Thermodynamics Questions and Answers AP Inter 1st Year Chemistry Chapter 5

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 5th Lesson Thermodynamics Class 11 Textbook Exercise Questions and Answers.

Thermodynamics Class 11 Questions and Answers AP Inter 1st Year Chemistry 5th Lesson

I. Multiple Choice Questions (1 Mark)

Question 1.
According to the third law of thermodynamics which one of the following quan- titles for a perfectly crystalline solid is zero at absolute zero (-273.15°C) ?
(1) Gibbs energy
(2) Entropy
(3) Enthalpy
(4) Internal energy
Answer:
(2) Entropy

Question 2.
The heat of combustion of carbon is -393.5 kJ/mol. The enthalpy change in the formation of 35.2g of CO₂ from carbon and oxygen gas is
(1) -315 kJ
(2) + 315 kJ
(3) -630 kJ
(4) -3.15 kJ
Answer:
(1) -315 kJ

Question 3.
For the reaction, N₂ + 3H₂ \(\text { ⇌ }\) 2NH₃, ∆H = ?
(1) ∆U + 2RT
(2) ∆U – 2RT
(3) ∆H = RT
(4) ∆U – RT
Answer:
(2) ∆U – 2RT

Question 4.
Which amongst the following options is the correct relation between change in enthalpy and Change in internal energy ?
(1) ∆H = AU + ∆n_gRT
(2) ∆H – ∆U = -∆nRT
(3) ∆H + ∆U = ∆nR
(4) ∆H = ∆U – ∆n_gRT
Answer:
(1) ∆H = AU + ∆n_gRT

Question 5.
A thermodynamic state function is a quantity
(1) used to determine heat changes
(2) whose value is independent of path
(3) used to determine pressure-volume work
(4) whose value depends. On temperature only
Answer:
(2) whose value is independent of path

Question 6.
For the process to occur under adiabatic conditions, the correct condition is:
(1) ∆T = 0
(2) ∆_p = 0
(3) q = 0
(4) w = 0
Answer:
(3) q = 0

Question 7.
The enthalpies of all elements in their standard states are:
(1) Unity
(2) Zero
(3) < 0
(4) Different for every element
Answer:
(2) Zero

Question 8.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 kJ/mol, -393.5 kJ mol^-1 and -285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄(g) will be
(1) -74.8 kJmol^-1
(2) 52.27 kJmol^-1
(3) +74.8 kJ mol^-1
(4) -52.26 kJ mol^-1
Answer:
(1) -74.8 kJmol^-1

Question 9.
A reaction, A + B → C + D + q, is found to have a positive entropy change. The reaction will be
(1) possible at high temperature
(2) possible only at low temperature
(3) not possible at any temperature
(4) possible at any temperature
Answer:
(4) possible at any temperature

Question 10.
Which of the following are not state functions ?
(1) work and entropy
(2) heat and work
(3) entropy and internal energy
(4) heat and entropy
Answer:
(2) heat and work

Question 11.
The Gibbs energy change (∆G) at constant temperature is equal to
(1) ∆H – TS
(2) ∆H + TS
(3) ∆H + T∆S
(4) ∆H – T∆S
Answer:
(4) ∆H – T∆S

Question 12.
For the reaction, 2Cl_(g) → Cl_2(g), the correct option is
(1) ∆_rH > 0 and ∆_rS < 0
(2) ∆_rH < 0 and ∆_rS > 0
(3) ∆_rH < 0 and ∆_rS < 0
(4) ∆_rH > 0 and ∆_rS > 0
Answer:
(3) ∆_rH < 0 and ∆_rS < 0

Question 13.
Which of the following is not an extensive property ?
(1) Mass
(2) Volume
(3) Temperature
(4)Enthalpy
Answer:
(3) Temperature

Question 14.
The correct thermodynamic conditions for the spontaneous reaction at all temperature is
(1) ∆_rH > 0 and ∆_rS > 0
(2) ∆_rH < 0 and ∆_rS > 0
(3) ∆_rH < 0 and ∆_rS < 0
(4) ∆_rH > 0 and ∆_rS > 0
Answer:
(2) ∆_rH < 0 and ∆_rS > 0

Question 15.
If a process is carried out at constant volume (∆V = 0), the correct expression is
(1) ∆U = q – P_ex∆V
(2) ∆U = q
(3) ∆U = q + P_ex∆V
(4) ∆U = w
Answer:
(2) ∆U = q

II. Fill in the Blanks (1 Mark)

Question 1.
q = w = -P_ext(v_f – v_i) is for irreversible …….. change.
Answer:
Isothermal

Question 2.
q = -w = nRT ln (v_f/v_i) is for isothermal ……… change.
Answer:
Reversible

Question 3.
For an isolated system, ∆U = 0, the value of ∆s will be ………..
Answer:
≥ 0 (zero))

Question 4.
In ………… system there is no exchange of matter and energy between the system and surroundings. Answer:
Isolated

Question 5.
For chemical reactions, ……… is used to measure the value of heat change at constant volume.
Answer:
bomb calorimeter

III. One Word Answer Questions (1 Mark)

Question 1.
Work is done by the system and ‘q’ amount of heat is supplied to the system. What type of system would it be ?
Answer:
When ‘q’ amount of heat is given to the system, then work is done by the system
i.e., it utilise heat but does not exchange matter. Thus system is closed system.
∴ ∆U = q – w.
∆U – Change in internal energy
q – heat supplied
w – work done by the system

Question 2.
The system loses ‘q’ amount of heat though no work is done on the system. What type of wall does the system have ?
Answer:
Given that system loses heat i.e., energy is transferred out as heat. No work done means no energy is transferred. So, the type of wall is Diathermic wall. Diather-mic wall allows heat to pass through but prevents work and does not allow mat-ter transfer.

Question 3.
What is the work done in the free expansion of an ideal gas in reversible and irreversible processes ?
Answer:
In the free expansion of an ideal gas, whether reversible or irreversible, the work done is zero. This is because, the external pressure against which the gas expands is zero during a free expansion.

Question 4.
In isothermal reversible change of an ideal gas, what is the value of q ?
Answer:
For an isothermal reversible expansion or compression of an ideal gas, tem- peratiire is constant. Internal energy of an ideal gas depends only on tempera-ture, so ∆U = 0.
From First law of thermodynamics,
∆U = q + w ⇒ 0 = q + w ⇒ q = -w
For reversible isothermal process, the work done by the gas is, W = nRT
\(\ln \left(\frac{V_2}{V_1}\right)\)
n = Number of moles
R = Gas constant
\(\frac{V_2}{V_1}\) = volume ratio.
∴ q = -w = nRT ln\(\left(\frac{V_2}{V_1}\right)\) = 2.303 nRT log\(\left(\frac{V_2}{V_1}\right)\).

Question 5.
For an adiabatic change in an ideal gas what is the relationship between its ∆U and W (adiabatic) ?
Answer:
For an adiabatic process, the system exchanges no heat with its surroundings (q = 0). From the first law of thermodynamics.
∆U = q + w
Since q = 0, ∆U = 0 + w = w
∴ Change in internal energy = work done in adiabatic process.

IV. Very Short Answer Questions (2 Marks)

Question 1.
Define a system.
Answer:
A system is a defined portion of the universe under study separated from its surroundings by a real or imaginary boundary. The system can exchange energy and/or matter with the surroundings depending on the type of boundary.

Question 2.
State the first law of the thermodynamics.
Answer:
The law of conservation of energy is taken as the first law of thermodynamics. It can be stated as below.

• Energy can neither be created nor destroyed, although it can be transferred from one from to another.
• It is impossible to construct perpetual motion machine of first kind.
• The total energy of the system and surroundings is constant.

Question 3.
State the second law of thermodynamics.
Answer:

• It is impossible for a self acting machine unaided by any external agency to convert heat from a body at low temperature to a body at higher temperature.
• All, spontaneous processes are thermodynamically irreversible and entropy of the system increases in all spontaneous processes.
• For any spontaneous process taking place in an isolate system, the change in entropy (Δs) is positive.

Question 4.
State the third law of thermodynamics.
Answer:
The third law of thermodynamics states that “The entropy of a perfect crystalline substance becomes zero as the temperature approaches absolute zero (0°K).
\(\lim _{T \rightarrow 0}\) S = 0
Where, S = Entropy
T = Temperature in Kelvin.

Question 5.
What are intensive and extensive properties ?
Answer:
Intensive properties: The properties which do not depend upon the quantity of matter present in the system are called Intensive properties.
Examples: Density, molar volume, molar heat capacity, molar entropy, surface tension, viscosity, pressure, temperature boiling point etc.

Extensive properties: The properties whose magnitude depend upon the quantity of matter present in the system.
Examples: Mass, volume, heat capacity, internal energy, entropy etc.

Question 6.
Define the state function. Give two examples.
Answer:
A state function is a property of a system that depends only on the current state of the system, but not on the path taken to reach that state.
Examples : Internal energy, Enthalpy etc.

Question 7.
The equilibrium constant for a reaction is 10. What will be value of Δg° ?
[R = 8.314 JK^-1 mol^-1, T = 300K]
Answer:
Standard Gibbs free energy change ΔG⁰ = -RT lnk
Given that equilibrium constant, k = 10
Gas constant, R = 8.314 Jk^-1 mol^-1
Temperature, T = 300K
∴ ΔG° = – RT ln k
⇒ ΔG = -(8.314 × 300 × ln(10))
⇒ ΔG = – (8.314 × 300 × 2.303)
⇒ ΔG = -5744.14 J mol^-1
⇒ ΔG = -5.74414 kJ mol^-1

Question 8.
Two litres of an ideal gas at a pressure of 10 atm expands isothermally into vacuum until its total volume is 20 litres. How much heat is absorbed and how much work is done in the expansion ?
Answer:
Given that the gas expands isothermally i.e., temperature remains constant. It expands into vacuum, so external pressure is zero. Therefore it is an irreversible isothermal free expansion.
Work Done: Since it is a free expansion, there is no external pressure.
∴ P_ext = 0.
Work done by the gas is given by
w = P_ext.Δv
⇒ w = 0 × (20 – 2) = 0 × 18 = 0
Heat absorbed: For isothermal expansion of ideal gas, ΔU = 0.
From the first, law of thermodynamics, ΔU = q + w ’
Since ΔU = 0, 0 = q + w ⇒ q = -w.

Question 9.
Calculate the entropy change in surroundings when 1.00 mole of H2O(1) is formed under standard conditions. Δ_rH° = -286kJ mol^-1.
Answer:
Given enthalpy change, Δ_rH° = -286kJ/mol
= -286000 J/mol
Temperature T = 298K
Entropy change in surroundings,
\(\Delta \mathrm{s}_{\text {surroundings }}\) = \(-\frac{\Delta \mathrm{H}_{\text {system }}}{\mathrm{T}}\)
= \(-\left(\frac{-286000}{298}\right)\)
= 959.73J/mol.k

Question 10.
Given N₂(g) + 3H₂(g) → 2NH₃(g); Δ_rH° = -92.4kJ. What is the standard enthalpy of the formation of NH₃ gas ?
Answer:
Given reaction, N₂(g) + 3H₂(g) → 2NH₃(g)
standard enthalpy change of reaction, Δ_rH° = -92.4kJ.
In the given reaction, the number of moles of NH₃ is 2. So, the enthalpy change of this reaction (-92.4 kJ) corresponds to the formation of 2 moles of NH₃.
∴ The enthalpy of formation of 1 mole of NH₃
Δ_fH° (NH₃) = \(\frac{-92.4}{2}\) = -46.2 kJ/mole

V. Short Answer Questions (4 Marks)

Question 1.
What are open, closed and isolated systems ? Give one example for each.
Answer:
The systems are classified into three types. These are :

a) Open system: A system which, can exchange matter as well as energy with the surroundings is called an open system.
Ex: Evaporation of water from a beaker presents as open system. Here vapours of water (matter) go into the atmosphere and heat (energy) required is absorbed by water from the surroundings.

b) Closed system: A system which can exchange energy but not matter with the surroundings is called a closed system.
Ex : Consider boiling of water in a closed metallic vessel. Here heat is transferred from the burner (surroundings to the system. Steam remains inside the vessel. Thus matter is not exchanged.

c) Isolated system: A system which can neither exchange matter nor energy with the surroundings is called an isolated system.
Ex: A thermos flask containing hot coffee. In this case, neither matter (coffee) nor energy (heat) can enter or leave the system.

Question 2.
State and explain the Hess’s law of constant Heat summation.
Answer:
Hess’s law states that the total amount of heat evolved or absorbed in a chemical reaction is always same whether the reaction is carried out in one step (or) in several steps.

Illustration : This means that the heat of reaction depends only on the initial and final stages and not on the intermediate stages through which the reaction is carried out. Let us consider a reaction in which A gives D. The reaction is brought out in one step and let the heat of reaction of ΔH.
A → D; ΔH
Suppose the same reaction is brought out in three stages as follows –

A → B; ΔH₁
A → C; ΔH₂
A → D; ΔH₃
The net heat of reaction is ΔH₁ + ΔH₂ + ΔH₃.
According to Hess law ΔH = ΔH₁ + ΔH₂ + ΔH₃.
Ex: Consider the formation of CO₂. It can be prepared in two ways.

1) Direct method: By heating carbon in excess of O₂.
C(s) + O₂(g) → CO₂(g); ΔH = -393.5 kJ

2) Indirect method : Carbon Can be converted into CO₂ in the following two steps.
C(s) + O₂(g) → CO₂; ΔH₁ = -110.5 kJ
CO(g) + \(\frac{1}{2}\)O₂(g) → CO₂(g); ΔH₂ = -283.02 kJ
Total ΔH = -393.52 kJ (ΔH₁ + ΔH₂)
The two ΔH values are same.

Question 3.
Define heat capacity. What are C_p and C_v ? Show that C_p – C_v = R.
Answer:
Heat capacity: Heat capacity of a substance (C) of a substance is defined as the amount of heat required to raise its temperature through one degree.
∴ C = \(\frac{\mathrm{q}}{\Delta \mathrm{~T}}\)
q-amount of heat absorbed

Heat capacity (C) is a state function. Hence, to evaluate C, the conditions such as volume or constant pressure have to be specified in order to define the path. Thus, there are two different types of heat capacities. These are Heat capacity at constant volume (C_v) and Heat capacity at constant pressure (C_p).

Heat capacity at constant volume (C_v): It may be defined as the amount of heat required to raise the temperature of a gas by 1 K while keeping the volume con-stant.

Heat capacity at constant pressure (C_p): It may be defined as the amount of heat required to raise the temperature of a gas by 1K while keeping the pressure con-stant.
Derivation of the relation C_p – C_v = R :
The equation for heat, q
At constant volume as q_v = C_VΔT = ΔU
At constant pressure as q_p = C_pΔT = ΔH
The difference between C_p and C_v can be derived for an ideal gas as follows.
For one mole of an ideal gas, ΔH = ΔU + Δ PV
We know that Ideal gas equation, PV = nRT
For one mole of an Ideal gas, n = 1. Then PV = RT
∴ ΔH = ΔU + Δ PV
⇒ ΔH = ΔU + Δ RT
⇒ ΔH = ΔU + R ΔT
Substitute the values of AH and AU in the above equation
∴ C_p ΔT = C_v ΔT + R ΔT
⇒ C_p = C_v + R
⇒ C_p – C_v = R

Question 4.
What is entropy? Explain with examples.
Answer:
Entropy (S): Entropy is taken as a measure of disorder of molecules (or) randomness of the system. Greater the disorder of molecules in a system, the higher is the entropy. Entropy is a state function. It depends on the temperature, pressure of the state.

Entropy chage, ΔS = \(\frac{\mathrm{q}_{\mathrm{rev}}}{\mathrm{~T}}\) [q_rev = heat absorbed by the system isorthermally and reversibly at T] For a spontaneous process in an isolated system the entropy change is positive.
(ΔS = positive)

Examples:

1) Melting of Ice : Solid ice has a well ordered molecular structure i.e, it has, low entropy. When it melts to become liquid water, the molecules move more freely i.e., it has higher entropy.

2) Mixing of gases: If we consider two separate gases, each have their own order. When mixed, the molecule become randomly distributed i.e., the mixture has higher disorder. They will increases entropy.

3) Burning of wood : Burning of wood, converts a structures object i.e., wood into ash, smoke and gases. In this process heat is released and particles disperse. This process greatly increase entropy.

Question 5.
Show that ΔH = ΔU + Δn_gRT.
Answer:
The energy change taking place at constant pressure and at a constant temperature is called Enthalpy change (ΔH).
We know that Enthalpy, H = U + PV
U – Internal energy
P – Pressure
V – Volume
Change in Enthalpy, ΔH = ΔU + ΔPV
ΔPV represents the change in the product pv from the initial to final state.
For Ideal gas, PV = nRT
n – Number of moles of gas
R – Universal gas constant
T – Temperature
∴ Δ(PV) = Δn_g.RT
Where, Δn_g is the difference in moles of gaseous products and gaseous reactants.
∴ Δn_g = n_g products – n_g reactants
Substitute Δ(PV) = Δn_gRT into the expression for ΔH
∴ ΔH = ΔU + ΔPV
⇒ ΔH = ΔU + Δn_gRT

Question 6.
What is enthalpy of a reaction? Explain the standard enthalpy of a reaction.
Answer:
Enthalpy is the amount of heat exchanged (absorbed or released) by a system with its surroundings at constant pressure.
If heat is released, the reaction is exothermic i.e., ΔH < 0. If heat is absorbed, the reaction is endothermic i.e., ΔH > 0.
Mathematically, ΔH = ΔU + PΔV
Where ΔU – Internal energy change.
Enthalpy is a state function. Thus the magnitude of enthalpy change depends only upon the enthalpies of the initial and final status.
∴ ΔH = H_products – H_Reactants
For gaseous reactions, ΔH = ΔU + Δn.RT

Standard enthalpy of a reaction (ΔH°) : The standard enthalpy of a reaction (ΔH°) is the enthalpy change when all the reactants and products, are in their standard states i.e., pressure at 1 atm and temperature 298K or 25°C.

Question 7.
Define and explain the standard enthalpy of formation (Δ_fH).
Answer:
The standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its constituent element in their standard states under standard conditions.
Standard state – More stable physical form for elements and pure, well defined state for compounds.
Temperature – 298k or 25°C
Pressure -1 atm or 1 bar.

Examples:

1. H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂OC(l), Δ_fH^θ = -285.8 kJ/mole
2. C(Graphite) + 2H₂(g) → CH₄(g), Δ_f\(\mathrm{H}^{\ominus}\) = -74.81 kJ/mole
3. S(Rhombic) + O₂(g) → SO₂(g), Δ_f\(\mathrm{H}^{\ominus}\) = – 297.5 kJ/mole

Question 8.
Define and explain the enthalpy of combustion (Δ_cH^θ)
Answer:
The standard enthalpy of combustion is the enthalpy change when 1 mole of a substance undergoes complete combustion in excess oxygen under standard conditions, i.e., 298k or 25°C temperature and 1 bar or 1 atm pressure.

Examples:

1) Cooking gas in cylinders contains mostly butane (C₄H₁₀). During complete combustion of 1 mole of butane, 2638 kJ of heat is released.
C₄H₁₀(g) + \(\frac{13}{2}\) O₂(g) → 4CO₂(g) + 5H₂O(l), Δ_c\(\mathrm{H}^{\ominus}\) = -2658.0 kJ/mole.

2) Combustion of glucose gives out 2802.0 kJ/mole of heat.
C₆H₁₂O₆(S) + 6O₂(g) → 6 CO₂(g) + 6 H₂O(l)
Δ_c\(\mathrm{H}^{\ominus}\) =-2802.0 kJ/mole.

Question 9.
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 ?
q = 701 J; w = -394J; ΔU = 307 J.
Answer:
Given that, heat absorbed by the system, q = +701J
Work done by the system, w = -394 J
From the first law of Thermodynamics,
ΔU = q + w
= 701 + (-394)
= 701 – 394
= 307J
∴ The change in internal energy for the process is 307J, This means, the internal energy of the system increases because, the heat absorbed (701J) exceeds the work done by the system (394J). This means the system gains energy.

Question 10.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24J mol^-1K^-1.
Answer:
Give that, mass of aluminium (Al) = 60.0g
Molar mass of Al = 27g/mol
Molar heat capacity = 24J/mol.k
Temperature change, ΔT = 55 – 35 = 20°C
Number of moles of Al = \(\frac{\text { Mass }}{\text { Molar Mass }}\) = \(\frac{60}{27}\) = 2.22 moles
Heat required, q = n × c × ΔT
Where, n – Number of moles of Al = 2.22 moles
C – Molar heat capacity = 24 J/mol.k
ΔT – Temperature change = 20
∴ q = 2.22 × 24 × 20
= 1065.6J
= 1.0656kJ

VI. Long Answer Questions (8 Marks)

Question 1.
Explain the experiment to determine the internal energy change of a chemical reaction.
Answer:
To determine experimentally the internal energy change (ΔU) of a chemical reaction, you would typically use a bomb calorimeter. This involves measuring the heat released or absorbed during the reaction at constant volume, which directly corresponds to the change in internal energy.

A bomb calorimeter is a closed, insulated system designed to measure heat changes at constant volume. It consists of a strong, sealed vessel called the bomb submerged in a water bath. The bomb is pressurized with excess oxygen gas. A known amount of the substance to be burned is placed in a platinum cup inside the bomb. The entire setup is insulated to minimise heat loss to the surroundings.

The temperature of the water bath is carefully measured before the reaction. The substance is ignited by an electrical spark. The reaction occurs rapidly and at constant volume within the sealed bomb.

The heat released by the reaction is absorbed by the water in the bath, causing its temperature to rise. The temperature of the water is carefully measured after the reaction is complete. The change in temperature (ΔT) is used to calculate the heat released (q) by the reaction.

Calculations:

Heat Absorbed by Water & Calorimeter (q)
q = (m_w. c_w + C_cat). ΔT
m_w = mass of water
c_w = specific heat capacity of water (~4.18 J/g°C)
C_cal = heat capacity of the calorimeter (determined via calibration)
ΔT = T₂ – T₁
T₁ – Initial temperature
T₂ – Final temperature
Internal Energy Change (ΔU): Since the reaction occurs at constant volume, the heat change (q) equals ΔU.
ΔU = q_v
If the reaction is exothermic, ΔU is negative.
If the reaction is endothermic, ΔU is positive.

Question 2.
Explain the experiment to determine the enthalpy change of a chemical reaction.
Answer:
To determine the enthalpy change (ΔH) of a chemical-reaction, experiments typically involve measuring the heat released obsorbed during the reaction.

This is done using a calorimeter, a device that isolates the reaction from its surroundings to minimize heat loss.

A simple calorimeter can be a nested Styrofoam cup or a more sophisticated one like a bomb calorimeter. Record the initial temperature of the reactants. Allow the reaction to occur within the calorimeter. Monitor and record the temperature change of the reactants as the reaction progresses.

Use the temperature change, the specific heat capacity of the system, and the mass of the reactants to calculate the amount of heat (q) exchanged during the reaction.

The enthalpy change (ΔH) is numerically equal to the heat exchanged (q) at constant pressure, allowing you to determine if the reaction is exothermic (heat released, ΔH < 0) or endothermic (heat absorbed, ΔH > 0).

Examples of enthalpy change experiments:
Neutralization reactions: Measuring the enthalpy change when an acid and a base are mixed.
Dissolution reactions: Measuring the enthalpy change when a solute dissolves in a solvent.
Combustion reactions: Measuring the enthalpy change when a fuel is burned.

Calculations
Heat Exchanged (q) at Constant Pressure
Q =m.c.ΔT
m = total mass of the solution (assume density ≈ 1 g/mL for water).
c = specific heat capacity (~4.18 J/g°C for water).
ΔT= temperature change.
Enthalpy Change (ΔH)
Since the experiment is done at constant pressure, ΔH=qp.
If heat is released (exothermic), ΔH is negative.
If heat is absorbed (endothermic), ΔH is positive.

Question 3.
Explain the spontaneity of a reaction interms of enthalpy change, entropy change and Gibb’s energy change.
Answer:
A reaction’s spontaneity is its ability to occur naturally without external energy input. It is determined by the Gibbs free energy change (ΔG), which combines enthalpy change (ΔH) and entropy change (ΔS) with temperature (T). A negative ΔG indicates a spontaneous reaction, while a positive ΔG indicates a non-spon-taneous reaction.

Gibbs Free Energy Change (ΔG): This thermodynamic property combines en-thalpy and entropy to predict spontaneity. The equation is ΔG = ΔH – TΔS.
ΔG – Gibbs free energy
T – Temperature in Kelvin
ΔH – Enthalpy change
ΔS – Entropy change
Enthalpy Change (ΔH): This represents the heat absorbed or released during a reaction. A negative ΔH indicates an exothermic reaction (heat released), and a positive ΔH indicates an endothermic reaction (heat absorbed).

Entropy Change (ΔS): This measures the degree of disorder or randomness in a system. The reactions with positive ΔS indicates an increase in disorder, and they are more likely to be spontaneous. The reactions with negative ΔS indicates a decrease in disorder and they are less likely to be spontaneous.

Temperature (T): The temperature plays a crucial role in determining the relative importance of enthalpy and entropy in driving the reaction.

Relation with spontaneity:

1. Exothermic Reactions (ΔH < 0): Reactions that release heat are generally spontaneous. If a reaction is exothermic and increases entropy (ΔS > 0), it will always be spontaneous, regardless of temperature, because AG will always be negative.

2. Endothermic Reactions (ΔH > 0): Reactions that absorb heat are not always spontaneous. For an endothermic reaction to be spontaneous, the increase in entropy (ΔS > 0) must be large enough to outweigh the positive enthalpy change, especially at high temperatures.

3. Entropy and Temperature : The temperature term (TΔS) in the Gibbs free energy equation is crucial. At higher temperatures, the entropy change becomes more important in determining spontaneity.

4. Spontaneous Reactions (ΔG < 0): If AG is negative, the reaction is sponta-neous in the forward direction. A negative ΔG means that the reaction will proceed naturally, without external energy input.

5. Non-Spontaneous Reactions (ΔG > 0) : If ΔG is positive, the reaction is non-spontaneous in the forward direction. The reaction requires energy input to proceed.