Question

Calculate ΔHrxn for the following reaction: Fe₂O₃(s)+3CO(g)→2Fe(s)+3CO₂(g) Use the following reactions and given ΔH′s. 2Fe(s)+3/2O₂(g)→Fe₂O₃(s), ΔH = -824.2 kJ CO(g)+1/2O₂(g)→CO₂(g), ΔH = -282.7 kJ

Answer 1

Answer : The enthalpy change of reaction is -23.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given final reaction is,

$Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $2Fe(s)+\frac{3}{2}O_2(g)\rightarrow Fe_2O_3(s)$    $\Delta H^o_1=-824.2kJ$

(2) $CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$    $\Delta H^o_2=-282.7kJ$

First we will reverse the reaction 1 and multiply equation 2 by 3 then adding both the equation, we get :

(1) $Fe_2O_3(s)\rightarrow 2Fe(s)+\frac{3}{2}O_2(g)$    $\Delta H^o_1=+824.2kJ$

(2) $3CO(g)+\frac{3}{2}O_2(g)\rightarrow 3CO_2(g)$    $\Delta H^o_2=3\times (-282.7kJ)=-848.1kJ$

The expression for final enthalpy is,

$\Delta H=\Delta H^o_1+\Delta H^o_2$

$\Delta H=(+824.2kJ)+(-848.1kJ)$

$\Delta H=-23.9kJ$

Therefore, the enthalpy change of reaction is -23.9 kJ