Question

consider the following thermochemical equations.
no(g) + o₃(g) → no₂(g) + o₂(g)
δh = -198.9 kj/mol
o₃(g) → 3/2 o₂(g)
δh = -142.3 kj/mol
o₂(g) → 2 o(g)
δh = +495 kj/mol
2 no₂(g) → 2 no(g) + o₂(g)
determine the enthalpy change, in kj/mol, for the reaction

Explanation:

Step1: Reverse the first equation

$$2NO(g)+O_{2}(g)\to 2NO_{2}(g)\quad \Delta H = + 198.9\space kJ/mol$$

Step2: Use the second equation as it is

$$O_{3}(g)\to\frac{3}{2}O_{2}(g)\quad \Delta H=-142.3\space kJ/mol$$

Step3: Reverse the third equation and divide by 2

$$O(g)\to\frac{1}{2}O_{2}(g)\quad \Delta H = - 247.5\space kJ/mol$$

Step4: Add the three equations

$$2NO(g)+O_{3}(g)\to 2NO_{2}(g)+\frac{1}{2}O_{2}(g)$$
$$\Delta H=(198.9 - 142.3-247.5)\space kJ/mol=-190.9\space kJ/mol$$

Answer:

$-190.9\space kJ/mol$