Question

consider these intermediate chemical equations. no(g) + o₃(g) → no₂(g) + o₂(g) δh₁ = -198.9kj ¹/₂o₂(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)? -304.1kj -305kj -93.7kj 588.7kj

Explanation:

Step1: Identify relevant equations

We have the equations:

1. $NO(g)+O_3(g)

ightarrow NO_2(g)+O_2(g)\quad\Delta H_1 = - 198.9kJ$

1. $\frac{1}{2}O_2(g)

ightarrow O_3(g)\quad\Delta H_2 = 142.3kJ$

1. $O(g)

ightarrow\frac{1}{2}O_2(g)\quad\Delta H_3=-247.5kJ$
We want $NO(g) + O(g)
ightarrow NO_2(g)$.

Step2: Manipulate equations

Add the three - given equations together.
When we add the equations:
$(NO(g)+O_3(g)
ightarrow NO_2(g)+O_2(g))+(\frac{1}{2}O_2(g)
ightarrow O_3(g))+(O(g)
ightarrow\frac{1}{2}O_2(g))$
The $O_3(g)$ and $\frac{1}{2}O_2(g)$ terms cancel out on the left - hand and right - hand sides, and we get $NO(g)+O(g)
ightarrow NO_2(g)$.

Step3: Calculate the overall enthalpy

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

Answer:

• 304.1kJ