Question

nitric oxide (no) can be formed from nitrogen, hydrogen and oxygen in two steps. in the first step, nitrogen and hydrogen react to form ammonia: \\(\ce{n_{2}(g) + 3h_{2}(g) <=> 2nh_{3}(g)} \quad \delta h = -92.\\,\text{kj}\\)\
in the second step, ammonia and oxygen react to form nitric oxide and water: \\(\ce{4nh_{3}(g) + 5o_{2}(g) <=> 4no(g) + 6h_{2}o(g)} \quad \delta h = -905.\\,\text{kj}\\)\
calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. round your answer to the nearest kj.

Explanation:

Step1: Adjust first reaction

Multiply $N_2(g) + 3H_2(g)
ightarrow 2NH_3(g)$ ($\Delta H=-92kJ$) by 2: $2N_2(g) + 6H_2(g)
ightarrow 4NH_3(g)$, $\Delta H=2\times(-92)=-184kJ$

Step2: Add adjusted reactions

Add $2N_2 + 6H_2
ightarrow 4NH_3$ ($\Delta H=-184kJ$) and $4NH_3 + 5O_2
ightarrow 4NO + 6H_2O$ ($\Delta H=-905kJ$):
$2N_2 + 6H_2 + 5O_2
ightarrow 4NO + 6H_2O$, $\Delta H=-184-905=-1089kJ$

Step3: Find per mole NO enthalpy

Divide total $\Delta H$ by 4: $\frac{-1089}{4}\approx-272kJ$

Answer:

-272