(A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.Concentrations at equilibrium are:$[H_2] = \frac{0.8 \ mol}{10 \ L} = 0.08 \ M$

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(A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.Concentrations at equilibrium are:$[H_2] = \frac{0.8 \ mol}{10 \ L} = 0.08 \ M$$[I_2] = \frac{0.8 \ mol}{10 \ L} = 0.08 \ M$$[HI] = \frac{2.4 \ mol}{10 \ L} = 0.24 \ M$The equilibrium constant $K_c$ is given by:$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.24)^2}{(0.08)(0.08)} = \frac{0.0576}{0.0064} = 9$.For the reverse reaction $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$,the equilibrium constant $K_c'$ is:$K_c' = \frac{1}{K_c} = \frac{1}{9} \approx 0.111$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

Medium

In the synthesis of $HI$,the amounts of $H_{2(g)}$,$I_{2(g)}$,and $HI_{(g)}$ at equilibrium were found to be $0.8 \ mol$,$0.8 \ mol$,and $2.4 \ mol$ respectively in a $10 \ L$ vessel. Calculate the equilibrium constant $(K_c)$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ and the equilibrium constant for the reverse reaction.

A
$K_c = 9, K_c' = 0.11$
B
$K_c = 0.11, K_c' = 9$
C
$K_c = 9, K_c' = 0.09$
D
$K_c = 0.09, K_c' = 9$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

For the reaction $P + Q \rightleftharpoons R + C$,the equilibrium constant $K_c$ is $10^{-2}$ and the forward rate constant $K_f$ is $10^{-1}$. The rate constant for the backward reaction $(K_b)$ will be:

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The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

MediumJEE MAIN 2021

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Equilibrium constant for the reaction $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$ is $81$. If the velocity constant of the forward reaction is $162 \ L \ mol^{-1} \ s^{-1}$,what is the velocity constant (in $L \ mol^{-1} \ s^{-1}$) for the backward reaction?

MediumTS EAMCET 2001

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At $444 \, ^\circ C$,the equilibrium constant for the reaction $HI \rightleftharpoons 1/2 H_2 + 1/2 I_2$ is $64$. What will be the equilibrium constant for the reaction $H_2 + I_2 \rightleftharpoons 2HI$?

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For the reaction $P + Q \rightleftharpoons R + C$,the equilibrium constant $K_c$ is $10^{-2}$ and the forward rate constant $K_f$ is $10^{-1}$. The rate constant for the backward reaction $(K_b)$ will be:

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$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

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At $444 \, ^\circ C$,the equilibrium constant for the reaction $HI \rightleftharpoons 1/2 H_2 + 1/2 I_2$ is $64$. What will be the equilibrium constant for the reaction $H_2 + I_2 \rightleftharpoons 2HI$?

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$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).