Question 6.9: Calculating the Enthalpy of Reaction from Standard Enthalpie...
Calculating the Enthalpy of Reaction from Standard Enthalpies of Formation
Large quantities of ammonia are used to prepare nitric acid. The first step consists of the catalytic oxidation of ammonia to nitric oxide, NO.
4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \stackrel{\mathrm{Pt}}{\longrightarrow} 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)
What is the standard enthalpy change for this reaction? Use Table 6.2 for data.
PROBLEM STRATEGY You record the values of \Delta H_{f}^{\circ} under the formulas in the equation, multiplying them by the coefficients in the equation. You calculate \Delta H^{\circ} by subtracting values for the reactants from values for the products.
Here is the equation with the \Delta H_{f}^{\circ} ‘s recorded beneath it:
\begin{array}{ccccc} 4 \mathrm{NH}_{3}(g)+&5 \mathrm{O}_{2}(g) & \longrightarrow & 4 \mathrm{NO}(g)+&6 \mathrm{H}_{2} \mathrm{O}(g) \\ 4(-45.9) & 5(0) && 4(90.3) & 6(-241.8) & (\mathrm{kJ}) \end{array}
The calculation is
\begin{aligned} \Delta H^{\circ} & \left.=\sum n \Delta H_{f}^{\circ} \text { (products }\right)-\sum m \Delta H_{f}^{\circ}(\text { reactants }) \\ & =\left[4 \Delta H_{f}^{\circ}(\mathrm{NO})+6 \Delta H_{f}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]-\left[4 \Delta H_{f}^{\circ}\left(\mathrm{NH}_{3}\right)+5 \Delta H_{f}^{\circ}\left(\mathrm{O}_{2}\right)\right] \\ & =[4(90.3)+6(-241.8)] \mathrm{kJ}-[4(-45.9)+5(0)] \mathrm{kJ} \\ & =-906 \mathbf{k J} \end{aligned}
Be very careful of arithmetical signs-they are a likely source of mistakes. Also pay particular attention to the state of each substance. Here, for example, you must use the \Delta H_{f}^{\circ} for \mathrm{H}_{2} \mathrm{O}(g), not for \mathrm{H}_{2} \mathrm{O}(l).