Enthalpy Change

Question

Given the following data:

2 O3 (g) →3 O2 (g) ∆H=−427 kJ

O2 (g) →2O (g) ∆H= 495 kJ

NO (g) + O3 (g) →NO2 (g) + O2 (g) ∆H= −199 kJ

calculate the enthalpy change for the reaction

NO (g) + O (g) →NO2 (g)

Answer 1

NO (g) + O (g) →NO2 (g) ... Target Equation

1). 2 O3 (g) →3 O2 (g) ∆H=−427 kJ

2). O2 (g) →2O (g) ∆H= 495 kJ

3). NO (g) + O3 (g) →NO2 (g) + O2 (g) ∆H= −199 kJ

reverse #1 and divide by 2: 1.5O2 ==> O3 ∆H = +213.5 kJ

reverse #2 and divide by 2: O ===> 0.5O2 ∆H = -247.5 kJ

copy #3 as it is: NO + O3 ==> NO2 + O2 ∆H = -199 kJ

Add these together and cancel like terms on opposite sides to get the target equation:

1.5O2 + O + NO + O3 ==> O3 + 0.5O2 + NO2 + O2

NO + O ==> NO2 ∆H = 213.5 - 247.5 - 199 = -233 kJ <-Answer