Question

calculate \\( \delta h^{\circ} \\) for the reaction shown below given the following data:

\\( 2 \mathrm{na}(\mathrm{s})+2 \mathrm{h}_{2} \mathrm{o}(\mathrm{l}) \to 2 \mathrm{naoh}(\mathrm{aq})+\mathrm{h}_{2}(\mathrm{~g}) \\)

substance	\\( \delta h_{\mathrm{f}}^{\circ} \mathrm{kj} / \mathrm{mol} \\)
\\( \mathrm{h}_{2} \mathrm{o}(\mathrm{l}) \\)	\\( -285.5 \\)

\\( \bigcirc-312.4 \mathrm{~kj} \\)
\\( \bigcirc-282.6 \mathrm{~kj} \\)
\\( \bigcirc+211.1 \mathrm{~kj} \\)
\\( \bigcirc-853.6 \mathrm{~kj} \\)
\\( \bigcirc-571.2 \mathrm{~kj} \\)

Explanation:

Step1: Recall the formula for $\Delta H^{\circ}$

$$\Delta H^{\circ}=\sum n\Delta H_{f}^{\circ}(\text{products})-\sum m\Delta H_{f}^{\circ}(\text{reactants})$$
where \(n\) and \(m\) are the stoichiometric coefficients.

Step2: Find $\Delta H_{f}^{\circ}$ for reactants and products

• For products: \(2\) moles of \(NaOH(aq)\) and \(1\) mole of \(H_{2}(g)\). The \(\Delta H_{f}^{\circ}\) of \(H_{2}(g)\) (element in standard state) is \(0\) kJ/mol. So \(\sum n\Delta H_{f}^{\circ}(\text{products})=(2\times(- 426.8))+(1\times0)=-853.6\) kJ/mol.
• For reactants: \(2\) moles of \(Na(s)\) (element in standard state, \(\Delta H_{f}^{\circ}=0\) kJ/mol) and \(2\) moles of \(H_{2}O(l)\). So \(\sum m\Delta H_{f}^{\circ}(\text{reactants})=(2\times0)+(2\times(-285.5))=-571\) kJ/mol.

Step3: Calculate $\Delta H^{\circ}$

$$\Delta H^{\circ}=-853.6-(-571)=- 282.6\ \text{kJ}$$

Answer:

-282.6 kJ