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

• A. 6.4
• B. 0.64
• C. 1.6
• D. 16.0

Answer 1

The correct option is D

16.0

Suppose 1 mole of A and B each taken then 0.8 mole/litre of C and D each

formed remaining concentration of A and B will be $(1-0.8)$ = $0.2mole/litre$

each.

Equilibrium constant = $\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{0.8\times 0.8}{0.2\times 0.2}$ = $16$