Question

Calculate the enthalpy of the reaction: 2NO(g) + O₂(g) --> 2NO₂(g) given the following reactions and enthalpies of formation:

1/2 N₂(g) + O₂(g) --> NO₂(g), delta H*A = 33.2 kJ
1/2 N₂(g) + 1/2O₂(g) --> NO(g), delta H*B= 90.2 kJ

Answer 1

The enthalpy change for the reaction 2NO(g) + O₂(g) → 2NO₂(g) is calculated using given enthalpy of formations to be -114 kJ.

Step-by-step explanation:

To calculate the enthalpy of the reaction 2NO(g) + O₂(g) → 2NO₂(g), we need to use the provided reactions and their respective enthalpies of formation. We have the following data:

• ½N₂(g) + O₂(g) → NO₂(g), ΔH*A = 33.2 kJ
• ½N₂(g) + ½O₂(g) → NO(g), ΔH*B = 90.2 kJ
• N₂(g) + O₂(g) → 2NO₂(g), ΔH = 66.4 kJ

To find the enthalpy change for 2NO(g) formation, we double the second equation, as enthalpy change is extensive. This gives:

N₂(g) + O₂(g) → 2NO(g), ΔH = 2 * 90.2 kJ = 180.4 kJ.

We then add the enthalpy of formation for 2NO(g) with that for 2NO₂(g) and subtract the enthalpy for N₂(g) + O₂(g) → 2NO(g) to find the enthalpy of the given reaction:

ΔHreaction = ΔHformation(2NO₂) - ΔHformation(2NO) = 66.4 kJ - 180.4 kJ = -114 kJ.

Therefore, the enthalpy change for the reaction 2NO(g) + O₂(g) → 2NO₂(g) is -114 kJ.