Question

Determine the heat of the following reaction:
$FeO(s)+F{{e}_{2}}{{O}_{3}}(s)\to F{{e}_{3}}{{O}_{4}}(s)$
Given information:

& 2Fe(s)+{{O}_{2}}(g)\to 2FeO(s);\Delta {{H}^{o}}=-544kJ \\\ & 4Fe(s)+3{{O}_{2}}(g)\to 2F{{e}_{2}}{{O}_{3}}(s);\Delta {{H}^{o}}=-1648.4kJ \\\ & F{{e}_{3}}{{O}_{4}}(s)\to 3Fe(s)+2{{O}_{2}}(g);\Delta {{H}^{o}}=+1118.4kJ \\\ \end{aligned}$$ a.) -1074kJ b.) -22.2kJ c.) -249.8kJ d.) -2214.6kJ

Answer 1

Hint: We know that enthalpy of a reaction is the enthalpy of products minus enthalpy of reactant. In this question many intermediate reactions and their heat are given, from these reactions we will solve our problem.

Complete step by step solution:
We know that heat of reaction, the amount of heat that must be added or removed during a chemical reaction in order to keep all of the substances present at the same temperature.
Reversing all the given reaction and multiplying last reaction by 2:
$2FeO(s)\to 2Fe(s)+{{O}_{2}}(g);\Delta {{H}^{o}}=+544kJ$
$2F{{e}_{2}}{{O}_{3}}(s)\to 4Fe(s)+3{{O}_{2}}(g);\Delta {{H}^{o}}=+1648.4kJ$
$2\times (3Fe(s)+2{{O}_{2}}(g)\to F{{e}_{3}}{{O}_{4}}(s);\Delta {{H}^{o}}=-1118.4kJ)$
Now we will add all three reactions:

& \underline{\begin{aligned} & 2FeO(s)\to 2Fe(s)+{{O}_{2}}(g);\Delta {{H}_{1}}^{o}=+544kJ \\\ & 2F{{e}_{2}}{{O}_{3}}(s)\to 4Fe(s)+3{{O}_{2}}(g);\Delta {{H}_{2}}^{o}=+1648.4kJ \\\ & 2\times (3Fe(s)+2{{O}_{2}}(g)\to F{{e}_{3}}{{O}_{4}}(s);\Delta {{H}_{3}}^{o}=-1118.4kJ) \\\ \end{aligned}} \\\ & 2FeO(s)+2F{{e}_{2}}{{O}_{3}}(s)\to 2F{{e}_{3}}{{O}_{4}}(s);\Delta {{H}_{4}}^{o}=? \\\ \end{aligned}$$ When we add all three equations then we will also add enthalpies of all three equations: $$\Delta {{H}_{4}}^{o}=\Delta {{H}_{1}}^{o}+\Delta {{H}_{2}}^{o}+2\times \Delta {{H}_{3}}^{o}$$ $$\Delta {{H}_{4}}^{o}$$= 544 + 1648 + 2 x (-1118.4) = -44.8kJ But for solving this problem we have to divide the final reaction by two and also we will divide final enthalpy by 2. So, the reaction is: $$FeO(s)+F{{e}_{2}}{{O}_{3}}(s)\to F{{e}_{3}}{{O}_{4}}(s)$$ And Final heat of reaction = $$\dfrac{\Delta {{H}_{4}}^{o}}{2}$$ = $$\dfrac{-44.8kJ}{2}$$ = -22.4kJ So, from the above explanation and calculation we can say that the correct option is “B”. Note: Whenever we add any number of reactions then their enthalpies will also be added and if we subtract then enthalpies also get subtracted. And when we multiply any reaction by anything then enthalpies also get multiplied.