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Chemistry Question on Thermodynamics

The following two reactions are known Fe2O3(s)+3CO(g)−>2Fe(s)+3CO2(g);ΔH=−26.8 kJ{Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g); \Delta H = - 26.8 \, kJ } FeO(s)+CO(g)−>Fe(s)+CO2(g);ΔH−16.5 kJ{FeO(s) + CO (g) -> Fe(s) + CO2(g); \Delta H - 16.5 \, kJ} The value of ΔH\Delta H for the following reaction Fe2O3(s)+CO(g)−>2FeO(s)+CO2(g){Fe2O3(s) + CO(g) -> 2FeO(s) + CO2(g) } is

A

- 43.3 kJ

B

-10.3 kJ

C

+6.2 kJ

D

+10.3 kJ

Answer

+6.2 kJ

Explanation

Solution

Given,
(I)Fe2O(s)+3CO(g)→2Fe(s)+3CO2(g);F{{e}_{2}}O(s)+3CO(g)\xrightarrow[{}]{{}}2Fe(s)+3C{{O}_{2}}(g);
ΔH=−26.8 kJ\Delta H=-26.8\,\text{kJ}
(II)FeO(s)+CO(g)→Fe(s)+CO2(g);FeO(s)+CO(g)\xrightarrow[{}]{{}}Fe(s)+C{{O}_{2}}(g);
ΔH=−16.5 kJ\Delta H=-16.5\,\text{kJ}
On multiplying Eq (II) with 2, we get
(III) 2FeO(s)+2CO(g)→2Fe(s)+2CO2(g);2FeO(s)+2CO(g)\xrightarrow[{}]{{}}2Fe(s)+2C{{O}_{2}}(g);
ΔH=−33 kJ\Delta H=-33\,\text{kJ}
On subtracting Eq (III) from I, we get
Fe2O3(s)+CO(g)→2FeO(s)+CO2(g);F{{e}_{2}}{{O}_{3}}(s)+CO(g)\xrightarrow[{}]{{}}2FeO(s)+C{{O}_{2}}(g);
ΔH=−26.8−(−33)\Delta H=-26.8-(-33)
=+6.2 kJ=+6.2\,\text{kJ}