Question

The space shuttle orbital utilizes the oxidation of methylhydrazine by dinitrogen tetroxide for propulsion:

$\mathbf{4}{\mathbf{N}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{5}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{12}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{9}{\mathbf{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Calculate$\mathbf{∆}{\mathbf{H}}^{\mathbf{o}}$for this reaction

Short answer

The enthalpy of the reaction is$-4594\mathrm{kJ}/\mathrm{mol}$.

Step-by-step solution

Step 1: Calculation of enthalpy of reaction

$\begin{array}{rcl}4{\mathrm{N}}_2{\mathrm{H}}_3{\mathrm{CH}}_3\left(\mathrm{l}\right)+5{\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{l}\right)& \to & 12{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+9{\mathrm{N}}_2\left(\mathrm{g}\right)+4{\mathrm{CO}}_2\left(\mathrm{g}\right)\\ \mathbf{∆}{\mathbf{H}}_{\mathbf{r}}^{\mathbf{o}}& \mathbf{=}& \mathbf{\sum }\mathbf{∆}{\mathbf{H}}_{\mathbf{P}}^{\mathbf{o}}\mathbf{-}\mathbf{\sum }\mathbf{∆}{\mathbf{H}}_{\mathbf{R}}^{\mathbf{o}}\\ \mathbf{∆}{\mathbf{H}}_{\mathbf{r}}^{\mathbf{o}}& \mathbf{=}& \mathbf{[}\mathbf{12}\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{}\mathbf{for}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{+}\mathbf{9}\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{}\mathbf{for}\mathbf{}{\mathbf{N}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{4}\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{}\mathbf{for}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{]}\mathbf{-}\\ & & \mathbf{[}\mathbf{4}\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{}\mathbf{for}\mathbf{}{\mathbf{N}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{3}}\left(l\right)\mathbf{+}\mathbf{5}\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{o}}\mathbf{}\mathbf{for}\mathbf{}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\left(l\right)\mathbf{]}\mathbf{}\end{array}$

The substitute the values of enthalpies of reactant and product:

$\begin{array}{rcl}& \mathbf{=}& \left[\left(12\times -242\mathrm{kJ}/\mathrm{mol}\right)+\left(9\times 0\mathrm{kJ}/\mathrm{mol}\right)+\left(4\times 393.5\mathrm{kJ}/\mathrm{mol}\right)\right]\mathbf{-}\\ & & \left[\left(4\times 54\mathrm{kJ}/\mathrm{mol}\right)+\left(5\times -20\mathrm{kJ}/\mathrm{mol}\right)\right]\\ & \mathbf{=}& \left(-2904\mathrm{kJ}/\mathrm{mol}+0\mathrm{kJ}/\mathrm{mol}+\left(-1574\mathrm{kJ}/\mathrm{mol}\right)\right)\mathbf{-}\left(216\mathrm{kJ}/\mathrm{mol}+\left(-100\mathrm{kJ}/\mathrm{mol}\right)\right)\\ & \mathbf{=}& \mathbf{-}\mathbf{4478}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{-}\mathbf{116}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}\\ & \mathbf{=}& \mathbf{-}\mathbf{4594}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}\end{array}$

Result:

Hence the enthalpy of the reaction is $\mathbf{-}\mathbf{4594}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$.