Question

what is \\( \delta h_{\mathrm{rxn}}^{\circ} \\) for the following reaction?\\( \mathrm{no}_{2}(g)+\mathrm{co}(g) \
ightarrow \mathrm{co}_{2}(g)+\mathrm{no}(g) \\)

Explanation:

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

$\Delta H^{\circ}_{rxn}=\sum n\Delta H^{\circ}_{f}(\text{products})-\sum n\Delta H^{\circ}_{f}(\text{reactants})$

Step2: Identify the products and reactants

For the reaction $\text{NO}_{2}(g)+\text{CO}(g)\to\text{CO}_{2}(g)+\text{NO}(g)$:

• Products: $\text{CO}_{2}(g)$ ($n = 1$, $\Delta H^{\circ}_{f}=-393.5\ \text{kJ/mol}$) and $\text{NO}(g)$ ($n = 1$, $\Delta H^{\circ}_{f}=+90.4\ \text{kJ/mol}$)
• Reactants: $\text{NO}_{2}(g)$ ($n = 1$, $\Delta H^{\circ}_{f}=+33.85\ \text{kJ/mol}$) and $\text{CO}(g)$ ($n = 1$, $\Delta H^{\circ}_{f}=-110.5\ \text{kJ/mol}$)

Step3: Substitute into the formula

$$\begin{align*} \Delta H^{\circ}_{rxn}&=(1\times(- 393.5)+1\times90.4)-(1\times33.85 + 1\times(-110.5))\\ &=(-393.5 + 90.4)-(33.85-110.5)\\ &=-303.1-(-76.65)\\ &=-303.1 + 76.65\\ &=-226.45\approx - 226.5\ \text{kJ} \end{align*}$$

Answer:

-226.5 kJ