Question 10.7: Given the following thermochemical equations, NO(g) + O3(g) ...
Given the following thermochemical equations,
\text{NO}(\text{g}) + \text{O}_3(\text{g}) → \text{NO}_2(\text{g}) + \text{O}_2(\text{g}) ∆H = –198.9 \text{kJ/mol}
\text{O}_3(\text{g}) → \frac{3}{2} \text{O}_2(\text{g}) ∆H = –142.3 \text{kJ/mol}
\text{O}_2(\text{g}) → 2\text{O}(\text{g}) ∆H = +495 \text{kJ/mol}
determine the enthalpy change for the reaction
\text{NO}(\text{g}) + \text{O(g)} → \text{NO}_2(\text{g})
Strategy Arrange the given thermochemical equations so that they sum to the desired equation. Make the corresponding changes to the enthalpy changes, and add them to get the desired enthalpy change
Setup The first equation has \text{NO} as a reactant with the correct coefficient, so we use it as is
\text{NO(g)} + \text{O}_3(\text{g}) \text{NO}_2(\text{g}) + \text{O}_2(\text{g}) ∆H = –198.9 \text{kJ/mol}
The second equation must be reversed so that the \text{O}_3 introduced by the first equation will cancel (\text{O}_3 is not part of the overall chemical equation). We also must change the sign on the corresponding ∆H value.
\frac{3}{2} \text{O}_2(\text{g}) → \text{O}_3 (\text{g}) ∆H = +142.3 \text{kJ/mol}
These two steps sum to give the following:
\text{NO}(\text{g}) + \cancel{\text{O}_3(\text{g})} → \text{NO}_2(\text{g}) + \cancel{\text{O}_2(\text{g})} ∆H = –198.9 \text{kJ/mol}
\underline{+ \frac{1}{2}\text{O}_2(\text{g}) \frac{3}{2} \cancel{\text{O}_2(\text{g})} → \cancel{\text{O}_3 (\text{g})} ∆H = +142.3 \text{kJ/mol}}
\text{NO(g)} + \frac{1}{2} \text{O}_2(\text{g}) → \text{NO}_2(\text{g}) ∆H = –56.6 \text{kJ/mol}
We then replace the \frac{1}{2}\text{O}_2 on the left with \text{O} by incorporating the last equation. To do so, we divide the third equation by 2 and reverse its direction. As a result, we must also divide its ∆H value by 2 and change its sign.
\text{O}(\text{g}) → \frac{1}{2} \text{O}_2(\text{g}) ∆H = –247.5 \text{kJ/mol}
Finally, we sum all the steps and add their enthalpy changes.
\text{NO}(\text{g}) + \cancel{\text{O}_3(\text{g})} → \text{NO}_2(\text{g}) + \cancel{\text{O}_2(\text{g})} ∆H = –198.9 \text{kJ/mol}
\frac{3}{2} \cancel{\text{O}_2(\text{g})} → \cancel{\text{O}_3 (\text{g})} ∆H = +142.3 \text{kJ/mol}
\underline{+\text{O}(\text{g}) → \frac{1}{2} \cancel{\text{O}_2(\text{g})} ∆H = –247.5 \text{kJ/mol}}
\text{NO(g)} + \text{O}(\text{g}) → \text{NO}_2(\text{g}) ∆H = –304 \text{kJ/mol}