# Consider the reaction 2 Al2O3(s) -> 2 Al(s) + 3 O2(g). The heat of formation of aluminum oxide is -1669.8 kJ/mol. Calculate the heat of reaction?

## Consider the reaction 2 Al2O3(s) -> 2 Al(s) + 3 O2(g). The heat of formation of aluminum oxide is -1669.8 kJ/mol. Calculate the heat of reaction.

May 11, 2016

#### Answer:

$\Delta {H}_{r x n}^{o} = \Sigma n {\left(\Delta {H}_{f}^{o}\right)}_{P r o \mathrm{du} c t s} - \Sigma n {\left(\Delta {H}_{f}^{o}\right)}_{R e a c \tan t s}$

#### Explanation:

$\Delta {H}_{r x n}^{o} = \Sigma n {\left(\Delta {H}_{f}^{o}\right)}_{P r o \mathrm{du} c t s} - \Sigma n {\left(\Delta {H}_{f}^{o}\right)}_{R e a c \tan t s}$

$\Delta {H}_{r x n}^{o} = \left(3 \Delta {H}_{f}^{o} {O}_{2} \left(g\right) + 2 \Delta {H}_{f}^{o} A l \left(s\right)\right) - 2 \Delta {H}_{f}^{o} A {l}_{2} {O}_{3} \left(s\right)$

We know that the standard enthalpy of formation $\left(\Delta {H}_{f}^{o}\right)$of the element in its most stable form is equal to zero.

$\Delta {H}_{r x n}^{o} = \left(\cancel{3 \Delta {H}_{f}^{o}} {O}_{2} \left(g\right) + \cancel{2 \Delta {H}_{f}^{o}} A l \left(s\right)\right) - 2 \Delta {H}_{f}^{o} A {l}_{2} {O}_{3} \left(s\right)$

$\Delta {H}_{r x n}^{o} = \left(0 + 0\right) - 2 \Delta {H}_{f}^{o} A {l}_{2} {O}_{3} \left(s\right)$

$\Delta {H}_{r x n}^{o} = - 2 m o l . \left(- 1669.8 \frac{k J}{m o l}\right)$

$\Delta {H}_{r x n}^{o} = + 3339.6 \text{ } k J$