**Question 1**

Choose the correct answer

A thermodynamic state junction is a quantity

(i) used to determine heat changes

(ii) whose value is independent of path

(iii) used to determine pressure volume work

(iv) whose value depends on temperature only

(ii) whose value is independent of path **(Correct Option)**

In thermodynamics, function of state, state quantity, or state variable is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state.

A state function describes the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of a path.

Functions like p, V, T etc. depend only on the state of a system and not on the path.

**Question 2**

For the process to occur under adiabatic conditions, the correct condition is

(i) ∆T= 0

(ii) ∆p = 0

(iii) ∆q = 0

(iv) w = 0

Adiabatic Condition means that there’s no transfer of heat between the system and surroundings. Therefore option **(iii) ∆q = 0** is correct

**Question 3**

The enthalpies of all elements in their standard states are

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

By definition, the enthalpy of formation of elements in their standard state is taken as zero.

Therefore, The enthalpies of all elements in their standard states is zero irrespective of the element.

Thus option **(ii) zero** is correct.

**Question 4**

ΔU^{⊖} of combustion of methane is – X kJ mol^{-1}

The value of ΔH^{⊖} is

(i) = ΔU^{⊖}

(ii) > ΔU^{⊖}

(iii) < ΔU^{⊖}

(iv) 0

Option **(iii) < ΔU ^{⊖}** is correct

We know that

ΔH

^{⊖}= ΔU

^{⊖}+ Δn

_{g}RT

Chemical Equation for combustion of methane is

CH

_{4}(g) + 2O

_{2}(g) → CO

_{2}(g) + 2H

_{2}O (l)

Δn

_{g}= Number of gaseous moles in product side – number of gaseous moles in reactant side

Number of gaseous moles in product side = 1

Number of gaseous moles in reactant side = 3

Δn

_{g}= 1 – 3 = – 2

Δn

_{g}= – 2

ΔU

^{⊖}= – X

Replacing these values in equation ΔH

^{⊖}= ΔU

^{⊖}+ Δn

_{g}RT

ΔH

^{⊖}= – X – 2RT

Therefore

ΔH

^{⊖}< ΔU

^{⊖}

Thus (iii) < ΔU

^{⊖}is correct option

**Question 5**

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 KJ mol^{-1}, – 393.5 KJ mol^{-1} and

– 285.8 KJ mol^{-1} respectively.

Enthalpy of formation of CH_{4}(g) will be

(i) – 74.8 KJ mol^{-1 }

(ii) – 52.27 KJ mol^{-1}

(iii) + 74.8 KJ mol^{-1}

(iv) + 52.26 KJ mol^{-1}

**Question 6**

A reaction, A + B → C + D + q is found to have a positive entropy change.

The reaction will be

(i) possible at high temperature

(ii) possible only at low temperature

(iii) not possible at any temperature

(iv) possible at any temperature

**(iv) possible at any temperature** is correct option

As per Gibbs Free Energy Equation

ΔG = ΔH – TΔS

Where

ΔG = Change in Gibbs Free Energy

ΔH = Change in Enthalpy

T = Temperature

ΔS = Change in Entropy

For reaction

A + B → C + D + q

ΔS > 0

ΔH < 0 (Because heat is being released see q on right hand side of chemical equation)

Therefore as per equation ΔG = ΔH – TΔS will be always ΔG < 0 no matter what is value of temperature T.

**Question 7**

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system.

What is the change in internal energy for the process?

Change in internal energy **ΔU = q + w**

As per question its given that

q = 701 J

w = – 394 J (negative because work is done by system)

Plugging these values in equation **ΔU = q + w**

ΔU = 701 – 394 = 307 J**ΔU = 307 J**

Therefore if in a process 701 J of heat is absorbed by a system and 394 J of work is being done by system then change in internal energy during this process will be **307 J**

**Question 8**

The reaction of cyanamide NH_{2}CN(s) with dioxygen was carried out in a bomb calorimeter and ∆U was found to be -742.7 KJ^{ }mol^{-1 }at 298 K.

Calculate the enthalpy change for the reaction at 298 K

NH_{2}CN (s) + 3/2 O_{2}(g) → N_{2}(g) + CO_{2}(g) + H_{2}O(l)

We know that

ΔH^{⊖} = ΔU^{⊖} + Δn_{g}RT

Chemical Equation for combustion of methane is

NH_{2}CN (s) + 3/2 O_{2}(g) → N_{2}(g) + CO_{2}(g) + H_{2}O(l)

Δn_{g} = Number of gaseous moles in product side – number of gaseous moles in reactant side

Number of gaseous moles in product side = 2

Number of gaseous moles in reactant side = 3/2

Δn_{g} = 2 – 3/2 = (4 – 3)/2 = 1/2

Δn_{g} = 1/2

ΔU^{⊖} = – 742.7 kJ mol^{-1}

R = 8.314 × 10^{-3} kJ K^{-1} mol^{-1}

T = 298 K

Replacing these values in equation ΔH^{⊖} = ΔU^{⊖} + Δn_{g}RT

ΔH^{⊖} = – 742.7 + (1/2) × 8.314 × 10^{-3} × 298

ΔH^{⊖} = – 742.7 + 4.157 × 10^{-3} × 298

ΔH^{⊖} = – 742.7 + 1238.786 × 10^{-3}

ΔH^{⊖} = – 742.7 + 1.238 = −741.462

ΔH^{⊖} = −741.462 = – 741.5 kJ**ΔH ^{⊖} = – 741.5 kJ**

**Question 9**

Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from 35 °C to 55 °C.

Molar heat capacity of Al is 24 J mol^{-1} K^{-1}

Relationship between molar heat capacity and heat is **q = mc(T _{2} – T_{1})**

Where

q = heat

m = amount of substance (moles)

c = molar heat capacity

T

_{1}= initial temperature

T

_{2}= final temperature

c = 24 J mol

^{-1}K

^{-1}

m = 60/27 moles

T

_{1}= 35 + 273 = 308 K

T

_{2}= 55 + 273 = 328 K

Plugging all these values in equation

**q = mc(T**

_{2}– T_{1})q = (60/27) × 24 × (328 – 308)

q = 2.22 × 24 × 20 = 1065.6 J

**q = 1.0656 kJ**

**Question 10**

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 °C to ice at – 10.0°C

Δ_{fus}H = 6.03 kJ mol^{-1} at 0°C

Cp [H_{2}O (l)] = 75.3 J mol^{-1} K^{-1}

Cp [H_{2}O (l)] = 36.8 J mol^{-1} K^{-1}

**Question 11**

Enthalpy of combustion of carbon to carbon dioxide is – 393.5 J mol^{-1}

Calculate the heat released upon formation of 35.2 g of CO_{2 } from carbon and oxygen gas

Heat of combustion for 1 mole of CO_{2} is – 393.5 kJ

Number of moles in 35.2 g of CO_{2} = 35.2/44 = 0.8 mole

Thus heat of combustion for 0.8 mole of CO_{2} is – 393.5 × 0.8 = **– 315 kJ**

Therefore heat released upon combustion of 35.2 g of CO_{2} from carbon and oxygen gas is **– 315 kJ**

**Question 12**

Calculate the enthalpy of the reaction

N_{2}O_{4}(g) + 3CO(g) → N_{2}O(g) + 3CO_{2}(g)

Given that

∆_{f}H CO(g) = – 110 kJ mol^{-1}

∆_{f}H CO_{2}(g) = – 393 kJ mol^{-1}

∆_{f}H N_{2}O(g) = 81 kJ mol^{-1}

∆_{f}H N_{2}O_{4}(g) = 9.7 kJ mol^{-1}

Enthalpy of reaction = total heat of formation for products – total heat of formation for reactants

= [81 + 3(- 393)] – [9.7 + 3(- 110)]

= [81 – 1179] – [9.7 – 330]

= – 1098 – (- 320.3)

= – 1098 + 320.3

= – 777.7 kJ mol^{-1}

Therefore **enthalpy of reaction is – 777.7 kJ mol ^{-1}**

**Question 13**

Given N_{2}(g) + H_{2}(g) → 2 NH_{3} (g)

∆_{r}H^{⊖} = – 92.4 kJ mol^{-1}

What is the standard enthalpy of formation of NH_{3} gas?

Standard enthalpy of formation of NH_{3} gas = ∆_{r}H^{⊖}/2 = – 92.4/2 = – 46.2 kJ mol^{-1}

Therefore standard enthalpy of formation of NH_{3} gas is **– 46.2 kJ mol ^{-1}**

**Question 14**

Calculate the standard enthalpy of formation of CH_{3}OH (l) form the following data.

CH_{3}OH (l) + 3/2 O_{2} (g) → CO_{2} (g) + 2 H_{2}O (l) ∆_{r}H^{⊖} = – 726 kJ mol^{-1}

C (s) + O_{2} (g) → CO_{2} (g) ∆_{c}H^{⊖} = – 393 kJ mol^{-1}

H_{2} (g) + O_{2} (g) → H_{2}O (l) ∆_{f}H^{⊖} = – 286 kJ mol^{-1}

Therefore standard enthalpy of formation of CH_{3}OH (l) is **– 239 kJ mol ^{-1}**

**Question 15**

Calculate the enthalpy change for the process

CCl_{4} (g) → C (g) + 4 Cl (g) and calculate bond enthalpy of C—Cl in CC14 (g)

Given

∆_{vap} H° (CCl_{4}) = 30.5 kJ mol^{-1}

∆_{f}H°(CCl_{4}) = – 135.5 kJ mol^{-1}

∆_{a}H° (C) = 715.0 kJ mol^{-1} where ∆_{a} H° is enthalpy of atomisation

∆_{a}H° (Cl_{2}) = 242 kJ mol^{-1}

The chemical equations implying to the given values of enthalpies are:

(i) CCl_{4(l)} → CCL_{4(g)} Δ_{vap}H**°** = 30.5 kJ mol^{-1}

(ii) C_{(s) } → C_{(g)} Δ_{a}H**°** = 715.0 kJ mol^{-1}

(iii) Cl_{2(g)} → 2Cl_{(g) } Δ_{a}H**°** = 242 kJ mol^{-1}

(iv) C_{(g) } + 4Cl_{(g)} → CCl_{4(g)} Δ_{f}H = -135.5 kJ mol^{-1}

Enthalpy change for the given process C_{(g) } + 4Cl_{(g)} → CCl_{4(g)} can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

ΔH = Δ_{a}H**°**(C) + 2Δ_{a}H**°** (Cl_{2}) – Δ_{vap}H**°** – Δ_{f}H

= (715.0 kJ mol^{-1}) + 2(242 kJ mol^{-1}) – (30.5 kJ mol^{-1}) – (-135.5 kJ mol^{-1})

∴ΔH = 1304 kJ mol^{-1}

**Bond enthalpy of C-Cl bond in CCl _{4(g)} = 326 kJ mol^{-1}**

**Question 16**

For an isolated system ΔU = 0

What will be ΔS?

A system that is isolated from its surroundings will have no change in its internal energy, denoted by the symbol U, since it will not be exchanging any energy with its surroundings.

However, entropy has a tendency to rise whenever there is a spontaneous process. Therefore, S must be greater than 0 or positive.

**Question 17**

For a reaction at 298 K

2A + B → C

∆H = 400 kJ mol^{-1} and ∆S = 0.2 kJ mol^{-1}

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?

Gibbs-Helmholtz Equation

∆G = ∆H – T∆S

For reaction to be spontaneous ∆G = 0

Therefore

0 = ∆H – T∆S

∆H = T∆S

T = ∆H/∆S

Replacing ∆H = 400 kJ mol^{-1} and ∆S = 0.2 kJ mol^{-1}

T = 400/0.2 = 2000 K**T = 2000 K**

Therefore this reaction will be in a state of equilibrium at temperature 2000 K and will be spontaneous above this temperature. So for reaction to be spontaneous temperature should be greater than 2000 K.

**Question 18**

For the reaction 2 Cl (g) → Cl_{2} (g) what will be the signs of ∆H and ∆S?

Reaction 2 Cl (g) → Cl_{2} (g) is showing formation of chlorine molecule from chlorine atoms, which means bonds are being created therefore energy is released. Thus ∆H for reaction 2 Cl (g) → Cl_{2} (g) is negative.

Also in the reaction 2 Cl (g) → Cl_{2} (g) two atoms are combining to form one molecule therefore randomness of atoms is decreasing therefore spontaneity is decreasing as chlorine atoms combing to form chlorine molecule. Therefore ∆S for the reaction 2 Cl (g) → Cl_{2} (g) is negative.

Thus for reaction 2 Cl (g) → Cl_{2} (g) signs of both ∆H and ∆S are negative.

**Question 19**

For the reaction

2 A(g) + B(g) → 2D(g)

ΔU**°** = – 10.5 kJ

ΔS**°** = – 44.1 J K^{–1} mol^{-1}

Calculate ΔG^{0} for the reaction, and predict whether the reaction may occur spontaneously.

For reaction 2 A(g) + B(g) → 2D(g)

ΔU° = – 10.5 kJ

ΔS° = – 44.1 J K^{–1} mol^{-1}

value of ΔG° is + 5.12 kJ therefore reaction will not occur spontaneously.

**Question 20**

Equilibrium constant for a reaction is 10

Calculate the value of ΔG^{⊖} given that R = 8 J K^{-1} mol^{-1} and T = 300 K

Equation for calculating value of ΔG^{⊖} is

ΔG^{⊖} = – RT ln K

Where

ΔG^{⊖} = Gibbs Free Energy

R = Gas Constant

T = Temperature

K = Equilibrium Constant

Replacing

R = 8 J K^{-1} mol^{-1}

T = 300 K

K = 10

Plugging all these values in equation ΔG^{⊖} = – RT ln K

ΔG^{⊖} = – 8 × 300 × ln 10

Value of **ln 10 = 2.302**

ΔG^{⊖} = – 8 × 300 × 2.302 = – 5524.8 J mol^{-1}

ΔG^{⊖} = – 5.5248 kJ mol^{-1}

Therefore if equilibrium constant for a reaction is 10 at temperature 300 K then value of Gibbs Free Energy will be **– 5.5248 kJ mol ^{-1}**

**Question 21**

Comment on the thermodynamic stability of NO(g) and NO_{2}(g) given

1/2 NO_{2} (g) + 1/2 O_{2} (g) → NO (g) Δ_{r}H^{⊖} = 90 kJ mol^{-1}

NO (g) + 1/2 O_{2} (g) → NO_{2} (g) Δ_{r}H^{⊖} = – 74 kJ mol^{-1}

If Δ_{r}H^{⊖} is positive then it means that products have more energy as compared to reactants, therefore products are unstable.

If Δ_{r}H^{⊖} is negative then it means that products have less energy as compared to reactants, therefore products are more stable as compared to reactants.

Based on these facts

For reaction 1/2 NO_{2} (g) + 1/2 O_{2} (g) → NO (g) value of Δ_{r}H^{⊖} is positive therefore NO(g) is unstable in nature.

For reaction NO (g) + 1/2 O_{2} (g) → NO_{2} (g) value of Δ_{r}H^{⊖} is negative therefore NO_{2}(g) is stable in nature.

**Question 22**Calculate the entropy change in surroundings when 1.0 mole of H

_{2}O (l) is formed under standard conditions.

Given that Δ

_{f}H

^{⊖}= – 286 kJ mol

^{-1}

It is given that 286 kJ mol^{-1}of heat is evolved on the formation of 1 mol of H_{2}O (l).

Thus, an equal amount of heat will be absorbed by the surroundings.

q_{surr }= +286 kJ mol^{-1}

Entropy change (ΔS_{surr}) for the surroundings = q_{surr }**/** T

= 286/298

**ΔS _{surr }= 0.959 J mol^{-1}K^{-1}**