The values of K_{P_1} and K_{P_2} for the rea | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For reaction $(1)$: $X \rightleftharpoons Y + Z$Initial moles: $1, 0, 0$Equilibrium moles: $(1-\alpha), \alpha, \alpha$. Total moles = $1+\alpha$.

Vedclass · Accepted Answer

$36 : 1$

Vedclass · Answer

(A) For reaction $(1)$: $X ightleftharpoons Y + Z$Initial moles: $1, 0, 0$Equilibrium moles: $(1-\alpha), \alpha, \alpha$. Total moles = $1+\alpha$.$P_X = \frac{1-\alpha}{1+\alpha} P_1, P_Y = \frac{\alpha}{1+\alpha} P_1, P_Z = \frac{\alpha}{1+\alpha} P_1$$K_{P_1} = \frac{P_Y P_Z}{P_X} = \frac{\alpha^2 P_1}{1-\alpha^2}$For reaction $(2)$: $A ightleftharpoons 2B$Initial moles: $1, 0$Equilibrium moles: $(1-\alpha), 2\alpha$. Total moles = $1+\alpha$.$P_A = \frac{1-\alpha}{1+\alpha} P_2, P_B = \frac{2\alpha}{1+\alpha} P_2$$K_{P_2} = \frac{P_B^2}{P_A} = \frac{(2\alpha)^2 P_2^2}{(1+\alpha)^2} 	imes \frac{(1+\alpha)}{(1-\alpha) P_2} = \frac{4\alpha^2 P_2}{1-\alpha^2}$Given $\frac{K_{P_1}}{K_{P_2}} = \frac{9}{1}$.$\frac{\alpha^2 P_1 / (1-\alpha^2)}{4\alpha^2 P_2 / (1-\alpha^2)} = \frac{P_1}{4P_2} = \frac{9}{1}$$\frac{P_1}{P_2} = \frac{36}{1}$.

The values of $K_{P_1}$ and $K_{P_2}$ for the reactions
$X \rightleftharpoons Y + Z$ --- $(1)$
$A \rightleftharpoons 2B$ --- $(2)$
are in the ratio of $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the total pressure at equilibrium for $(1)$ and $(2)$ are in the ratio:

Similar Questions

For the reaction $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$,$A$ is $33 \%$ dissociated at a total pressure $P$. The correct relation between $P$ and $K_{p}$ is

$A$ $1 \, M$ solution of glucose reaches dissociation equilibrium according to the equation $C_6H_{12}O_6 \rightleftharpoons 6HCHO$. What is the concentration of $HCHO$ at equilibrium if the equilibrium constant $K_c$ for the formation of glucose from formaldehyde is $6 \times 10^{22}$?

$3.00 \ mol$ of $PCl_5$ kept in $1 \ L$ closed reaction vessel was allowed to attain equilibrium at $380 \ K$. If $1.59 \ mol$ of reactant was converted into the product at equilibrium,then $K_c$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The values of $K_{P_1}$ and $K_{P_2}$ for the reactions$X \rightleftharpoons Y + Z$ --- $(1)$$A \rightleftharpoons 2B$ --- $(2)$are in the ratio of $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the total pressure at equilibrium for $(1)$ and $(2)$ are in the ratio: