Question

Equlibrium constant of a reaction h2 +i2 =2hi is 64 if the volume of container is reduced to one half of its original value the value of equilibrium constant

Answer 1

Answer:

The value of new equilibrium constant will be 64.

Explanation:

$H_2+I_2\rightleftharpoons 2HI$

Equilibrium constant of the reaction  =$K_c=64$

$K_c=64=\frac{[HI]^2}{[H_2][I_2]}$

$[concentration]=\frac{n}{v}$

$64=\frac{(\frac{n_{HI}}{V})^2}{\frac{n_{H_2}}{V}\times \frac{n_{I_2}}{V}}$

On reducing the volume of container = V' = 0.5 V

The new equilibrium constant is given by:

$K_c'=\frac{[\frac{n_{HI}}{0.5V}]^2}{[\frac{n_{H_2}}{0.5V}]\times [\frac{n_{I_2}}{0.5V}]}$

$K_c'=1\times K_c=1\times 64$

The value of new equilibrium constant will be 64.

Answer 2

Answer:

64 ....plz support me guyz..