Question

Calculate ΔG∘rxn for the following reaction:4CO(g)+2NO2(g)→4CO2(g)+N2(g).Use the following reactions and given ΔG∘rxn values:A) 2NO(g)+O2(g)→2NO2(g), ΔG∘rxn= - 72.6 kJB) 2CO(g)+O2(g)→2CO2(g), ΔG∘rxn= - 514.4 kJC) 12O2(g)+12N2(g)→NO(g), ΔG∘rxn= 87.6 kJ

Answer 1

Answer : The   value of $\Delta G^o_{rxn}$ for the reaction is -1131.4 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,

$4CO(g)+2NO_2(g)\rightarrow 4CO_2(g)+N_2(g)$    $\Delta G^o_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $2NO(g)+O_2(g)\rightarrow 2NO_2(g)$     $\Delta G^o_1=-72.6kJ$

(2) $2CO(g)+O_2(g)\rightarrow 2CO_2(g)$     $\Delta G^o_2=-514.4kJ$

(3) $\frac{1}{2}O_2(g)+\frac{1}{2}N_2(g)\rightarrow NO(g)$     $\Delta G^o_3=87.6kJ$

Now we will reverse the reaction 1, multiply the reaction 2 by 2, reverse and half the reaction 3 and then adding all the equations, we get :

(1) $2NO_2(g)\rightarrow 2NO(g)+O_2(g)$     $\Delta G^o_1=72.6kJ$

(2) $4CO(g)+2O_2(g)\rightarrow 4CO_2(g)$     $\Delta G^o_2=2\times (-514.4kJ)=-1028.8kJ$

(3) $2NO(g)\rightarrow O_2(g)+N_2(g)$     $\Delta G^o_3=-2\times 87.6kJ=-175.2kJ$

The expression for $\Delta G^o_{rxn}$ will be,

$\Delta G^o_{rxn}=\Delta G^o_{1}+\Delta G^o_{2}+\Delta G^o_{3}$

$\Delta G^o_{rxn}=(72.6kJ)+(-1028.8kJ)+(-175.2kJ)$

$\Delta G^o_{rxn}=-1131.4kJ$

Therefore, the value of $\Delta G^o_{rxn}$ for the reaction is -1131.4 kJ