Consider the reaction below:

N2(g) + 2 O2(g) 2 NO2(g) ΔH = +66.4 kJ

What would the enthalpy change for the following reaction be?

4 NO2(g) 2 N2(g) + 4 O2(g) ΔH = ???

a. +66.4 kJ
b. -66.4 kJ
c. +132.8 kJ
d. -132.8 kJ
e. -33.4 kJ

D.

Explanation:

The forward reaction is endothermic. This is shown by the positive enthalpy change (+ 66.4 kJ). The reverse reaction will, therefore, be exothermic (-66.4 kJ).

However, because in the reverse reaction we are dealing with double the moles as in the forward reaction, the enthalpy change will also be double.

66.4 * 2 = 132.8

= - 132.8 kJ

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