Question

If the following reaction:

NO(g) + 1/2O2(g) --> NO2(g)
has the following enthalpy change:
H° = -56kJ/mol
What is the enthalpy of the decomposition reaction of 2 moles of NO2?
A. 112 kJ
B. -112 kJ
C. -56 kJ
D. 56 kJ

Answer 1

The enthalpy of the decomposition reaction of 2 moles of NO2 is +112 kJ, which is option A.

Step-by-step explanation:

The enthalpy change for the reaction NO(g) + 1/2O2(g) --> NO2(g) is ΔH° = -56kJ/mol. This means that -56 kJ of energy is released when 1 mole of NO2(g) is formed from NO(g) and O2(g). To find the enthalpy of the decomposition reaction of 2 moles of NO2, we reverse the reaction and multiply the enthalpy change by 2 (since enthalpy is a state function and directly proportional to the amounts of substances involved). Therefore, the reverse reaction 2NO2(g) --> 2NO(g) + O2(g) has an enthalpy change of +112 kJ(opposite sign as the formation, and doubled for 2 moles).