Question

The standard Gibbs free energy change at $300K$for the reaction $2A \rightleftharpoons B + C$ is $2494.2J$. At a given time, the composition of the given mixture is $[A] = 1/2$ , $[B] = 2$ and $[C] = 1/2$. The reaction proceeds in $[R = 8.314J/K/mol,e = 2.718]$,
A. Forward direction because $Q > {K_e}$
B. Reverse direction because $Q > {K_e}$
C. Forward direction because $Q < {K_e}$
D. Reverse direction because $Q < {K_e}$

Answer 1

Hint: $Q$ is the reaction quotient and $K$ is the equilibrium constant. If $Q$ is greater than $K$ equilibrium shifts in the reverse direction, and if $Q$ less than $K$ equilibrium shifts in forward direction. And if $Q$ equals to $K$ equilibrium is attained.

Complete step by step answer:
Consider a reversible reaction;
$aA + bB \leftrightarrows cC + dD …………………………….. (1)$
Where, $A$ and $B$ are reactants, $C$ and $D$are products, and $a$ ,$b$ ,$c$ and $d$ are coefficient. Now the reaction quotient is the ratio of concentration of reactants upon concentration of product, taking exponents of their respective coefficient i.e.
$Q = {[c]^c}{[D]^d}/{[A]^a}{[B]^b} ……………………….. (2)$
Now, according to the values given in the question;
$2A \leftrightarrows B + C$
$A = 1/2$ , $B = 2$ ,$C = 1/2$
Substituting the values given in equation $2$ we get
$Q = [B][C]/{[A]^2}$
$Q = (2*1/2)/{(1/2)^2}$
$Q = 1/(1/4)$
$Q = 4$.
Now , we know that relation between standard gibbs free energy and equilibrium constant is
$\Delta G = - 2.303RT\log K - (3)$
Where, $R = $ gas constant ,
$K = $ equilibrium constant,
$\Delta G = $ standard gibbs free energy
$T = $ temperature.
According to the question,
$R = 8.314J/K/mol$
$\Delta G = 2494.2J$
$T = 300K$
Substituting above values in equation $(3)$ we get
$2494.2 = - 2.303*8.314*300\log K$
$2494.2 = - 5744.143\log K$
$\log K = - 0.4342$
Taking $anti\log $ both the side we get
$anti\log (\log K) = anti\log ( - 0.4342)$
$K = 0.37$
Now, Q is referred, to see the direction change in the attained equilibrium ,
If $K$ is greater than $Q$, then reactants convert into products which creates a forward shift in reaction to attain equilibrium.
If $K$ is less than $Q$, then products convert into reactants which creates a reverse shift in reaction to attain equilibrium .
If $K$ is equal to $Q$, then the equilibrium is attained.
Now, by the calculated results above
$Q = 4$ and $K = 0.37$
 We conclude that $Q > K$ , therefore reaction shifts in reverse direction.
Hence, the correct answer is, ‘B. Reverse direction because $Q > {K_e}$’.

Note: Since $Q$ is the ratio of products upon reactants. If $Q$ is greater than one , products are greater hence reaction moves in reverse direction. If $Q$ is less than one then reactants are greater and the reaction moves in a forward direction. And if $Q$ is equals to zero then the reaction is at equilibrium.