Question

Calculate the enthalpy change for the reaction, $Fe_{2}O_{3}+3CO \rightarrow 2Fe+3CO_{2};$ from the following data:
$2Fe+\frac{3}{2}O_{2} \rightarrow Fe_{2}O_{3};ΔH=-177.1\,kcal$
$C+\frac{1}{2}O_{2} \rightarrow CO;ΔH=-32.8\,kcal$
$C+O_{2} \rightarrow CO_{2};ΔH=-94.3\,kcal$
Express the magnitude of your answer by rounding off to one significant figure.

Answer 1

Let the enthalpy of the given reaction be $\triangle H_{r}$
$( Fe )_{2} O _{3}(s)+3 CO (g) \rightarrow 2 Fe (s)+3(C O)_{2}(g) \ldots(i)$
Given equation:
$2 Fe ( s )+\frac{3}{2} O _{2}( g ) \rightarrow( Fe )_{2} O _{3}( s ) ; \Delta H =-177.1 \,kcal . \ldots$(ii)
$C(s)+\frac{1}{2} O_{2}(g) \rightarrow C O(g) ; \Delta H=-32.8 \,kcal . \ldots(i i i)$ $C(s)+O_{2}(g) \rightarrow(C O)_{2}(g) ; \Delta H=-94.3 \,kcal \ldots(i v)$
Enthalpy is an extensive property.
Equation (i) can be obtained from equation (ii), (iii) and (iv) as given below:
Equation $(iv) \times 3-$ Equation $(iii) \times 3-$ Equation $(ii)$
$\equiv$ Equation (i)
Thus, according to Hess' law,
$[(-98.3) \times 3]-[(-32.8) \times 3]-[-177.1]=\Delta H_{r}$
$\Delta H_{r}=-282.9+98.4+177.1=-282.9+275.5$
$\Delta H_{r}=-7.4 \,kcal$
Thus, the enthalpy of equation (i) is $-7.4\,kcal$ .