Question

At a temperature T, a compound $A{B}_4\left(g\right)$ dissociates as $2A{B}_4\rightleftharpoons A_2\left(g\right)+4B_2\left(g\right)$ with a degree of dissociation x, which is small compared with unity. The expression of $K_p$ in terms of x and total pressure P is?

• A. $8P^3{x}^5$
• B. $256P^3{x}^5$
• C. $4P{x}^2$
• D. None of these

Answer 1

Answer: (A)

$K_p$ is the pressure constant of equilibrium and its formula is same as that of equilibrium constant, only in the place of concentration, partial pressure of reactants are used.

$2A{B}_4\leftrightharpoons A_2\left(g\right)+4B_2\left(g\right)$

$t=0$ $1$ $0$ $0$

$t=equilibrium$ $1-2\alpha$ $\alpha$ $4\alpha$

where,$2\alpha =x$

Degree of dissociation is x.(small compared to unity)

Total pressure is P.

$K_p=\frac{P.P_{A_2}\times (P.P_{B_2}{)}^4}{(P.P_{A{B}_4}{)}^2}$

Total number of moles is equal to $1+\frac{3x}{2}$

Here $x$ is very very small than 1.

$K_p=\frac{\frac{x}{2}\times P\times (2x{)}^4\times P^4}{(1-x{)}^2\times P^2}$

$K_p=8P^3{x}^5$

Hence, option (A) is correct.