Question

For the reaction $a+b\rightleftharpoons c+d$, initial concentrations of a and b are equal and at equilibrium, the concentration of d will twice of that of a. What will be the equilibrium constant for the reaction?

• A. $2$
• B. $9$
• C. $4$
• D. $3$

Answer 1

Answer: (C)

$4$

$a+b\rightleftharpoons c+d$

Initial concentrations of 'a' and 'b' are equal let it be 'x'.

At equilibrium, the concentration will be twice that of 'a' i.e. 2x.

Since concentration of c and d are equal thus the concentration of c at equilibrium will be 2x.

$\therefore$ The equilibrium constant for the reaction $K_c=\frac{2x\times 2x}{x\times x}=\frac{x^2}{x^2}$

$\therefore K_c=4.$