Question

consider these intermediate chemical equations.
no (g) + o₃ (g) → no₂ (g) + o₂ (g) δh₁ = -198.9kj
³/₂o₂ (g) → o₃ (g) δh₂ = 142.3kj
o (g) → ¹/₂o₂ (g) δh₃ = -247.5kj
what is the enthalpy of the overall chemical equation no (g) + o (g) → no₂ (g)?
-93.7kj
-304.1kj
-305kj
588.7kj

Explanation:

Step1: Identify the target reaction

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

Step2: Manipulate given reactions

The first reaction is $NO(g)+O_3(g)
ightarrow NO_2(g)+O_2(g)$ with $\Delta H_1 = - 198.9kJ$.
The second reaction $\frac{3}{2}O_2(g)
ightarrow O_3(g)$ with $\Delta H_2=142.3kJ$.
The third reaction $O(g)
ightarrow\frac{1}{2}O_2(g)$ with $\Delta H_3=-247.5kJ$.
Adding these three reactions together:
$(NO(g)+O_3(g)
ightarrow NO_2(g)+O_2(g))+(\frac{3}{2}O_2(g)
ightarrow O_3(g))+(O(g)
ightarrow\frac{1}{2}O_2(g))$
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 $NO(g)+O(g)
ightarrow NO_2(g)$.

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:

$-304.1kJ$