Question

At ${\text{320 K}}$ , a gas ${{\text{A}}_2}$ is {\text{20% }} dissociated to A(g). The standard free energy change at ${\text{320 K}}$ and 1 atm in ${\text{J mo}}{{\text{l}}^{ - 1}}$ is approximately:
${\text{[R }} = 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}};\ln 2 = 0.693;\ln 3 = 1.098]$
A.4763
B.2068
C.4281
D.1844

Answer 1

For calculation of standard free energy we have to use the formula for the same. The value of equilibrium constant can be calculated by given dissociation percentage. Start by considering X as the initial amount. Equilibrium constant is the ratio of concentration of product to the time raised to their stoichiometry.

Formula used: $\Delta {\text{G}}^\circ = - {\text{RT}}\ln {\text{K}}$
Here $\Delta {\text{G}}^\circ$ is standard free energy change, R is universal gas constant, T is temperature and K is rate constant.

Complete step by step answer:
Given that a gas ${{\text{A}}_2}$ dissociates into A. Since, it does not dissociate completely equilibrium will form an equilibrium reaction as:
${{\text{A}}_2} \to 2{\text{A}}$
It is given that {\text{20% }} of ${{\text{A}}_2}$ is dissociated and hence {\text{80% }} of ${{\text{A}}_2}$ is the remaining amount and {\text{2}} \times {\text{20% }} is the amount of gas A forms because 2 moles of gas are formed after dissociation of 1 mole of ${{\text{A}}_2}$ . Let us assume the initial amount as X. Then amount of
${\text{[}}{{\text{A}}_2}] = \dfrac{{80}}{{100}} \times {\text{X}}$ and ${\text{[A}}] = 2 \times \dfrac{{20}}{{100}} \times {\text{X}}$
$\left[ {{{\text{A}}_2}} \right] = 0.8{\text{X}}$ and ${\text{[A}}] = 0.4{\text{X}}$
Now we will calculate the equilibrium constant as:
${\text{K}} = \dfrac{{{{[{\text{A}}]}^2}}}{{[{{\text{A}}_2}]}}$
$\Rightarrow {\text{K}} = \dfrac{{{{(0.4)}^2}}}{{0.8}} = 0.2$
We have put square over concentration of A gas because 2 is the stoichiometry of has A in the equation.
Now substituting the values in the formula for standard free energy:
$\Delta {\text{G}}^\circ = - 8.314 \times 320 \times \ln 0.2$
$\Rightarrow \Delta {\text{G}}^\circ = - 4281{\text{ J mo}}{{\text{l}}^{ - 1}}$

Hence, the correct option is C.

Note:
In the above question {\text{20% }} is actually the degree of dissociation given to us. Degree of dissociation is the ratio of amount of substance dissociated to the amount of reactant initially present. When we use a degree of dissociation, we can take the initial concentration as 1.