Question

For the reaction,

$A\left(s\right)+B\left(g\right)+Heat\leftrightharpoons 2C\left(s\right)+2D\left(g\right)$

At equilibrium, if the pressure of $B$ is doubled, then to re-establish the equilibrium, the factor by which concentration of $D$ changes will be:

• A. $2$
• B. $\sqrt{2}$
• C. $3$
• D. $\sqrt{3}$

Answer 1

The correct option is B $\sqrt{2}$

$A\left(s\right)+B\left(g\right)+Heat\rightleftharpoons 2C\left(s\right)+2D\left(g\right)$

$K_p=\frac{(P_D{)}^2}{P_B}$ ..... equation 1

When the pressure of B is doubled,

$P_B^{\prime }=2P_B$

$K_p=\frac{(P_D^{\prime }{)}^2}{2P_B}$ ........ equation 2

On comparing equation-1 and equation-2, we get-

$P_D^{\prime }=\sqrt{2}{P}_D$

Hence, option B is correct.