1. Choose the Correct Answer. a Thermodynamic State Function 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. 

Ans: A thermodynamic state function is a quantity whose value is independent of the path. Functions like p, V, T etc. depend only on the state of a system and not on the path. 

Hence, alternative (ii) is correct. 

2. For the Process to Occur Under Adiabatic Conditions, the Correct Condition Is: 

(i)  Δ T=0 Δ T=0 

(ii)  Δ p=0 Δ p=0 

(iii) q=0q=0 

(iv) w=0w=0 

Ans: A system is said to be under adiabatic conditions if there is zero exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q=0𝑞=0. Therefore, alternative (iii) is correct. 

3. The Enthalpies of All Elements in Their Standard States Are: 

(i) unity 

(ii) zero 

(iii) < 0 

(iv) different for each element 

Ans: The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct.

4.  Δ U θ  Δ U 𝜃  of combustion of methane is -XkJ mol-1-XkJ mol-1. The value of  Δ Hθ Δ H𝜃is 

(i) =  Δ U θ  Δ U 𝜃 

(ii) >  Δ U θ  Δ U 𝜃  

(iii) <  Δ U θ  Δ U 𝜃  

(iv) = 0 

Ans: Since 

ΔHθ = ΔUθ + ΔngRTΔ𝐻𝜃 = Δ𝑈𝜃 + Δ𝑛𝑔𝑅𝑇

and

ΔUθ=−XkJ mol−1Δ𝑈𝜃=−𝑋𝑘𝐽 𝑚𝑜𝑙−1

, 

ΔHθ = (−X) + ΔngRTΔ𝐻𝜃 = (−𝑋) + Δ𝑛𝑔𝑅𝑇

⇒ΔHθ<ΔUθ⇒Δ𝐻𝜃<Δ𝑈𝜃

Therefore, alternative (iii) is correct.

5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol-1kJ mol-1 ,-393.5 kJ mol-1kJ mol-1, and -285.8 kJ mol-1kJ mol-1 respectively. Enthalpy of formation of CH4(g)CH4(g) will be 

(i) -74.8 kJ mol-1kJ mol-1

(ii) -52.27 kJ mol-1kJ mol-1  

(iii) +74.8 kJ mol-1kJ mol-1

(iv) +52.26 kJ mol-1kJ mol-1  . 

Ans: According to the question, 

(i) CH4(g)+2O2(g)→CO2(g)+2H2O(l);ΔcHΘ=−890.3kJmol−1𝐶𝐻4(𝑔)+2𝑂2(𝑔)→𝐶𝑂2(𝑔)+2𝐻2𝑂(𝑙);Δ𝑐𝐻Θ=−890.3𝑘𝐽𝑚𝑜𝑙−1 

(ii) C(s)+2O2(g)→CO2(g);ΔcHΘ=−393.5kJmol−1𝐶(𝑠)+2𝑂2(𝑔)→𝐶𝑂2(𝑔);Δ𝑐𝐻Θ=−393.5𝑘𝐽𝑚𝑜𝑙−1 

(iii) 2H2(g)+O2(g)→2H2O(l);ΔcHΘ=−285.8kJmol−12𝐻2(𝑔)+𝑂2(𝑔)→2𝐻2𝑂(𝑙);Δ𝑐𝐻Θ=−285.8𝑘𝐽𝑚𝑜𝑙−1 

Thus, the desired equation is the one that represents the formation of CH4(g)𝐶𝐻4(𝑔)that is as follows:

C(s)+2H2(g)→CH4(g);ΔfHCH4=ΔcHc+2ΔcHH2−ΔcHCO2𝐶(𝑠)+2𝐻2(𝑔)→𝐶𝐻4(𝑔);Δ𝑓𝐻𝐶𝐻4=Δ𝑐𝐻𝑐+2Δ𝑐𝐻𝐻2−Δ𝑐𝐻𝐶𝑂2

Substituting the values in the above formula :

Enthalpy of formationCH4(g)𝐶𝐻4(𝑔)=(−393.5)+2×(−285.8)−(−890.3)=−74.8kJmol−1(−393.5)+2×(−285.8)−(−890.3)=−74.8𝑘𝐽𝑚𝑜𝑙−1 

Therefore, alternative (i) is correct.

6. A Reaction, A + B →C + D + qA + 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 

Ans: For a reaction to be spontaneous, ΔGΔ𝐺 should be negative ΔG = ΔH − TΔSΔ𝐺 = Δ𝐻 − 𝑇Δ𝑆 

According to the question, for the given reaction, 

ΔS =Δ𝑆 = positive 

ΔH=Δ𝐻= negative (since heat is evolved) 

That results in ΔG=Δ𝐺= negative

Therefore, the reaction is spontaneous at any temperature. 

Hence, alternative (iv) is correct. 

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? 

Ans: According to the first law of thermodynamics, 

ΔU= q + W….(i)Δ𝑈= 𝑞 + 𝑊….(𝑖)

Where, 

ΔUΔ𝑈 = change in internal energy for a process 

q = heat 

W = work 

Given, 

q = + 701 J (Since heat is absorbed) 

W = -394 J (Since work is done by the system)

Substituting the values in expression (i), we get 

ΔU= 701 J + (−394 J)Δ𝑈= 701 𝐽 + (−394 𝐽)

ΔU= 307 JΔ𝑈= 307 𝐽

Hence, the change in internal energy for the given process is 307 J.

8.The reaction of cyanamide, NH2CN(s)𝑁𝐻2𝐶𝑁(𝑠) with dioxygen was carried out in a bomb calorimeter and  Δ U Δ U was found to be -742.7 KJ mol−1𝐾𝐽 𝑚𝑜𝑙−1 at 298 K. Calculate the enthalpy change for the reaction at 298 K. 

NH4CN(g)+32O2(g)→N2(g)+CO2(g)+H2O(l)NH4CN(g)+32O2(g)→N2(g)+CO2(g)+H2O(l)

Ans: Enthalpy change for a reaction (ΔH)(Δ𝐻) is given by the expression, 

ΔH = ΔU + ΔngRTΔ𝐻 = Δ𝑈 + Δ𝑛𝑔𝑅𝑇

Where, 

ΔUΔ𝑈 = change in internal energy 

ΔngΔ𝑛𝑔 = change in number of moles 

For the given reaction, 

Δng=∑ngΔ𝑛𝑔=∑𝑛𝑔 (products) – ∑ng∑𝑛𝑔(reactants)

Δng=(2−1.5)Δ𝑛𝑔=(2−1.5)moles

Δng=+0.5Δ𝑛𝑔=+0.5moles

And, ΔUΔ𝑈= -742.7 kJ mol−1𝑘𝐽 𝑚𝑜𝑙−1 

T = 298 K 

R = 8.314×10−3kJmol−1K−18.314×10−3𝑘𝐽𝑚𝑜𝑙−1𝐾−1

Substituting the values in the expression of ΔH Δ𝐻 

ΔH = (−742.7 kJ mol−1) + (+0.5 mol) (298 K)8.314×10−3kJmol−1K−1Δ𝐻 = (−742.7 𝑘𝐽 𝑚𝑜𝑙−1) + (+0.5 𝑚𝑜𝑙) (298 𝐾)8.314×10−3𝑘𝐽𝑚𝑜𝑙−1𝐾−1

ΔH Δ𝐻  = -742.7 + 1.2 

ΔH = −741.5kJ mol−1Δ𝐻 = −741.5𝑘𝐽 𝑚𝑜𝑙−1

9. Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from 35∘C35∘𝐶 to 55 ∘ C55 ∘ C. Molar heat capacity of Al is 24J mol-1K-124J mol-1K-1. 

Ans: From the expression of heat (q), 

q = m. c. ΔT𝑞 = 𝑚. 𝑐. Δ𝑇

Where, 

c = molar heat capacity 

m = mass of substance 

ΔTΔ𝑇 = change in temperature 

Given, 

m = 60 g

c = 24J mol−1K−124𝐽 𝑚𝑜𝑙−1𝐾−1

DeltaT=(55−35)∘C𝐷𝑒𝑙𝑡𝑎𝑇=(55−35)∘𝐶

ΔT=(328−308)K=20KΔ𝑇=(328−308)𝐾=20𝐾

Substituting the values in the expression of heat:

q=(6027mol)(24Jmol−1K−1)(20K)𝑞=(6027𝑚𝑜𝑙)(24𝐽𝑚𝑜𝑙−1𝐾−1)(20𝐾)

q = 1066.7 J

q = 1.07 kJ

10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 ∘ C10.0 ∘ C  to ice at -10.0 ∘ C-10.0 ∘ C,  Δ fusH = 6.03 KJ mol-1 Δ fusH = 6.03 KJ mol-1  at 0 ∘ C0 ∘ C.

   Cp [H2O(l) ] = 75.3 J mol-1 K-1   Cp [H2O(l) ] = 75.3 J mol-1 K-1

   Cp [H2O(s) ] = 36.8 J mol-1 K-1   Cp [H2O(s) ] = 36.8 J mol-1 K-1

.

Ans: Total enthalpy change involved in the transformation is the sum of the following changes: 

  1. Energy change involved in the transformation of 1 mol of water at 10.0∘C10.0∘𝐶  to 1mol of water at0∘C0∘𝐶.
  2. Energy change involved in the transformation of 1 mol of water at 0∘C0∘𝐶to 1 mol of ice at 0∘C0∘𝐶. 
  3. Energy change involved in the transformation of 1 mol of ice at 0∘C0∘𝐶 to 1 mol of ice at 10∘C10∘𝐶. 

Total ΔH= Cp [H2O(l)] ΔT+ΔHfreezing+ Cp [H2O(s)] ΔTΔ𝐻= 𝐶𝑝 [𝐻2𝑂(𝑙)] Δ𝑇+Δ𝐻𝑓𝑟𝑒𝑒𝑧𝑖𝑛𝑔+ 𝐶𝑝 [𝐻2𝑂(𝑠)] Δ𝑇 

ΔH=(75.3 Jmol−1K−1)(0−10)K+(−6.03×103Jmol−1)+(36.8 Jmol−1K−1)(−10−0)KΔ𝐻=(75.3 𝐽𝑚𝑜𝑙−1𝐾−1)(0−10)𝐾+(−6.03×103𝐽𝑚𝑜𝑙−1)+(36.8 𝐽𝑚𝑜𝑙−1𝐾−1)(−10−0)𝐾

ΔH= −753 J mol−1 − 6030 J mol−1 − 368 J mol−1Δ𝐻= −753 𝐽 𝑚𝑜𝑙−1 − 6030 𝐽 𝑚𝑜𝑙−1 − 368 𝐽 𝑚𝑜𝑙−1

ΔH= −7151 J mol−1Δ𝐻= −7151 𝐽 𝑚𝑜𝑙−1

ΔH= −7.151 kJ mol−1Δ𝐻= −7.151 𝑘𝐽 𝑚𝑜𝑙−1

Hence, the enthalpy change involved in the transformation is  −7.151 kJ mol−1 −7.151 𝑘𝐽 𝑚𝑜𝑙−1 

11. Enthalpy of combustion of carbon to carbon dioxide is −393.5 kJ mol−1−393.5 𝑘𝐽 𝑚𝑜𝑙−1 Calculate the heat released upon formation of 35.2 g of CO2CO2 from carbon and dioxygen gas. 

Ans: Formation of CO2𝐶𝑂2 from carbon and dioxygen gas can be represented as 

C(s)+O2(g) toCO2(g);ΔH=−393.5kJ,mol−1𝐶(𝑠)+𝑂2(𝑔) 𝑡𝑜𝐶𝑂2(𝑔);Δ𝐻=−393.5𝑘𝐽,𝑚𝑜𝑙−1

(1mole=44g) 

Heat released in the formation of 44 g of CO2𝐶𝑂2 = 393.5 kJmoll−1𝑘𝐽𝑚𝑜𝑙𝑙−1 

Heat released in the formation of 35.2 g of 

CO2=(393.5kJ)×(35.2g)(44g)=314.8kJ𝐶𝑂2=(393.5𝑘𝐽)×(35.2𝑔)(44𝑔)=314.8𝑘𝐽

So, heat released upon formation of 35.2 g of CO2CO2from carbon and dioxygen gas is 314.8 kJ.

12. Enthalpies of formation of CO (g),CO2(g)CO2(g), N2O(g)N2O(g) and N2O4(g)N2O4(g)are -110 ,-393, 81 kJ and 9.7 kJ mol-1𝑘𝐽 𝑚𝑜𝑙-1 respectively. Find the value of Δ rH Δ rHfor the reaction: 

N2O4(g)+3CO(g)→N2O(g)+3CO2(g)N2O4(g)+3CO(g)→N2O(g)+3CO2(g)

Ans: ΔrHΔ𝑟𝐻 for a reaction is defined as the difference between ΔfHΔ𝑓𝐻value of products and ΔfHΔ𝑓𝐻value of reactants. 

ΔrHΔ𝑟𝐻= ∑ΔfH∑Δ𝑓𝐻 (product) – ∑ΔfH∑Δ𝑓𝐻(reactant)

For the given reaction, 

N2O4(g)+3CO(g)→N2O(g)+3CO2(g)𝑁2𝑂4(𝑔)+3𝐶𝑂(𝑔)→𝑁2𝑂(𝑔)+3𝐶𝑂2(𝑔)

ΔrH=[{ΔfH(NO2)+3ΔfH(CO2)}−{ΔfH(N2O)+3ΔfH(CO)}]Δ𝑟𝐻=[{Δ𝑓𝐻(𝑁𝑂2)+3Δ𝑓𝐻(𝐶𝑂2)}−{Δ𝑓𝐻(𝑁2𝑂)+3Δ𝑓𝐻(𝐶𝑂)}] 

Substituting the values of ΔfHΔ𝑓𝐻for CO (g),CO2(g)𝐶𝑂2(𝑔), N2O(g)𝑁2𝑂(𝑔) and N2O4(g)𝑁2𝑂4(𝑔)from the question, we get: 

ΔrH=[{81kJmol−1+3(−393)kJmol−1}−{9.7kJmol−1+3(−110)kJmol−1}]Δ𝑟𝐻=[{81𝑘𝐽𝑚𝑜𝑙−1+3(−393)𝑘𝐽𝑚𝑜𝑙−1}−{9.7𝑘𝐽𝑚𝑜𝑙−1+3(−110)𝑘𝐽𝑚𝑜𝑙−1}]

ΔrH=−777.7kJmol−1Δ𝑟𝐻=−777.7𝑘𝐽𝑚𝑜𝑙−1

Hence, the value of ΔrHΔ𝑟𝐻 for the reaction is −777.7kJmol−1−777.7𝑘𝐽𝑚𝑜𝑙−1

13. Given 

N2(g)+3H2(g)→2NH3(g); Δ rH θ =-92.4kJmol-1N2(g)+3H2(g)→2NH3(g); Δ rH 𝜃 =-92.4kJmol-1

What is the standard enthalpy of formation of NH3NH3 gas?

Ans: Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state. 

Re-writing the given equation for 1 mole of NH3(g)𝑁𝐻3(𝑔)is as follows:

12N2(g)+32H2(g)→2NH3(g)12𝑁2(𝑔)+32𝐻2(𝑔)→2𝑁𝐻3(𝑔)

Therefore, standard enthalpy of formation of NH3(g)𝑁𝐻3(𝑔)

= 1/2ΔrHθ1/2Δ𝑟𝐻𝜃 

1/2 (−92.4 kJ mol−1)1/2 (−92.4 𝑘𝐽 𝑚𝑜𝑙−1)

−46.2 kJ mol−1−46.2 𝑘𝐽 𝑚𝑜𝑙−1

14. Calculate the standard enthalpy of formation of CH3OH(ℓ)CH3OH(ℓ) from the following data: 

CH3OH(l)+32O2(g)→CO2(g)+2H2O(l), Δ rH θ =-726   kJ   mol-1CH3OH(l)+32O2(g)→CO2(g)+2H2O(l), Δ rH 𝜃 =-726   kJ   mol-1

C(g)+O2(g)→CO2(g); Δ cH θ =-393   kJ   mol-1C(g)+O2(g)→CO2(g); Δ cH 𝜃 =-393   kJ   mol-1

H2(g)+12O2(g)→H2O(l); Δ fH θ =-286kJ   mol-1H2(g)+12O2(g)→H2O(l); Δ fH 𝜃 =-286kJ   mol-1

Ans: The reaction that takes place during the formation of CH3OH(ℓ)𝐶𝐻3𝑂𝐻(ℓ)can be written as: 

C(s)+2H2O(g)+12O2(g)→CH3OH(ℓ)(1)𝐶(𝑠)+2𝐻2𝑂(𝑔)+12𝑂2(𝑔)→𝐶𝐻3𝑂𝐻(ℓ)(1)

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as: 

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

ΔfHθ[CH3OH(ℓ)]=ΔcHθ+2ΔfHθ[H2O(l)]−ΔrHθΔ𝑓𝐻𝜃[𝐶𝐻3𝑂𝐻(ℓ)]=Δ𝑐𝐻𝜃+2Δ𝑓𝐻𝜃[𝐻2𝑂(𝑙)]−Δ𝑟𝐻𝜃

= (-393kJmol−1𝑘𝐽𝑚𝑜𝑙−1 ) + 2(-286 kJmol−1𝑘𝐽𝑚𝑜𝑙−1) – (-726 kJmol−1𝑘𝐽𝑚𝑜𝑙−1) 

= (-393 – 572 + 726) kJmol−1𝑘𝐽𝑚𝑜𝑙−1

Therefore, ΔfHθ[CH3OH(ℓ)]=−239 kJmol−1Δ𝑓𝐻𝜃[𝐶𝐻3𝑂𝐻(ℓ)]=−239 𝑘𝐽𝑚𝑜𝑙−1 

15. Calculate the Enthalpy Change for the Process 

CCl4(g) → C(g) + 4Cl(g)CCl4(g) → C(g) + 4Cl(g)

and calculate bond enthalpy of C-Cl in CCl4(g)CCl4(g).

 Δ vap H θ (CCl4)=30.5   kJmol-1 Δ vap H 𝜃 (CCl4)=30.5   kJmol-1

 Δ fH θ (CCl4)=-135.5   kJmol-1 Δ 𝑓H 𝜃 (CCl4)=-135.5   kJmol-1

 Δ aH θ (C)=715.0   kJmol-1 Δ 𝑎H 𝜃 (C)=715.0   kJmol-1

where,  Δ aH θ  Δ 𝑎H 𝜃  is enthalpy of atomization

 Δ aH θ (Cl2)=242   kJmol-1 Δ 𝑎H 𝜃 (Cl2)=242   kJmol-1 

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

  1. CC4(l)→CCl4(g)ΔvapHθ=30.5kJmol−1𝐶𝐶4(𝑙)→𝐶𝐶𝑙4(𝑔)Δ𝑣𝑎𝑝𝐻𝜃=30.5𝑘𝐽𝑚𝑜𝑙−1 
  2. C(s)→C(g)ΔaHθ=715.0 kJ mol−1𝐶(𝑠)→𝐶(𝑔)Δ𝑎𝐻𝜃=715.0 𝑘𝐽 𝑚𝑜𝑙−1 
  3. Cl2(g)→2Cl(g)ΔaHθ=242 kJ mol−1𝐶𝑙2(𝑔)→2𝐶𝑙(𝑔)Δ𝑎𝐻𝜃=242 𝑘𝐽 𝑚𝑜𝑙−1 
  4. C(g)+4Cl(g)→CCl4(g)ΔfH=−135.5 kJ mol−1𝐶(𝑔)+4𝐶𝑙(𝑔)→𝐶𝐶𝑙4(𝑔)Δ𝑓𝐻=−135.5 𝑘𝐽 𝑚𝑜𝑙−1 

Enthalpy change for the given process CCl4(g)→C(g)+4Cl(g)𝐶𝐶𝑙4(𝑔)→𝐶(𝑔)+4𝐶𝑙(𝑔) can be calculated using the following algebraic calculations as: 

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

ΔH=ΔaHθ(C)+2ΔaHθ(Cl2)−Δvap Hθ−ΔfHΔ𝐻=Δ𝑎𝐻𝜃(𝐶)+2Δ𝑎𝐻𝜃(𝐶𝑙2)−Δ𝑣𝑎𝑝 𝐻𝜃−Δ𝑓𝐻

(715.0 kJ mol−1)+2(242 kJ mol−1)−(30.5 kJ mol−1)−(−135.5 kJ mol−1)(715.0 𝑘𝐽 𝑚𝑜𝑙−1)+2(242 𝑘𝐽 𝑚𝑜𝑙−1)−(30.5 𝑘𝐽 𝑚𝑜𝑙−1)−(−135.5 𝑘𝐽 𝑚𝑜𝑙−1)

Therefore, ΔH=1304 kJ mol−1Δ𝐻=1304 𝑘𝐽 𝑚𝑜𝑙−1 

Bond enthalpy of C-Cl bond in CCl4(g)𝐶𝐶𝑙4(𝑔) 

=13044kJmol−1=13044𝑘𝐽𝑚𝑜𝑙−1

=326 kJ mol−1=326 𝑘𝐽 𝑚𝑜𝑙−1

16. For an isolated system,  Δ U=0 Δ U=0 , what will be  Δ S Δ S?

Ans: ΔSΔ𝑆will be positive i.e., greater than zero. 

Since for an isolated system, ΔU=0Δ𝑈=0, hence ΔSΔ𝑆will be positive and the reaction will be spontaneous. 

17. For the reaction at 298 K, 2A+B→C2A+B→C 

 Δ H= 400 kJ mol-1 Δ H= 400 kJ mol-1 and  Δ S= 0.2 kJ K-1 mol-1 Δ S= 0.2 kJ K-1 mol-1 At what temperature will the reaction become spontaneous considering  Δ H Δ H and  Δ S Δ S to be constant over the temperature range?

Ans: From the expression, 

ΔG= ΔH−TΔSΔ𝐺= Δ𝐻−𝑇Δ𝑆

Assuming the reaction at equilibrium, ΔTΔ𝑇for the reaction would be: 

T=(ΔH−ΔG)1ΔS𝑇=(Δ𝐻−Δ𝐺)1Δ𝑆

=ΔHΔS=Δ𝐻Δ𝑆

(ΔGΔ𝐺 = 0 at equilibrium) 

=400 kJ mol−10.2 kJ K−1 mol−1=400 𝑘𝐽 𝑚𝑜𝑙−10.2 𝑘𝐽 𝐾−1 𝑚𝑜𝑙−1

T = 2000 K 

For the reaction to be spontaneous, ΔGΔ𝐺must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K. 

18. For the reaction, 2Cl(g)→Cl2(g)2Cl(g)→Cl2(g) What are the signs of ∆H and ∆S ? 

Ans: ΔHΔ𝐻 and ΔSΔ𝑆 are negative. 

The given reaction represents the formation of chlorine molecules from chlorine atoms. Here, bond formation is occurring. Therefore, energy is being released. Hence, ΔHΔ𝐻is negative. 

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔSΔ𝑆is negative for the given reaction. 

19. For the reaction 

2A(g) + B(g) → 2D(g)2A(g) + B(g) → 2D(g) 

 Δ U θ  Δ U 𝜃   = -10.5 kJ and  Δ S θ  Δ S 𝜃  = -44.1JK-1JK-1 . 

Calculate  Δ G θ  Δ G 𝜃  for the reaction, and predict whether there action may occur spontaneously. 

Ans: For the given reaction,

2A(g) + B(g) → 2D(g)2𝐴(𝑔) + 𝐵(𝑔) → 2𝐷(𝑔)

Δng=2−(3)Δ𝑛𝑔=2−(3) = -1 mole

Substituting the value of ΔUθΔ𝑈𝜃 in the expression ofΔHΔ𝐻:

ΔHθ=ΔUθ+ΔngRTΔ𝐻𝜃=Δ𝑈𝜃+Δ𝑛𝑔𝑅𝑇

=(−10.5 kJ)−(−1)(8.314×10−3 kJ K−1 mol−1)(298 K)=(−10.5 𝑘𝐽)−(−1)(8.314×10−3 𝑘𝐽 𝐾−1 𝑚𝑜𝑙−1)(298 𝐾)

−10.5kJ−2.48kJ−10.5𝑘𝐽−2.48𝑘𝐽

ΔHθ=−12.98kJΔ𝐻𝜃=−12.98𝑘𝐽

Substituting the values of ΔHθΔ𝐻𝜃 and ΔSθΔ𝑆𝜃 in the expression of ΔGθΔ𝐺𝜃: 

ΔGθ = ΔHθ −TΔSθΔ𝐺𝜃 = Δ𝐻𝜃 −𝑇Δ𝑆𝜃

=−12.98kJ−(298K)(−44.1)J K−1=−12.98𝑘𝐽−(298𝐾)(−44.1)𝐽 𝐾−1

=−12.98kJ+13.14kJ=−12.98𝑘𝐽+13.14𝑘𝐽

ΔGθΔ𝐺𝜃= + 0.16 kJ 

Since ΔGθΔ𝐺𝜃 for the reaction is positive, the reaction will not occur spontaneously.

20. The equilibrium constant for a reaction is 10. What will be the value of  Δ G θ  Δ G 𝜃 ? R = 8.314 text  JK-1 mol-1𝑡𝑒𝑥𝑡  JK-1 mol-1 , T = 300 K. 

Ans: From the expression, 

ΔGθ = −2.303 RTlogKeqΔ𝐺𝜃 = −2.303 𝑅𝑇𝑙𝑜𝑔𝐾𝑒𝑞

ΔGθΔ𝐺𝜃 for the reaction, 

=(2.303)(8.314 JK−1 mol−1)(300K)log10=(2.303)(8.314 𝐽𝐾−1 𝑚𝑜𝑙−1)(300𝐾)𝑙𝑜𝑔10

=−5744.14 Jmol−1=−5744.14 𝐽𝑚𝑜𝑙−1

=−5.744k Jmol−1=−5.744𝑘 𝐽𝑚𝑜𝑙−1

21. Comment on the thermodynamic stability of NO(g), given

12NO(g)+12O2(g)→NO2(g): Δ rH θ =90kJmol-112NO(g)+12O2(g)→NO2(g): Δ rH 𝜃 =90kJmol-1

NO(g)+12O2(g)→O2(g): Δ rH θ =-74   kJ   mol-1NO(g)+12O2(g)→O2(g): Δ rH 𝜃 =-74   kJ   mol-1

Ans: The positive value of ΔrHΔ𝑟𝐻 indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (N2𝑁2  andO2𝑂2). Hence, NO(g) is unstable. The negative value of ΔrHΔ𝑟𝐻indicates that heat is evolved during the formation of NO2(g)𝑁𝑂2(𝑔) from NO(g) and O2𝑂2(g). The product, NO2(g)𝑁𝑂2(𝑔)is stabilized with minimum energy. 

Hence, unstable NO(g) changes to unstable NO2(g)𝑁𝑂2(𝑔). 

22. Calculate the entropy change in surroundings when 1.00 mol of H2O(l)H2O(l) is formed under standard conditions.  Δ fH θ  Δ fH 𝜃  = -286 kJ mol-1kJ mol-1 . 

Ans: It is given that 286 kJ mol−1𝑘𝐽 𝑚𝑜𝑙−1 of heat is evolved on the formation of 1 mol of H2O(l)𝐻2𝑂(𝑙). Thus, an equal amount of heat will be absorbed by the surroundings.

qsurr=+286kJkJ mol−1𝑞𝑠𝑢𝑟𝑟=+286𝑘𝐽𝑘𝐽 𝑚𝑜𝑙−1

Entropy change (ΔSsurr)(Δ𝑆𝑠𝑢𝑟𝑟) for the surroundings

=qsurr7=𝑞𝑠𝑢𝑟𝑟7

=286kJmol−1298K=286𝑘𝐽𝑚𝑜𝑙−1298𝐾

Therefore, (ΔSsurr)(Δ𝑆𝑠𝑢𝑟𝑟)= 959.73Jmol−1K−1959.73𝐽𝑚𝑜𝑙−1𝐾−1.