Question

In the reversible reaction $A+B\rightleftharpoons C+D$, the concentration of each $C$ and $D$ at equilibrium was $0.8$ mole/litre, then the equilibrium constant $K_c$ will be (if initial concentration of $A$ and $B$ be will $1M$ each)-

• A. $6.4$
• B. $0.64$
• C. $1.6$
• D. $16.0$

Answer 1

The correct option is D $16.0$

The initial concentrations of A, B, C and D are 1M, 1M, 0M and 0M respectively.
The equilibrium concentrations of A, B, C and D are $1-0.8=0.2M$,$1-0.8=0.2M$, 0.8M and 0.8M respectively.
The expression for the equilibrium constant is $K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{0.8\times 0.8}{0.2\times 0.2}=\mathrm{16.0.}$