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Formula from Chapter 5: Enthalpy from heats of formation
\(ΔH_{\mathrm{reaction}} = \sum (n_{\mathrm{prod}} \cdot ΔH_{\mathrm{f,prod}}) - \sum (n_{\mathrm{react}} \cdot ΔH_{\mathrm{f,react}})\)
\(\ce{3NO2(g)}\)
\(\ce{ + }\)
\(\ce{H2O(l)}\)
\(\ce{->}\)
\(\ce{2HNO3(aq)}\)
\(\ce{ + }\)
\(\ce{NO(g)}\)
\(\ce{ }\)
\(\mathrm{ΔHf°}_{\mathrm{\ce{HNO3(aq)}}}\)
\(= -207.4\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\mathrm{ΔHf°}_{\mathrm{\ce{NO(g)}}}\)
\(= 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\mathrm{ΔHf°}_{\mathrm{\ce{NO2(g)}}}\)
\(= 33.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\mathrm{ΔHf°}_{\mathrm{\ce{H2O(l)}}}\)
\(= -285.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ν_{\mathrm{\ce{HNO3(aq)}}}\)
\(= 2\)
\(ν_{\mathrm{\ce{NO(g)}}}\)
\(= 1\)
\(ν_{\mathrm{\ce{NO2(g)}}}\)
\(= 3\)
\(ν_{\mathrm{\ce{H2O(l)}}}\)
\(= 1\)
\(ΔH°_{\mathrm{rxn}}\)
\(= \sum (\mathrm{ΔHf°}_{\mathrm{\ce{HNO3(aq)}}} \cdot ν_{\mathrm{\ce{HNO3(aq)}}}, \mathrm{ΔHf°}_{\mathrm{\ce{NO(g)}}} \cdot ν_{\mathrm{\ce{NO(g)}}}) - \sum (ν_{\mathrm{\ce{NO2(g)}}} \cdot \mathrm{ΔHf°}_{\mathrm{\ce{NO2(g)}}}, ν_{\mathrm{\ce{H2O(l)}}} \cdot \mathrm{ΔHf°}_{\mathrm{\ce{H2O(l)}}})\)
\(\ \ \ =\sum (-207.4\ \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 2, 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 1) - \sum (3 \cdot 33.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}, 1 \cdot (-285.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}))\)
\(\ \ \ =\sum (-414.80\ \frac{\mathrm{kJ}}{\mathrm{mol}}, 90.20\ \frac{\mathrm{kJ}}{\mathrm{mol}}) - \sum (99.60\ \frac{\mathrm{kJ}}{\mathrm{mol}}, -285.80\ \frac{\mathrm{kJ}}{\mathrm{mol}})\)
\(\ \ \ =-324.60\ \frac{\mathrm{kJ}}{\mathrm{mol}} - (-186.20\ \frac{\mathrm{kJ}}{\mathrm{mol}})\)
\(\ \ \ =-138.4\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)