Question

For the reaction $P+Q\rightleftharpoons R+S$, initially the concentration of P is equal to that of Q ($1$ molar) but at equilibrium, the concentration of R will be twice of that of P, then the equilibrium constant of the reaction is?

• A. $4/3$
• B. $4$
• C. $3/10$
• D. $8$

Answer 1

The correct option is B $4$

At equilibrium, the concentration of R is twice of that of P.

$P+Q\rightleftharpoons R+S$

Initial Conc. $1$ $1$ $0$ $0$

Final Conc. At equilibrium $1-x$ $1-x$ $2(1-x)$ $2(1-x)$

Now, $K_c$ for the given reaction will be:

$K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$

Substituting the values, we get:

$K_c=\frac{2(1-x)\times 2(1-x)}{(1-x)(1-x)}$

On simplifying the above, we get the value of $K_c=4$

Hence the correct option will be B.