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

Hint: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.