Question

Nitrogen dioxide, NO2(g) (Delta.Hf = 33.84 kJ/mol), is decomposed according to the following reaction:

2 upper N upper O subscript 2 (g) right arrow upper N subscript 2 (g) plus 2 upper O subscript 2 (g).

What is the enthalpy change when 2.50 mol of nitrogen dioxide decomposes?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
13.5 kJ of energy released
13.5 kJ of energy absorbed
84.6 kJ of energy released
84.6 kJ of energy absorbed

Answer 1

ΔH°rxn for 2 NO2(g) → N2(g) + 2 O2(g) = [0 + 2(0)] − [2(33.84 kJ/mol)] = −67.68 kJ per 2.00 mol NO2.

For 2.50 mol NO2: ΔH = −67.68 kJ × (2.50/2.00) = −84.6 kJ.

Therefore 84.6 kJ of energy is released.