Question

Consider the following intermediate chemical equations.

NO(g)+O(g) --> NO2(g)+O2(g) ΔH1= - 198.9kJ
3/2 O2(g)-->)3(g) ΔH2= 142.3kJ
O(g)-->1/2 O2(g) ΔH3= -247.5kJ

What is the enthalpy of the overall chemical equation NO(g) + O(g)-->NO2(g)?
A) -305 kJ
B) -304.1 kJ
C) -93.7 kJ
D) 588.7 kJ

Answer 1

Considering the first reaction there is no NO on the left and NO2 on the right we start with adding equation 1. To remove O3 we subtract equation 2, then here is 1/2O2 left which we remove by subtracting 1/2 equation 3.
That is; Eqn 1 - Eqn 2 - 1/2(Eqn 3)
NO(g) + O3(g) + 3/2 O2(g) + O(g) = NO2(g) + O2(g) + O3(g) + 1/2 O2(g)
This gives;
NO(g) + O(g) = NO2(g) as required, since O3(g) + 3/2O2(g) is on both sides and thus subtracts out.
Hence; ΔH = ΔH1-ΔH2-1/2 ΔH3
= (-198.9 + 142.3 - 0.5×495.0) kJ = -304.1 kJ

Answer 2

The answer is: B) -304.1 kJ.

Chemical equation 1: NO(g) + 3O(g) → NO₂(g) + O₂; ΔH₁ = -198.9 kJ.

Chemical equation 2: 3/2O₂(g) → 3O(g); ΔH₂ = 142.3 kJ.

Chemical equation 3: O(g) → 1/2O₂(g); ΔH₃ = -247.5 kJ.

Using Hess's law, the fouth reaction is sum of first three reactions:

ΔH₄ = ΔH₁ + ΔH₂ + ΔH₃.

ΔH₄ = -198.9 kJ + 142.3 kJ + (-247.5 kJ).

ΔH₄ = -304.1 kJ.