Question

Using Hess's law, what is ÎH°rxn for the following reaction?NO(g) + O(g) â NO2(g)NO(g) + O3(g) â NO2(g) + O2(g) ÎH°rxn = â198.8 kJ/molO3(g) â 3/2 O2(g) ÎH°rxn= â142.2 kJ/molO2(g) â 2O(g) ÎH°rxn= +498.8 kJ/mol

Answer 1

Answer:

The enthalpy of the reaction asked is -306 kJ/mol.

Explanation:

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g),\Delta H_1=-198.8 kJ/mol$..[1]

$O_3(g)\rightarrow \frac{3}{2} O_2(g) ,\Delta H_2= -142.2 kJ/mol$..[2]

$O2(g)\rightarrow 2O(g) \Delta H_3= 498.8 kJ/mol$..[3]

$NO(g) + O(g)\rightarrow NO_2(g),\Delta H_4=?$..[4]

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

[1] - [2]  - [3] × 0.5 = [4]

$\Delta H_4=\Delta H_1-Delta H_2-0.5\times \Delta H_3$(By using Hess's law)

$\Delta H_4=-198.8 kJ/mol-( -142.2 kJ/mol)-0.5\times (498.8 kJ/mol)$

$=-306 kJ/mol$

The enthalpy of the reaction asked is -306 kJ/mol.