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Thermodynamics question

2018 · 15 Apr · Shift 2 · Q20
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Thermodynamics question

2018 · 15 Apr · Shift 2 · Q20

JEE MainChemistryThermodynamicsMCQ+4 / −1
Given (i) 2Fe2O3Fe_2O_3Fe2​O3​(s) →\to→ 4FeFeFe(s) + 3O2O_2O2​(g); Δ\DeltaΔ rGo = + 1487.0 kJ mol-1 (ii) 2COCOCO(g) + O2O_2O2​(g) →\to→ 2CO2CO_2CO2​(g); Δ\DeltaΔ rGo = −-− 514.4 kJ mol-1 Free energy change, Δ\DeltaΔ rGo for the reaction 2Fe2O3Fe_2O_3Fe2​O3​(s) + 6COCOCO(g) →\to→ 4FeFeFe(s) + 6CO2CO_2CO2​(g) will be :
  1. A
    −-− 112.4 kJ mol-1
  2. B
    −-− 56.2 kJ mol-1
  3. C
    −-− 168.2 kJ mol-1
  4. D
    −-− 208.0 kJ mol-1
View written solutionFree

Correct answer: B

  1. We need the standard Gibbs free energy change for:

2Fe2O3(s)+6CO(g)→4Fe(s)+6CO2(g)2Fe_2O_3(s) + 6CO(g) \rightarrow 4Fe(s) + 6CO_2(g)2Fe2​O3​(s)+6CO(g)→4Fe(s)+6CO2​(g)

We are given:

(i)2Fe2O3(s)→4Fe(s)+3O2(g),ΔrG∘=+1487.0 kJ mol−1\text{(i)}\quad 2Fe_2O_3(s) \rightarrow 4Fe(s) + 3O_2(g), \qquad \Delta_r G^\circ = +1487.0\ \text{kJ mol}^{-1}(i)2Fe2​O3​(s)→4Fe(s)+3O2​(g),Δr​G∘=+1487.0 kJ mol−1

(ii)2CO(g)+O2(g)→2CO2(g),ΔrG∘=−514.4 kJ mol−1\text{(ii)}\quad 2CO(g) + O_2(g) \rightarrow 2CO_2(g), \qquad \Delta_r G^\circ = -514.4\ \text{kJ mol}^{-1}(ii)2CO(g)+O2​(g)→2CO2​(g),Δr​G∘=−514.4 kJ mol−1

  1. To match the target reaction, multiply reaction (ii) by 333:

6CO(g)+3O2(g)→6CO2(g)6CO(g) + 3O_2(g) \rightarrow 6CO_2(g)6CO(g)+3O2​(g)→6CO2​(g)

So,

ΔrG∘=3(−514.4)=−1543.2 kJ mol−1\Delta_r G^\circ = 3(-514.4) = -1543.2\ \text{kJ mol}^{-1}Δr​G∘=3(−514.4)=−1543.2 kJ mol−1

  1. Now add this to reaction (i):

2Fe2O3(s)→4Fe(s)+3O2(g)(ΔrG∘=+1487.0)2Fe_2O_3(s) \rightarrow 4Fe(s) + 3O_2(g) \qquad (\Delta_r G^\circ = +1487.0)2Fe2​O3​(s)→4Fe(s)+3O2​(g)(Δr​G∘=+1487.0)

6CO(g)+3O2(g)→6CO2(g)(ΔrG∘=−1543.2)6CO(g) + 3O_2(g) \rightarrow 6CO_2(g) \qquad (\Delta_r G^\circ = -1543.2)6CO(g)+3O2​(g)→6CO2​(g)(Δr​G∘=−1543.2)

Adding,

2Fe2O3(s)+6CO(g)→4Fe(s)+6CO2(g)2Fe_2O_3(s) + 6CO(g) \rightarrow 4Fe(s) + 6CO_2(g)2Fe2​O3​(s)+6CO(g)→4Fe(s)+6CO2​(g)

since 3O2(g)3O_2(g)3O2​(g) cancels on both sides.

  1. Therefore,

ΔrG∘=1487.0−1543.2=−56.2 kJ mol−1\Delta_r G^\circ = 1487.0 - 1543.2 = -56.2\ \text{kJ mol}^{-1}Δr​G∘=1487.0−1543.2=−56.2 kJ mol−1

  1. Option check:
  • A: −112.4-112.4−112.4  not correct
  • B: −56.2-56.2−56.2  correct
  • C: −168.2-168.2−168.2  not correct
  • D: −208.0-208.0−208.0  not correct

Hence, the correct answer is Option B.

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