Question

The standard enthalpy of combustion of ethene gas $\left[C_2{H}_4\left(g\right)\right]$ is$\mathbf{-}\mathbf{1411}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$ at 298 K. Given the following enthalpies of formation, calculate$\mathbf{∆}{\mathbf{H}}_{\mathbf{r}}^{\mathbf{o}}$ for ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\left(g\right)$

$$\begin{gathered} {\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{393}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol} \\ {\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{285}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol} \end{gathered}$$

Short answer

The standard enthalpy of formation$∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}$ for${\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)$is$52.5\mathrm{kJ}$.

Step-by-step solution

The balanced chemical equation the above process:

The standard enthalpy of combustion of ethene gas $\left[{\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)\right]$is$-1411.1\mathrm{kJ}/\mathrm{mol}$

Reaction: ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\left(g\right)\mathbf{+}\mathbf{3}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}{\mathbf{H}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{1411}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$

Now from the given data, we know that:

The enthalpy of formation of${\mathrm{CO}}_2\left(\mathrm{g}\right)=-393.5\mathrm{kJ}/\mathrm{mol}$ and${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)=-285.8\mathrm{kJ}/\mathrm{mol}$

Since, $\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{rxn}\mathbf{=}\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}\left(\mathrm{reactant}\right)\mathbf{-}\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}\left(\mathrm{product}\right)$

When we put these values in the formula we get:

$$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{rxn}}^{\mathrm{o}}=2\left(-393.5\right)\mathrm{kJ}+2\left(-285.8\right)\mathrm{kJ}-∆{\mathrm{H}}_{\mathrm{f}}\left({\mathrm{C}}_2{\mathrm{H}}_4\right) \\ -1411.1\mathrm{kJ}=-1358.6\mathrm{kJ}-∆{\mathrm{H}}_{\mathrm{f}}\left({\mathrm{C}}_2{\mathrm{H}}_4\right) \\ ∆{\mathrm{H}}_{\mathrm{f}}\left({\mathrm{C}}_2{\mathrm{H}}_4\right)=52.5\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Hence, enthalpies of formation,$\left(∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\right)$ for${\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)$is$52.5\mathrm{kJ}/\mathrm{mol}$.