Question

The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate as fuel. A possible reaction is

$\mathbf{3}\mathbf{Al}\left(s\right)\mathbf{+}\mathbf{3}{\mathbf{NH}}_{\mathbf{4}}{\mathbf{ClO}}_{\mathbf{4}}\left(s\right)\mathbf{\to }{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{Al}}_{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{3}\mathbf{NO}\left(g\right)\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)$

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

Short answer

The enthalpy of the reaction is -2677 kJ/mol

Step-by-step solution

Step 1: Calculation of enthalpy of reaction

$3\mathrm{Al}\left(\mathrm{s}\right)+3{\mathrm{NH}}_4{\mathrm{ClO}}_4\left(\mathrm{s}\right)\to {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+{\mathrm{AlCl}}_3\left(\mathrm{s}\right)+3\mathrm{NO}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

localid="1658137152125" $\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}& =& \sum ∆{\mathrm{H}}_{\mathrm{P}}^{\mathrm{o}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}^{\mathrm{o}}\\ ∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}& =& \sum ∆{\mathrm{H}}_{{\mathrm{fAl}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)}^{\mathrm{o}}+\sum ∆{\mathrm{H}}_{{\mathrm{fAlCl}}_3\left(\mathrm{s}\right)}^{\mathrm{o}}+3\sum ∆{\mathrm{H}}_{\mathrm{fNO}\left(\mathrm{g}\right)}^{\mathrm{o}}+6\sum ∆{\mathrm{H}}_{{\mathrm{fH}}_2\mathrm{O}\left(\mathrm{g}\right)}^{\mathrm{o}}-3\sum ∆{\mathrm{H}}_{\mathrm{fAl}\left(\mathrm{s}\right)}^{\mathrm{o}}-3\sum ∆{\mathrm{H}}_{{\mathrm{fNHClO}}_4\left(\mathrm{s}\right)}^{\mathrm{o}}\\ & =& \left[\left(-1676\right)+\left(-704\right)+3\times \left(90\right)+6\times \left(-242\right)\right]-\left[\left(3\times 0\right)+3\times \left(-295\right)\right]\\ & =& \left(-3562\right)+885\\ & =& -2677\mathrm{kJ}/\mathrm{mol}\end{array}$

Result:

Hence the enthalpy of the reaction is $\mathbf{-}\mathbf{2677}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$.