Given the Data:

N2(g) + O2g) = 2NO(g) delta H = 180.7kj, 2NO(g) + O2(g) = 2NO2(g) delta H = 0113.1kj, 2N2O(g) = 1N2(g) + O2(g) delta H = -163.2kj.
Using Hess’s Law calculate delta H for the reaction:
N2O(g) + NO2(g) = 3NO(g)

See your earlier post.

To calculate ΔH for the reaction N2O(g) + NO2(g) = 3NO(g) using Hess's Law, we need to manipulate the given equations to get the desired reaction as the sum of those equations. Here's a step-by-step process:

1. Reverse the first equation to get N2O(g). Since the reaction is reversed, the sign of ΔH also changes:
N2(g) + O2(g) → 2NO(g) (ΔH = -180.7 kJ)

2. Multiply the second equation by 2 to get 2NO2(g):
2NO(g) + O2(g) → 2NO2(g) (ΔH = -113.1 kJ)

3. Multiply the third equation by 3 to get 3NO(g):
6N2O(g) → 3N2(g) + 3O2(g) (ΔH = -489.6 kJ)

4. Now, add the above three equations together to balance the reaction:
N2O(g) + NO2(g) + 2NO(g) + O2(g) + 6N2O(g) = 3NO(g) + 2NO2(g) + 3N2(g) + 3O2(g)

5. Simplify the equation:
7N2O(g) + NO2(g) = 5NO(g) + 2NO2(g) + 3N2(g) + 3O2(g)

6. The ΔH for the reaction is the sum of ΔH values from the manipulated equations:
ΔH = (-180.7 kJ) + (-113.1 kJ) + (-489.6 kJ)
ΔH = -783.4 kJ

Therefore, the ΔH for the reaction N2O(g) + NO2(g) = 3NO(g) is -783.4 kJ.