consider these intermediate chemical equations.\nno (g) + o₃ (g) → no₂ (g) + o₂ (g) δh₁ = -198.9kj\n³/₂o₂…

Explanation:

Step1: Identify the target reaction

We want to find $\Delta H$ for $NO(g)+O(g)\rightarrow NO_2(g)$.

Step2: Manipulate given reactions

The first reaction is $NO(g)+O_3(g)\rightarrow NO_2(g)+O_2(g)$ with $\Delta H_1 = - 198.9kJ$. The second reaction $\frac{3}{2}O_2(g)\rightarrow O_3(g)$ with $\Delta H_2=142.3kJ$. The third reaction $O(g)\rightarrow\frac{1}{2}O_2(g)$ with $\Delta H_3=-247.5kJ$. Adding these three reactions together: $(NO(g)+O_3(g)\rightarrow NO_2(g)+O_2(g))+(\frac{3}{2}O_2(g)\rightarrow O_3(g))+(O(g)\rightarrow\frac{1}{2}O_2(g))$ The $O_3(g)$ and $\frac{3}{2}O_2(g)$ and $\frac{1}{2}O_2(g)$ cancel out on the left - hand and right - hand sides, leaving $NO(g)+O(g)\rightarrow NO_2(g)$.

Step3: Calculate the overall $\Delta H$

According to Hess's law, $\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$. $\Delta H=-198.9 + 142.3-247.5$ $\Delta H=-304.1kJ$

Answer:

$-304.1kJ$