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Thermodynamics question

2021 · 27 Aug · Shift 2 · Q22
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Thermodynamics question

2021 · 27 Aug · Shift 2 · Q22

JEE MainChemistryThermodynamicsNumerical+4 / −1
Data given for the following reaction is as follows :

FeO(s) + C(graphite) →\to→ Fe(s) + CO(g)

Substance ΔH∘\Delta H^\circΔH∘
(kJ mol −1^{ - 1}−1)
ΔS∘\Delta S^\circΔS∘
(J mol −1^{ - 1}−1 K −1^{ - 1}−1)
FeO(s)Fe{O_{(s)}}FeO(s)​ −266.3- 266.3−266.3 57.49
C(graphite){C_{(graphite)}}C(graphite)​ 0 5.74
Fe(s)F{e_{(s)}}Fe(s)​ 0 27.28
CO(g)C{O_{(g)}}CO(g)​ −110.5- 110.5−110.5 197.6


The minimum temperature in K at which the reaction becomes spontaneous is ‾\underline{\hspace{2cm}}​. (Integer answer)
Numerical answer
View written solutionFree

Correct answer: 965

  1. Given reaction

FeO(s)+C(graphite)→Fe(s)+CO(g)\text{FeO}(s) + C(\text{graphite}) \to \text{Fe}(s) + \text{CO}(g)FeO(s)+C(graphite)→Fe(s)+CO(g)

For spontaneity,

ΔG∘=ΔH∘−TΔS∘<0\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ < 0ΔG∘=ΔH∘−TΔS∘<0

The minimum temperature for spontaneity is when

ΔG∘=0⇒T=ΔH∘ΔS∘\Delta G^\circ = 0 \Rightarrow T = \frac{\Delta H^\circ}{\Delta S^\circ}ΔG∘=0⇒T=ΔS∘ΔH∘​

So first compute ΔH∘\Delta H^\circΔH∘ and ΔS∘\Delta S^\circΔS∘ for the reaction.


  1. Calculate standard enthalpy change

Using

ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants})ΔHrxn∘​=∑ΔHf∘​(products)−∑ΔHf∘​(reactants)

Products:

ΔHf∘(Fe)=0\Delta H^\circ_f(\text{Fe}) = 0ΔHf∘​(Fe)=0 ΔHf∘(CO)=−110.5 kJ mol−1\Delta H^\circ_f(\text{CO}) = -110.5\ \text{kJ mol}^{-1}ΔHf∘​(CO)=−110.5 kJ mol−1

Reactants:

ΔHf∘(FeO)=−266.3 kJ mol−1\Delta H^\circ_f(\text{FeO}) = -266.3\ \text{kJ mol}^{-1}ΔHf∘​(FeO)=−266.3 kJ mol−1 ΔHf∘(C)=0\Delta H^\circ_f(C) = 0ΔHf∘​(C)=0

Therefore,

ΔHrxn∘=[0+(−110.5)]−[(−266.3)+0]\Delta H^\circ_{\text{rxn}} = [0 + (-110.5)] - [(-266.3) + 0]ΔHrxn∘​=[0+(−110.5)]−[(−266.3)+0]

ΔHrxn∘=−110.5+266.3=155.8 kJ mol−1\Delta H^\circ_{\text{rxn}} = -110.5 + 266.3 = 155.8\ \text{kJ mol}^{-1}ΔHrxn∘​=−110.5+266.3=155.8 kJ mol−1


  1. Calculate standard entropy change

Using

ΔSrxn∘=∑S∘(products)−∑S∘(reactants)\Delta S^\circ_{\text{rxn}} = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})ΔSrxn∘​=∑S∘(products)−∑S∘(reactants)

Products:

S∘(Fe)=27.28 J mol−1K−1S^\circ(\text{Fe}) = 27.28\ \text{J mol}^{-1}\text{K}^{-1}S∘(Fe)=27.28 J mol−1K−1 S∘(CO)=197.6 J mol−1K−1S^\circ(\text{CO}) = 197.6\ \text{J mol}^{-1}\text{K}^{-1}S∘(CO)=197.6 J mol−1K−1

Reactants:

S∘(FeO)=57.49 J mol−1K−1S^\circ(\text{FeO}) = 57.49\ \text{J mol}^{-1}\text{K}^{-1}S∘(FeO)=57.49 J mol−1K−1 S∘(C)=5.74 J mol−1K−1S^\circ(C) = 5.74\ \text{J mol}^{-1}\text{K}^{-1}S∘(C)=5.74 J mol−1K−1

Thus,

ΔSrxn∘=(27.28+197.6)−(57.49+5.74)\Delta S^\circ_{\text{rxn}} = (27.28 + 197.6) - (57.49 + 5.74)ΔSrxn∘​=(27.28+197.6)−(57.49+5.74)

ΔSrxn∘=224.88−63.23=161.65 J mol−1K−1\Delta S^\circ_{\text{rxn}} = 224.88 - 63.23 = 161.65\ \text{J mol}^{-1}\text{K}^{-1}ΔSrxn∘​=224.88−63.23=161.65 J mol−1K−1


  1. Find threshold temperature

Convert enthalpy into J mol−1^{-1}−1:

ΔH∘=155.8×103=155800 J mol−1\Delta H^\circ = 155.8\times 10^3 = 155800\ \text{J mol}^{-1}ΔH∘=155.8×103=155800 J mol−1

Now,

T=ΔH∘ΔS∘=155800161.65T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{155800}{161.65}T=ΔS∘ΔH∘​=161.65155800​

T≈964.74 KT \approx 964.74\ \text{K}T≈964.74 K

For the reaction to become spontaneous,

T>964.74 KT > 964.74\ \text{K}T>964.74 K

Hence the minimum temperature is approximately

965 K\boxed{965\ \text{K}}965 K​

If integer truncation is expected, one may write 964964964, but physically the smallest whole-number temperature at which spontaneity starts is 965 965\,965K.


  1. Comparison with stored answer

Stored correct answer = 964964964

My calculated threshold is 964.74 964.74\,964.74K, so the least integer temperature making ΔG∘<0\Delta G^\circ < 0ΔG∘<0 is 965 965\,965K. Therefore I do not agree with the stored answer.

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