Given
(i) 2Fe2O3(s) → 4Fe(s) + 3O2(g); \(\Delta_r G^\circ = +1487.0\ \mathrm{kJ\ mol^{-1}}\)
(ii) 2CO(g) + O2(g) → 2CO2(g); \(\Delta_r G^\circ = -514.4\ \mathrm{kJ\ mol^{-1}}\)
Free energy change, \(\Delta_r G^\circ\) for the reaction
2Fe2O3(s) + 6CO(g) → 4Fe(s) + 6CO2(g) will be :
(i) 2Fe2O3(s) → 4Fe(s) + 3O2(g); \(\Delta_r G^\circ = +1487.0\ \mathrm{kJ\ mol^{-1}}\)
(ii) 2CO(g) + O2(g) → 2CO2(g); \(\Delta_r G^\circ = -514.4\ \mathrm{kJ\ mol^{-1}}\)
Free energy change, \(\Delta_r G^\circ\) for the reaction
2Fe2O3(s) + 6CO(g) → 4Fe(s) + 6CO2(g) will be :
Show Hint
- \(-112.4\ \mathrm{kJ\ mol^{-1}}\)
- \(-56.2\ \mathrm{kJ\ mol^{-1}}\)
- \(-168.2\ \mathrm{kJ\ mol^{-1}}\)
- \(-208.0\ \mathrm{kJ\ mol^{-1}}\)
The Correct Option is B
Solution and Explanation
Take (i) as written: 2Fe2O3 → 4Fe + 3O2, \(\Delta G_1 = +1487.0\).
Take 3 × (ii): 6CO + 3O2 → 6CO2, \(\Delta G_2 = 3(-514.4) = -1543.2\).
Adding the two, 3O2 cancels and we get the target reaction.
\(\Delta_r G^\circ = 1487.0 - 1543.2 = -56.2\ \mathrm{kJ\ mol^{-1}}\).
Answer: (2) \(-56.2\ \mathrm{kJ\ mol^{-1}}\)
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