Question

How would you calculate the standard enthalpy change for the following reaction at ${25 ^\circ }C$ : ${H_2}O(g) + C(graphite)(s) \to {H_2}(g) + CO(g)$ ?

Answer 1

Hint: To solve this question, first we will find the standard enthalpies of each of the given formations of the reaction separately. And then apply the formula to find the standard enthalpy change for the given reaction.

Complete step by step answer:
We can calculate the standard enthalpy change of reaction by using the standard enthalpies of formation of the species that take part in the reaction, i.e. the reactants and the products.
We can find the standard enthalpies of formation, $\Delta {H ^\circ }_f$ :
In this case, we have-
For ${H_2}O(g)$ : $ - 241.82kJ/mol$
For $C(s)$ : $0kJ/mol$
For ${H_2}$ : $0kJ/mol$
For $CO$ : $ - 110.53kJ/mol$
To find the standard enthalpy change of reaction, $\Delta {H ^\circ }_{rxn}$ , using the standard enthalpies of formation, we need to take into account the fact that, for a chemical reaction, the change in enthalpy is independent of the pathway taken.
Since, the enthalpy changes for those respective reactions are the standard enthalpies of formation, it follows that we can say:
$\Delta {H ^\circ }_{rxn} = \sum (n \times \Delta {H ^\circ }_{f\,products}) - \sum (m \times \Delta {H ^\circ }_{f\,reac\tan ts})$
here, $n$ and $m$ represents the stoichiometric coefficients of the products and of the reactants respectively.
As we can see, standard enthalpies of formation are given per mole, so we need to take into account how many moles of each compound we have.
Therefore, we have:
$
  \Delta {H ^\circ }_{rxn} = [1mole.( - 241.82\dfrac{{kJ}}{{mole}}) + 1mole.(0\dfrac{{kJ}}{{mole}})] - [1mole.0\dfrac{{kJ}}{{mole}} + 1mole.( - 110.53\dfrac{{kJ}}{{mole}})] \\
   \Rightarrow \Delta {H ^\circ }_{rxn} = - 241.82kJ - ( - 110.53kJ) \\
 = - 131.3kJ \\
$
Hence, the standard enthalpy change for the given reaction at ${25 ^\circ }C$ is $ - 131.3kJ$ .

Note: All elements are written in their standard states, and one mole of product is formed. This is true for all enthalpies of formation. The standard enthalpy of formation is measured in units of energy per amount of substance, usually stated in kilojoule per mole , but also in kilocalorie per mole , joule per mole or kilocalorie per gram .