Question

get this
ate \\( \delta h \\) for the following reaction:

\\( \frac{1}{2} \mathrm{~n}_{2}(g)+\frac{1}{2} \mathrm{o}_{2}(g) \
ightarrow \mathrm{no}(g) \\)

the following information:

\\( \operatorname{rxn}(i) \mathrm{n}_{2}(g)+2 \mathrm{o}_{2}(g) \longrightarrow 2 \mathrm{no}_{2}(g) \quad \delta h=66.4 \mathrm{~kj} \\)
\\( \operatorname{rxn}(\mathrm{ii}) 2 \mathrm{no}(g)+\mathrm{o}_{2}(g) \longrightarrow 2 \mathrm{no}_{2}(g) \quad \delta h=-114.1 \mathrm{~kj} \\)

\\( \mathrm{kj} \\)

Explanation:

Step1: Reverse reaction (ii)

$$2NO_{2}(g)\to 2NO(g)+O_{2}(g)\quad \Delta H = 114.1\,\text{kJ}$$

Step2: Subtract reaction (ii) (reversed) from reaction (i)

$$\begin{align} N_{2}(g)+2O_{2}(g)&\to 2NO_{2}(g)\quad \Delta H = 66.4\,\text{kJ}\\ 2NO_{2}(g)&\to 2NO(g)+O_{2}(g)\quad \Delta H = 114.1\,\text{kJ}\\ \hline N_{2}(g)+O_{2}(g)&\to 2NO(g)\quad \Delta H=(66.4 + 114.1)\,\text{kJ}=180.5\,\text{kJ} \end{align}$$

Step3: Divide the resulting reaction by 2

$$\frac{1}{2}N_{2}(g)+\frac{1}{2}O_{2}(g)\to NO(g)\quad \Delta H=\frac{180.5}{2}\,\text{kJ}=90.25\,\text{kJ}$$

Answer:

$90.25$