Question

Use Hess's law to calculate ΔG°rxn using the following information. NO(g) + O(g) → NO2(g) ΔG°rxn = ? 2 O3(g) → 3 O2(g) ΔG°rxn = +489.6 kJ O2(g) → 2 O(g) ΔG°rxn = +463.4 kJ NO(g) + O3(g) → NO2(g) + O2(g) ΔG°rxn = -199.5 kJ

Answer 1

Answer : The value of $\Delta G^o_{rxn}$ is, -676 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The given main reaction is,

$NO(g)+O(g)\rightarrow NO_2(g)$    $\Delta G^o_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $2O_3(g)\rightarrow 3O_2(g)$     $\Delta G^o_1=489.6kJ$

(2) $O_2(g)\rightarrow 2O(g)$    $\Delta G^o_2=463.4kJ$  

(3) $NO(g)+O_3(g)\rightarrow NO_2(g)+O_2(g)$    $\Delta G^o_3=-199.5kJ$

Now we are reversing the reaction 1 and 2 and multiplying reaction 3 by 2 and then adding all the equations, we get :

(1) $3O_2(g)\rightarrow 2O_3(g)$     $\Delta G^o_1=-489.6kJ$

(2) $2O(g)\rightarrow O_2(g)$    $\Delta G^o_2=-463.4kJ$  

(3) $2NO(g)+2O_3(g)\rightarrow 2NO_2(g)+2O_2(g)$    $\Delta G^o_3=2\times -199.5kJ=-399kJ$

The overall reaction is:

$2NO(g)+2O(g)\rightarrow 2NO_2(g)$

$\Delta G^o_{rxn}=\Delta G^o_1+\Delta G^o_2+\Delta G^o_3$

$\Delta G^o_{rxn}=(-489.6kJ)+(-463.4kJ)+(-399kJ)$

$\Delta G^o_{rxn}=-1352kJ$

Now we are dividing the overall reaction by 2, we get:

$NO(g)+O(g)\rightarrow NO_2(g)$

$\Delta G^o_{rxn}=\frac{-1352kJ}{2}=-676kJ$

Thus, the value of $\Delta G^o_{rxn}$ is, -676 kJ