Example 5.15: Using Hess’s Law
What is the standard enthalpy change for the reaction:
\(\ce{3NO2(g)}\)\(\ce{ + }\)\(\ce{H2O(l)}\)\(\ce{->}\)\(\ce{2HNO3(aq)}\)\(\ce{ + }\)\(\ce{NO(g)}\)\(\ce{ }\) \(ΔH°\)\( = \)?
Solution: Using the Equation
Use the special form of Hess’s law given previously:
\(ΔH°_{\mathrm{rxn}} = \sum (ν_{\mathrm{product}} \cdot \mathrm{ΔHf°}_{\mathrm{product}}) - \sum (ν_{\mathrm{reactant}} \cdot \mathrm{ΔHf°}_{\mathrm{reactant}})\)
\(\mathrm{ΔHf°}_{\mathrm{\ce{NO(g)}}}\) \(= 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\mathrm{ΔHf°}_{\mathrm{\ce{NO2(g)}}}\) \(= 33.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\mathrm{ΔHf°}_{\mathrm{\ce{H2O(l)}}}\) \(= -285.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ν_{\mathrm{\ce{HNO3(aq)}}}\) \(= 2\)
\(ν_{\mathrm{\ce{NO(g)}}}\) \(= 1\)
\(ν_{\mathrm{\ce{NO2(g)}}}\) \(= 3\)
\(ν_{\mathrm{\ce{H2O(l)}}}\) \(= 1\)
\(ΔH°_{\mathrm{rxn}}\) \(= \sum (\mathrm{ΔHf°}_{\mathrm{\ce{HNO3(aq)}}} \cdot ν_{\mathrm{\ce{HNO3(aq)}}}, \mathrm{ΔHf°}_{\mathrm{\ce{NO(g)}}} \cdot ν_{\mathrm{\ce{NO(g)}}}) - \sum (ν_{\mathrm{\ce{NO2(g)}}} \cdot \mathrm{ΔHf°}_{\mathrm{\ce{NO2(g)}}}, ν_{\mathrm{\ce{H2O(l)}}} \cdot \mathrm{ΔHf°}_{\mathrm{\ce{H2O(l)}}})\)
\(\ \ \ =\sum (-207.4\ \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 2, 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 1) - \sum (3 \cdot 33.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}, 1 \cdot (-285.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}))\)
\(\ \ \ =\sum (-414.80\ \frac{\mathrm{kJ}}{\mathrm{mol}}, 90.20\ \frac{\mathrm{kJ}}{\mathrm{mol}}) - \sum (99.60\ \frac{\mathrm{kJ}}{\mathrm{mol}}, -285.80\ \frac{\mathrm{kJ}}{\mathrm{mol}})\)
\(\ \ \ =-324.60\ \frac{\mathrm{kJ}}{\mathrm{mol}} - (-186.20\ \frac{\mathrm{kJ}}{\mathrm{mol}})\)
\(\ \ \ =-138.4\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
Solution: Supporting Why the General Equation Is Valid
Alternatively, we can write this reaction as the sum of the decompositions of
\(\ce{3NO2(g)}\) and
\(\ce{1H2O(l)}\) into their constituent elements, and the formation of
\(\ce{2HNO3(aq)}\) and
\(\ce{1NO(g)}\) from their constituent elements. Writing out these reactions, and noting their relationships to the
\(\mathrm{ΔHf°}\) values for these compounds (from
Appendix G ), we have:
\(\ce{3NO2(g)}\)\(\ce{->}\)\(\ce{3/2N2(g)}\)\(\ce{ + }\)\(\ce{3O2(g)}\)\(\ce{ }\) \(ΔH1°\)\( = - 99.6\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ce{H2O(l)}\)\(\ce{->}\)\(\ce{H2(g)}\)\(\ce{ + }\)\(\ce{1/2O2(g)}\)\(\ce{ }\) \(ΔH2°\)\( = + 285.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ce{H2(g)}\)\(\ce{ + }\)\(\ce{N2(g)}\)\(\ce{ + }\)\(\ce{1/2O2(g)}\)\(\ce{->}\)\(\ce{2HNO3(aq)}\)\(\ce{ }\) \(ΔH3°\)\( = - 414.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ce{1/2N2(g)}\)\(\ce{ + }\)\(\ce{1/2O2(g)}\)\(\ce{->}\)\(\ce{NO(g)}\)\(\ce{ }\) \(ΔH4°\)\( = + 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ΔH2°\) \(= 285.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ΔH3°\) \(= -414.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ΔH4°\) \(= 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ce{3NO2(g)}\)\(\ce{ + }\)\(\ce{H2O(l)}\)\(\ce{->}\)\(\ce{2HNO3(aq)}\)\(\ce{ + }\)\(\ce{NO(g)}\)\(\ce{ }\)
Summing their enthalpy changes gives the value we want to determine:
\(\mathrm{ΔHrxn°}\) \(= ΔH1° + ΔH2° + ΔH3° + ΔH4°\)
\(\ \ \ =-99.6\ \frac{\mathrm{kJ}}{\mathrm{mol}} + 285.8\ \frac{\mathrm{kJ}}{\mathrm{mol}} + -414.8\ \frac{\mathrm{kJ}}{\mathrm{mol}} + 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ \ \ =186.20\ \frac{\mathrm{kJ}}{\mathrm{mol}} + -414.8\ \frac{\mathrm{kJ}}{\mathrm{mol}} + 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ \ \ =-228.60\ \frac{\mathrm{kJ}}{\mathrm{mol}} + 90.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ \ \ =-138.4\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
So the standard enthalpy change for this reaction is ΔH° = -138.4 kJ.
Note that this result was obtained by (1) multiplying the
\(\mathrm{ΔHf°}\) of each product by its stoichiometric coefficient and summing those values, (2) multiplying the
\(\mathrm{ΔHf°}\) of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.