# Question #b6523

Dec 19, 2014

The answer is ${32.4}^{\circ} C$.

$2 N {a}_{2} {O}_{2} \left(s\right) + 2 {H}_{2} {O}_{\left(l\right)} \to 4 N a O {H}_{\left(s\right)} + {O}_{2} \left(g\right)$

In order to determine the heat of reaction, $\Delta H$, we need the standard formation enthalpy values ($\Delta {H}_{f}^{\circ}$) for the reactants and the products and the number of moles that react

$N {a}_{2} {O}_{2}$: $\Delta {H}_{f}^{\circ} = - 504.6 \frac{k J}{m o l}$;
${H}_{2} O$: $\Delta {H}_{f}^{\circ} = - 285.8 \frac{k J}{m o l}$;
$N a O H$: $\Delta {H}_{f}^{\circ} = - 470.1 \frac{k J}{m o l}$;
${O}_{2}$: $\Delta {H}_{f}^{\circ} = 0 \frac{k J}{m o l}$.

We know that $N {a}_{2} {O}_{2}$'s molar mass is $78.0 \frac{g}{m o l}$, which means that the number of moles of $N a O H$ is

${n}_{N {a}_{2} {O}_{2}} = {m}_{N {a}_{2} {O}_{2}} / \left(m o l a r m a s s\right) = \frac{7.800 g}{78.0 \frac{g}{m o l}} = 0.100$ moles

We can determine the number of molesof water by using its density of approximately $1.00 \frac{g}{m L}$, its molar mass of $18.0 \frac{g}{m o l}$, and its given volume

${n}_{{H}_{2} O} = {m}_{{H}_{2} O} / \left(m o l a r m a s s\right) = \frac{\rho \cdot V}{m o l a r m a s s} \to$
${n}_{{H}_{2} O} = \frac{1.00 \frac{g}{m L} \cdot 110.00 m L}{18.0 \frac{g}{m o l}} = 6.11$ moles

We can see that $N {a}_{2} {O}_{2}$ is the limiting reagent, which means that the number of moles of water that will react will be ${n}_{{H}_{2} O} = {n}_{N {a}_{2} {O}_{2}} = 0.100$.

Therefore, $\Delta H$ can be calculated to be (keeping in mind the mole-to-mole ratios)

$\Delta H = 0.200 \cdot \left(- 470.1\right) - \left(0.100 \cdot \left(- 285.8\right) + 0.100 \cdot \left(- 504.6\right)\right) = - - 14.9 k J$

Since $\Delta H < 0$, we are dealing with an exothermic reaction; this means that the heat absorbed by the water will be $q = - \Delta H = 14.9 k J$. The change in temperature can then be calculated to be ( ${c}_{{H}_{2} O}$ - water's specific heat):

$q = {m}_{{H}_{2} O} \cdot {c}_{{H}_{2} O} \cdot \Delta T \to \Delta T = \frac{q}{m \cdot c}$

$\Delta T = \frac{14.9 \cdot {10}^{3} J}{110.00 g \cdot 4.18 \frac{J}{{g}^{\circ} C}} = {32.4}^{\circ} C$

The temperature of the water increased by ${32.4}^{\circ} C$.