Answer:
ΔH for the the reaction NO(g) + O(g) ⇒ NO₂(g) is ΔH= -304.1 kJ
Explanation:
<u>The complete question is:</u>
Consider the chemical equations shown here.
NO(g) + O₃(g) ⇒ NO₂(g) + O₂(g) (ΔH= -198.9 kJ
)
1.5 O₂(g) ⇒ O₃(g) (ΔH= 142.3 kJ
)
O(g) ⇒ 0.5 O₂(g) (ΔH= -247.5 kJ)
What is ΔH for the reaction shown below?
NO(g) + O(g) ⇒ NO₂(g)
Solution:
We have to use the Hess's Law: if a series of reagents react to give a series of products, the heat of reaction released or absorbed is independent of whether the reaction is carried out in one, two or more stages. That means enthalpy changes are additive.
NO(g)+ O₃(g) ⇒ NO₂(g) + O₂(g) (ΔH₁= -198.9 kJ
)
+
1.5 O₂(g) ⇒ O₃(g) (ΔH₂= 142.3 kJ
)
+
O(g) ⇒ 0.5 O₂(g) (ΔH₃= -247.5 kJ)
=
NO(g) + O₃(g) + 1.5 O₂(g) + O(g) ⇒ NO₂(g) + O₂(g) + O₃(g) + 0.5 O₂(g)
We remove the compounds that are in both members of the reaction:
NO(g) + O(g) ⇒ NO₂(g)
We only have to add the reactions so we add the value of each enthalpy change.
ΔH for the the reaction is given by:
ΔH= ΔH₁ + ΔH₂ + ΔH₃= -198.9 kJ +142.3 kJ -247.5 kJ= -304.1 kJ
