For the reactions (1) and (2) :A rightlefthar | vedclass

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

Question

(B) For reaction $(1)$: $A \rightleftharpoons B + C$Initial moles: $1, 0, 0$Equilibrium moles: $(1 - \alpha), \alpha, \alpha$Total moles at equilibriu

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) For reaction $(1)$: $A \rightleftharpoons B + C$Initial moles: $1, 0, 0$Equilibrium moles: $(1 - \alpha), \alpha, \alpha$Total moles at equilibrium: $1 + \alpha$$K_{P_1} = \frac{p_B \cdot p_C}{p_A} = \frac{(\frac{\alpha}{1+\alpha} P_1)(\frac{\alpha}{1+\alpha} P_1)}{\frac{1-\alpha}{1+\alpha} P_1} = \frac{\alpha^2}{1-\alpha^2} P_1$For reaction $(2)$: $D \rightleftharpoons 2E$Initial moles: $1, 0$Equilibrium moles: $(1 - \alpha), 2\alpha$Total moles at equilibrium: $1 + \alpha$$K_{P_2} = \frac{p_E^2}{p_D} = \frac{(\frac{2\alpha}{1+\alpha} P_2)^2}{\frac{1-\alpha}{1+\alpha} P_2} = \frac{4\alpha^2}{1-\alpha^2} P_2$Given $\frac{K_{P_1}}{K_{P_2}} = \frac{9}{1}$.Substituting the expressions: $\frac{\frac{\alpha^2}{1-\alpha^2} P_1}{\frac{4\alpha^2}{1-\alpha^2} P_2} = \frac{P_1}{4P_2} = 9$Therefore,$\frac{P_1}{P_2} = 36 : 1$.

Similar Questions

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,the equilibrium constant $K_c = 1.8 \times 10^{-6}$ at $184 \, ^\circ C$. Comparing $K_p$ and $K_c$ at $184 \, ^\circ C$,we find that:

At $T(K)$,$K_{c}$ value for $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$ is $16$. In a closed $1 \ L$ flask,one mole each of $AO_2, BO_2, AO_3$ and $BO$ are taken and heated to $T(K)$. What is the concentration (in $mol \ L^{-1}$) of $AO_3$ at equilibrium?

For the reaction system $A_{(g)} \rightleftharpoons B_{(g)} + 2C_{(g)}$,if the equilibrium concentration of $B$ is doubled,the equilibrium concentration of $C$ will become ....

$K_{c}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $39.0$. In a closed one litre flask,one mole of $A_{2(g)}$ was heated to $T \ K$. What are the concentrations of $A_{2(g)}$ and $B_{2(g)}$ (in $mol \ L^{-1}$) respectively at equilibrium?

Vedclass Test Series

Exam Paper Generator

Online Exam Module