What is the change in enthalpy of the first reaction below, given the enthalpies

of the other two reactions?

2NO (g) + O2 (g) + 2NO2 (g)

N, (8) + O2 (g) → NO (g)

N, (8) + O2 (g) + NO2 (g)

AH = 90 kJ/mol

AH° = 34 kJ/mol

A. 56 kJ

B. - 112 kJ

C. 124 kJ

D. - 248 kJ

The correct option is;

B. - 112 kJ

Explanation:

The parameters given are;

N (g) + O₂ (g) → NO (g) ΔH = 90 kJ/mol

N (g) + O₂ (g) → NO₂ (g) ΔH = 34 kJ/mol

The required chemical reaction is given as follows;

2NO (g) + O₂ (g) → 2NO₂ (g)

Therefore, the heat of formation of 2 moles of NO = 2 * 90 = 180 kJ

The heat of formation of 2 moles of NO₂ = 2 * 34 = 68 kJ

Hence, given that the heat of formation of O₂ at room temperature = 0 kJ/mol, we have;

Change in enthalpy of the chemical reaction = Heat of formation of the products - Heat of formation of the reactants

Change in enthalpy of the chemical reaction = 68 kJ - 180 kJ = - 112 kJ

Change in enthalpy of the chemical reaction = - 112 kJ.