ∆H0 = -285.8KJ/mole
(ii) C(s) + O2(g) → CO2(g), ∆H0 = -393.5KJ/mole

The heat content of the products is less than those of the reactants. All these reactions are accompanied by evolution of heat. Therefore, these reactions are spontaneous because they are accompanied by decrease of energy.
∴ ∆H might be responsible for a reaction to be spontaneous.

But a number of reactions are known which are endothermic (∆H is positive )but still spontaneous.
Ex: CaCO3(S) → CaO(S) + CO2(g)
∆H = +177.8kJ/mol
∴ ∆H cannot be the sole criterion for predicting the spontaneity of a reaction.

Question 33.
What is entropy? Explain with examples.
Entropy (S):
Entropy is a measure of disorder or randomness of molecules of the substance. [AP 16,22][TS 17]

1. The greater the disorder in a system the higher is the entropy.
2. Entropy is a thermodynamic property.
3. Entropy is an extensive property.
4. Entropy change (∆S) is a state function.
5. Order of Entropy:
Svapour > Sliquid > Ssolid
6. For a spontaneous change AS>0.
7. If the system is not isolated, the total entropy change (∆S total) must be positive. .
∆STotal = ∆Ssystem + ∆Ssurrounciings > O
8. Entropy change (∆S) between any two states is ∆S = $$\frac{q_{rev}}{T}$$. Here, qrev is the heat absorbed by the system isothermally and reversibly at “T” during the state change.
9. The units of entropy change(∆S) is J/K.

Question 34.
Is increase in enfropy a criterion for spontaneity? Explain.
In exothermic reactions heat released by the reaction increases the disorder of the surroundings and overall entropy change is positive which makes the reaction spontaneous.

All spontaneous process are thermo dynamically irreversible and entropy of the system increases in all spontaneous processes.

But positive ∆S is not a necessary and sufficient condition for the spontaneous nature of a reaction.

For the spontaneity of a reaction, ∆G must be negative. Eventhough entropy doesnot increase if the ∆H is more negative than TAs in the equation ∆G = ∆H – T∆S. If the value of ∆G becomes negative and the reaction becomes spontaneous.

Question 35.
Can ∆C and ∆S discriminate between irreversible and reversible processes? Explain.
∆U does not discriminate between irreversible and reversible process. For isothermal process involving ideal gas ‘temperature(T)’ is constant. Hence ∆U = 0 for both reversible and irreversible process.

But ∆S discriminates the irreversible and reversible process.
In an isothermal reversible process ∆S(Total) = 0
But, in the irreversible process ∆S > 0
Thus ∆S discriminates a reversible and an irreversible process.

Question 36.
In which of the following processes entropy increases?
a) A liquid evaporates to vapour.
b) Temperature of a crystalline solid lowered from 115K to OK.
c) CaCO3(s) → CaO(s) + CO2(g)
d) Cl2(g) → 2Cl
a) Entropy increases, because evaporation of liquid is a spontaneous process and it is leads to increase of randomness.

b) Entropy decreases, because at OK, there is a perfect order in constituent particles.

c) Entropy increases, because the reactant is solid whereas one of the products is a gaseous substance.

d) Cl2(g) → 2Cl(g) : Here both are gases.
But one Cl2 molecule converts into 2 Cl atoms due to which randomness increases. So entropy increases.

Question 37.
For the oxidation of iron 4Fe(s) + 3O2(g) → 2Fe(2)O3(s) the entropy change is -549.45 JK-1 mol-1 at 298K. Though it has negative entropy change the reaction is spontaneous. Why?
(∆rHθ = -1648 × 10³ Jml-1) As ∆STotal is positive, the reaction is spontaneous.

Question 38.
Which formulae in the following are correct? a) Correct
b) Correct
c) Correct
d) Correct
e) Correct. Question 39.
Calculate ∆Gθ for conversion of oxygen to ozone $$\frac{3}{2}$$O2(g) → O3(g) at 298K. Kp for the reaction is 2.43 × 10-29.
Given data: T= 298K, Kp = 2.43 × 10-29
R = 8.314 Jmol-1K-1
We know ∆G°= -2.303 RT log Kp
= -2.303 × 8.314 × 298 × (log 2.43 × 10-29)
= 163229 = 163.2 kJ mol-1

Question 40.
State and explain second law of ther-modynamics and explain it.
Second law of thermodynamics is stated in different forms

1. Heat cannot How from a colder body to a hotter body on its own.
2. Heat cannot be converted completely into work without causing some permanent changes in the system or in the surroundings.
3. It is impossible to construct a perpetual motion machine of second kind.
4. All spontaneous processes are thermodynamically irreversible and entropy of the system increases in all spontaneous processes.

Explanation : 2nd law of thermodynamics is useful in predicting

1. Whether a process is spontaneous or not in the specified direction.
2. What fraction of one form of energy is converted into another form of energy in a transformation.

Question 41.
State the third law of thermodynamics. What do you understand by it?
Third law of Thermodynamics: [TS 16, 17]
“The entropy of a pure and perfectly crystalline substance approaches zero when the temperature approaches absolute zero.”

A zero entropy means a perfect order or least disorder.

Third law of thermodynamics imposes a limitation on the value of entropy.

The entropy of a pure substance increases with increase in temperature and decreases with decrease in temperature.

Third law of thermodynamics is useful for calculating the entropy .(S) of a substance

at any temperature if temperature dependence Cp is known in evaluating the absolute value of entropy.
Mathematically, = J Ajr-dt

Question 42.
Explain “Entropy” concept.
Entropy (S):
Entropy is a measure of disorder or randomness of molecules of the substance. [AP 16,22][TS 17]

1. The greater the disorder in a system the higher is the entropy.
2. Entropy is a thermodynamic property.
3. Entropy is an extensive property.
4. Entropy change (∆S) is a state function.
5. Order of Entropy:
Svapour > Sliquid > Ssolid
6. For a spontaneous change AS>0.
7. If the system is not isolated, the total entropy change (∆S total) must be positive. .
∆STotal = ∆Ssystem + ∆Ssurrounciings > O
8. Entropy change (∆S) between any two states is ∆S = $$\frac{q_{rev}}{T}$$. Here, qrev is the heat absorbed by the system isothermally and reversibly at “T” during the state change.
9. The units of entropy change(∆S) is J/K.

Question 43.
Explain spontaneity of a process in terms of Gibbs energy.
Gibbs Energy:
Gibbs energy is a thermodynamic function. This is the difference in the enthalpy (H) and the product of entropy (S) and absolute temperature (T) of the system.
G = H-TS

Gibbs energy is the amount of energy available from a system which can be put to useful work at constant temperature and pressure. .

The change in Gibbs energy for the system
∆Gsystem at constant temperature is
∆Gsystem = ∆Hsystem – T∆Gsystem
If ∆Gsystem is negative (<0) the process is spontaneous.
If ∆Gsystem is positive (>0) the process is non-spontaneous.
If ∆Gsystemis zero the system has attained equilibrium.

Question 44.
The sign and magnitude of Gibbs energy change of a chemical process tells about its spontaneity and useful work that could be extracted from it. Explain.
Gibbs formula: ∆G = ∆H – T∆S
To sum up, the criteria for spontaneity of a process in terms of ∆G is as follows.
i) If ∆G is negative, the process is spontaneous.
ii) If ∆G is zero, the system is in equilibrium.
iii) If ∆G is positive, the process does not occur in the forward direction, i.e non-spontaneous Question 45.
In a process 701J of heat is absorbed by a system and 394J of work is done by the system. What is the change in internal energy for the process?
Given q = 701J, w = -394J
From first law of Thermodynamics
∆U = q + w = 701 – 394 = 307 J
Change in internal energy = 307 J Question 46.
The reaction of cyanamide (s) with dioxygen was carried out in a bomb calorimeter and ∆U was found to be – 742.7Kjmol-1 at 298K. Calculate the enthalpy change for the reaction at 298K.
NH2CN(s) + $$\frac{3}{2}$$O2(g) → N2(g) + CO2(g) + H2O(l)
NH2CN(s) + $$\frac{3}{2}$$O2(g) → N2(g) + CO2(g) + H2O(l)
Here, ∆n = np – nR = 2 – $$\frac{3}{2}=\frac{1}{2}$$
R = 8.314 × 10-3 KJmol-1K-1 T = 298K
∆U = -724.7 KJmol-1K-1
We know ∆H = ∆U + ∆nRT = -724.7 + $$\frac{1}{2}$$ × 8.314 × 10-3 x 298
= -724.7 + 1.238 = -723.47 kJmol-1.

Question 47.
Calculate the number of kJ of heat necessary to rise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity’ of aluminium is 24J moHK-1.
Formula: q = msdT
q = heat liberated
m = mass of aluminium
s = molar heat capacity of aluminium
dT = change in temperature
∴ q = $$\frac{60}{27}$$ × 24 × 20 = 1.09kJ.

Question 48.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C.
fusH = 6.03kJmol-1
Cp[H2O(l)] = 75.3 J mol-1 K-1
Cp[H2O(s)J = 36.8 Jmol-1K-1
Total enthalpy change involved in the transformation is the sum of the following changes:
a) Energy change involved the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.
b) Energy change involved in the transformation of 1 mole of water at 0°C 1 mol of ice at 0°C
c) Energy change involved iri the transformation of 1 mole of ice at 0°C at 1 mole of ice at -10°C.
Total = ∆H = nCp[H2O(l)]∆T + ∆Hfreezing + Cp[H2O(s)]∆T
= (75.3)(0 – 10) + (-6.03 × 10³) + (36.8)(-10 – 0)
= -753 – 6030 – 368 = -7151 J mol-1
= -7.151 kJ mol-1

Question 49.
Enthalpy of combustion of carbon to CO2 is -393.5 kJmol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. [TS 15]
C + O2 → CO2, ∆H = -393.5kJmol-1
Heat released upon formation of 44g of CO2 = 393.5kJmol-1.
Heat released upon the formation of 35.2g
of CO2 = $$\frac{-393.5\times35.2}{44}$$ = -314.8 kJ.

Question 50.
Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are -110, -393, 81 and 9.7 kJmol-1 respectively. Find the value of ∆H for the reaction.
N2O4(g) + 3CO(g) → N2O(g) + 3CO2 (g)
N2O4 (g) + 3CO(g) → N2O(g) + 3CO2(g)
∆H = Total enthalpies of products – Total enthalpies of reactants.
= (HN2O + 3HCO2)-(HN2O4 + 3HCO)
= ( 81 + 3 × -393) – (9.7 + 3 × -110)
= ( 81 – 1179) – (9.7 – 330) = -777.7 kJ mol-1.

Question 51.
Given
N2(g) + 3H2(g) → 2NH3(g); ∆rHθ = -92.4kJmol-1
What is the standard enthalpy of formation of NH3 gas?
The heat of reaction ∆rHθ is -92.4 kJmol-1.
This is the heat of formation of 2 moles of ammonia.
The enthalpy of formation of 1 mol of NH3 = $$\frac{-92.4}{2}$$ = -46.2kJ
∴ Standard enthalpy of formation of ammonia = -46.2 kJ

Question 52.
Calculate the standard enthalpy of formation of CH3OH (I) from the following data: .
CH3OH(l) + $$\frac{3}{2}$$O2(g) → CO2(g) + 2H2O(l) ; ∆H0r = – 726k.lmol-1.
C(graphite) + O2(g) → CO2(g); ∆H0c = -393kJmol-1
H2(g) + $$\frac{1}{2}$$O2(g) → H2O(l); ∆H0f = -286kJmol-1.
CH3OH + $$\frac{3}{2}$$O2 → CO2 + 2H2O, ∆H0r = -726 kJmol-1. ——— (1)
C(graphite) + O2(g) → CO2(g), ∆H0c = -393kJmol-1. ——- (2)
H2 + $$\frac{1}{2}$$O2 → H2O. ∆H0f = -286 kJmol-1. ——- (3)
Required equation is
C(graphite) + 2H2 + $$\frac{1}{2}$$O2 → CH3OH,
Eqn (2) + 2 × Eqn(3) – Eqn (1) ⇒
∆H = (-393) + 2(-286) – (-726kJmol-1)
= -393 – 572 + 726 = -239kJmol-1
∴ ∆H = -239 kJmol-1

Question 53.
Calculate the enthalpy change for the process
CCl4(g) → C(g) + 4Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g).
vapHθ(CCl4) = 30.5 kJmol-1
fHθ(CCl4) = -135.5 kJmol-1
0Hθ(C) = 715.0 kJmol-1, where ∆aHθ
is enthalpy of atomisation.
aHθ(Cl2) = 242 kJ mol-1
C(s) + 2Cl2(g) → CCl4(l)
∴ ∆H = 715 + (2 × 242) – 4eC-Cl – ∆Hvap -135.5 = 119 – 4eC-Cl – 30.5
4eC-Cl = 1304
CCl4 → C + 4Cl = ∆H = 1304 kJ
Bond energy of C – Cl = $$\frac{1304}{4}$$ = 326kJ Question 54.
For an isolated system, ∆U = 0 what will be ∆S?
Since ∆U = 0, ∆S will be positive and the reaction will be spontaneous.

Question 55.
For the reaction at 298K.
2A + B → C
∆H = 400kJmol-1 and ∆S = 0.2kJmol-1 At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range.
∆G = ∆H – T∆S
Since ∆H & ∆S are constant we have ∆G = 0
∴ ∆H = T∆S ⇒ T = $$\frac{\Delta H}{\Delta S}=\frac{400}{0.2}$$ = 2000K
Thus the reaction will be in a state of equilibrium at 2000K and will be spontaneous above this temperature.

Question 56.
For the reaction 2Cl(g) → Cl2(g), what are the signs of ∆H and ∆S?
∆H: Negative. Because energy is released in bond formation.

∆S: Negative. Because entropy decreases when atoms combine to form molecules.

Question 57.
For the reaction
2A(g) + B(g) → 2D(g)
∆Uθ = -10.5 kJ and ∆Sθ = -44.1 kJ
Calculate ∆Gθ for the reaction, and predict whether the reaction can occur spontaneously or not.
Formula: ∆H = ∆U + ∆ngRT
Now, ∆H = -10.5 + (-1) × 8.314 × 10-3 × 298 = -12.97 kJ .
∴ ∆G = ∆H – T∆S
= -12.97 – 298(-44.1 × 10-3) = 0.164 kJ
Here ∆G>0. Hence the reaction is non-spontaneous.

Question 58.
The equilibrium constant for a reaction is 10. What will be the value of AG°?
R = 8.314 JK-1mol-1 T = 300K.
Formula: ∆G0 = -RTlnK
∆G0 = -2.303 RTlog K
∆G0 = – 2. 303 RTlog K AG° =-2.303 × 8.314 × 300 × log 10
= – 5744J/mole = – 5.744 kJ mol-1

Question 59.
State the first law of thermodynamics. Explain its mathematical notation.
First law of thermodynamics(Law of conservation of energy):
Energy can neither be created nor be destroyed but energy in a process may be converted from one form to another form. Mathematically first law of thermodynamics can be represented as ∆U = q + w
q = Amount of heat absorbed by the system ∆U= Increase in internal energy of the system
w = Work done on a system
For infinitesimally small changes q = dU +w

According to first law of thermodynamics a part of the amount of heat (q) absorbed by the system is used for increasing the internal energy (∆U) of the system and the remaining part is used for doing work (w). Heat absorbed by the system is given + sign, heat given out by the system is given – sign. Work done by a system is given -sign and work done on a system is given +sign. Question 60.
State the 2nd law of thermodynamics in any two ways
Second law of thermodynamics is stated in different forms

1. Heat cannot How from a colder body to a hotter body on its own.
2. Heat cannot be converted completely into work without causing some permanent changes in the system or in the surroundings.
3. It is impossible to construct a perpetual motion machine of second kind.
4. All spontaneous processes are thermodynamically irreversible and entropy of the system increases in all spontaneous processes.

Explanation : 2nd law of thermodynamics is useful in predicting

1. Whether a process is spontaneous or not in the specified direction.
2. What fraction of one form of energy is converted into another form of energy in a transformation.

Question 61.
Explain Gibb’s energy.
Gibb’s energy(G):
Gibb’s energy is a thermodynamic quantity of a system. Gibbs energy is amount of energy available from a system which can be put to useful work at constant temperature and pressure.

It gives us an idea of spontaneity of reaction.
∆G = ∆H – T∆S
If ∆G is negative(∆G <0), then the process is spontaneous. If ∆G is positive (∆G>0) for non-spontaneous reactions
If ∆G is zero the system has attained equilibrium.

Question 62.
Explain the spontaneity of a reaction in terms of Gibbs energy?
Gibbs formula: ∆G = ∆H – T∆S
To sum up, the criteria for spontaneity of a process in terms of ∆G is as follows.
i) If ∆G is negative, the process is spontaneous.
ii) If ∆G is zero, the system is in equilibrium.
iii) If ∆G is positive, the process does not occur in the forward direction, i.e non-spontaneous Question 1.
State and explain“Hess law of constant heat summation” with example.

Hess law :
“The total heat change in a reaction is the same, whether the chemical reaction takes place in a single step or in several steps”.
It is based on the I law of thermodynamics.

Explanation :
Consider a reaction A → D. Suppose this reaction proceeds in two paths. Path – I : A → D; ∆H
Path – II: A → B; ∆H1
B → C; ∆H2
C → D; ∆H3
Total heat change in path – II is
∆H1 + ∆H2 + ∆H3
Now, from the Hess law, we have
∆H = ∆H1 + ∆H2 + ∆H3
Ex: CO2 can be obtained from C(graphite) and O2(g) in two different ways.

Path – I : C(graphite) + O2(g) → CO2(g);
Here, ∆H = -393.5 KJ mol-1.

Path – II: C(graphite) + 1/2O2(g) → CO(g);
Here, ∆H1 =-110.5 KJ mol-1
CO(g) + 1/2O2(g) → CO2(g);
Here, ∆H2 = -283.02 KJ mol-1
Total heat change in path-II = ∆H1 + ∆H2
= (-110.5) + (-283.02) = -393.52 KJ mol-1.
∆H = ∆H1 + ∆H2
Thus Hess law is proved. Question 2.
Explain the experiment of determine the internal energy change of a chemical reaction.

1. The experimental technique used for the calculation of heat changes in a chemical process is called Calorimetry.
2. The device used for this purpose is called Bomb Calorimeter.
3. Calorimeter consists of a steel vessel called ‘Bomb’ which is immersed in a water bath.
4. The combustile substance of known mass (m) is burnt in the steel vessel, by passing pure oxygen at constant volume.
5. Heat evolved during this process is measured from the rise in temperature (∆t). 6. The heat capacity of the calorimeter is taken as C.
7. Now the heat change at constant volume is measured using the formula qv = C × m × ∆t
8. At constant volume, no work is done. Hence qv gives the required ∆U.

Question 3.
Explain the experiment to determine the enthalpy change of a chemical reaction.

1. The change of enthalpy ∆H of a reaction can be measured using a calorimeter as shown in the figure.
2. Here, the calorimeter is kept open to the atmosphere.
3. 4. The calorimeter is immersed in an insulated water bath fitted with stirrer and thermometer.
5. The temperature of the bath is recorded in the beginning and after the end of the reaction and change in temperature(∆t) is recorded.
6. Knowing the heat capacity of water bath and calorimeter (C) and also the change in temperature, the heat absorbed or evolved in the reaction is calculated using the formula: qp = C × m × ∆t
7. The heat change at constant pressure gives the required ∆H.

Question 4.
Explain the spontaneity of a reaction in terms of enthalpy change, entropy change and Gibbs energy change.
Gibbs formula: ∆G = ∆H – T∆S
To sum up, the criteria for spontaneity of a process in terms of ∆G is as follows.
i) If ∆G is negative, the process is spontaneous.
ii) If ∆G is zero, the system is in equilibrium.
iii) If ∆G is positive, the process does not occur in the forward direction, i.e non-spontaneous Multiple Choice Questions

Question 1.
1) energy changes involved in a chemical reaction.
2) the extent to which a chemical reaction proceeds.
3) the rate at which a reaction proceeds.
4) the feasibility of a chemical reaction.
3) the rate at which a reaction proceeds.

Question 2.
Which of the following statements is correct?
1) The presence of reacting species in a covered beaker is an example of open system.
2) There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
3) The presence of reactants in a closed vessel made up of copper is an example’of a closed system.
4) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.
3) The presence of reactants in a closed vessel made up of copper is an example’of a closed system.

Question 3.
The state of a gas can be described by quoting the relationship between
1) pressure, volume, temperature
2) temperature, amount, pressure
3) amount, volume, temperature
4) pressure, volume, temperature, amount
4) pressure, volume, temperature, amount

Question 4.
The volume of gas is reduced to half from its original volume. The specific heat will be ___
1) reduce to half
2) be doubled
3) remain constant
4) increase four times
3) remain constant

Question 5.
Enthalpy of sublimation of a substance is equal to
1) enthalpy of fusion+enthalpy of vapourisation
2) enthalpy of fusion
3) enthalpy of vapourisation
4) twice the enthalpy of vapourisation
1) enthalpy of fusion+enthalpy of vapourisation Question 6.
Which one among the following is the correct option for right relationship between CP and CV for one mole of ideal
1) CV = RCP
2) CP + CV = R
3) CP – CV = R
4) CP = RCV
3) CP – CV = R

Question 7.
For a given reaction, ∆H=35.5 kJ mol-1 and ∆S = 83.6JK-1 mol-1. The reaction is spontaneous at (Assume that ∆H and ∆S do not vary with temperature)
1) T>425 K
2) all temperature
3) T>298 K
4) T<425K Answer: 1) T>425 K

Question 8.
In which case change in entropy is negative?
1) 2H(g) → H2(g)
2) Evaporation of water
3) Expansion of a gas at constant temperature
4) Sublimation of solid to gas
1) 2H(g) → H2(g)

Question 9.
For irreversible expansion of an ideal gas under isothermal condition, the correct option is
1) ∆U ≠ 0, ∆Stotal = 0
2) ∆U = 0, ∆Stotal = 0
3) ∆U ≠ 0, ∆Stotal ≠ 0
4) ∆U = 0, ∆Stotal ≠ 0
4) ∆U = 0, ∆Stotal ≠ 0

Question 10.
The bond dissociation energies of X2, Y2 and XY are in the ratio of 1:0.5:1 . ∆H for the formation of XY is -200kJmol-1. The bond dissociation energy of X2 will be.
1) 200 kJ mol-1
2) 100 kJmol-1
3) 800 kJ mol-1
4) 400 kJ mol-1
3) 800 kJ mol-1

Question 11.
An ideal gas expands isothermally from 10-3 m³ to 10-2 m³ at 300K against a constant pressure of 105 Nm-2. The work done on the gas is
1) +270kJ
2) -900 J
3) +900 kJ
4) -900 kJ
2) -900 J

Question 12.
Under isothermal conditions, a gas at 300K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is
[Given that 1 L bar = 100J]
1) 30 J
2) -30 J
3) 5 kJ
4) 25 J
2) -30 J Question 13.
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ∆U of the gas in joule will be
1) -500 J
2) -505 J
3) +505 J
4) 11136.25 J
2) -505 J

Question 14.
1) q = 0, ∆T = 0 and w = 0
2) q = 0, ∆T < 0 and w > 0
3) q < 0, ∆T = 0 and w = 0 4) q > 0, ∆T > 0 and w > 0
1) q = 0, ∆T = 0 and w = 0

Question 15.
Which of the following is not correct?
1) ∆G is zero for a reversible reaction
2) ∆G is positive for a spontaneous reaction
3) ∆G is negative for a spontaneous reaction
4) ∆G is positive for a non-spontaneous reaction
2) ∆G is positive for a spontaneous reaction

16. In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.
1) q = 0, ∆T ≠ 0, w = 0
2) q ≠ 0, ∆T = 0, w = 0
3) q = 0, ∆T = 0, w = 0
4) q = 0, ∆T < 0, w ≠ 0
3) q = 0, ∆T = 0, w = 0

Question 17.
fUΘ of formation of CH4(g) at certain temperature is -393 kJ mol-1. The value of ∆fHΘ. is
1) zero
2) < ∆fUΘ
3) > ∆fUΘ
4) equal to > ∆fUΘ
2) < ∆fUΘ

Question 18.
The entropy change can be calculated by using the expression ∆S = $$\frac{q_{rev}}{T}$$. When water freezes in a glass beaker, choose the correct statement amongst the following :
1) ∆S(system) decreases but ∆S(surroundings) remains the same.
2) ∆S(system) increases but ∆S(surroundings) decreases.
3) ∆S(system) decreases but ∆S(surroundings) increases.
4) ∆S(system)decreases and ∆S(surroundings) also decreases.
3) ∆S(system) decreases but ∆S(surroundings) increases.

Question 19.
The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
1) is always negative
2) is always positive
3) may be positive or negative
4) is never negative 