(A) For a reversible reaction at equilibrium,the equilibrium constant $K_c$ is defined as the ratio of the rate constant of the forward reaction ($k_f

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(A) For a reversible reaction at equilibrium,the equilibrium constant $K_c$ is defined as the ratio of the rate constant of the forward reaction ($k_f$ or $k_1$) to the rate constant of the backward reaction ($k_b$ or $k_2$).Given:$k_f = 2.1 \times 10^{-3} \ s^{-1}$$k_b = 4.2 \times 10^{-4} \ s^{-1}$Using the formula:$K_c = \frac{k_f}{k_b}$Substituting the values:$K_c = \frac{2.1 \times 10^{-3}}{4.2 \times 10^{-4}}$$K_c = \frac{21 \times 10^{-4}}{4.2 \times 10^{-4}}$$K_c = \frac{21}{4.2} = 5.0$

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

Medium

For the reversible reaction in equilibrium
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
If the rate constant for the forward reaction is $k_1 = 2.1 \times 10^{-3} \ s^{-1}$ and for the backward reaction is $k_2 = 4.2 \times 10^{-4} \ s^{-1}$,then the equilibrium constant $K_c$ for the above reaction is:

A
$5.0$
B
$2.0$
C
$0.5$
D
$2.5$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

For the reaction $CO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons CO_{2(g)}$,the value of $K_P/K_C$ is

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The value of $K_P / K_C$ for the reaction at $T(K)$ is:
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$

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For the reaction $A + B \rightleftharpoons C + D$,the initial concentrations of $A$ and $B$ are equal. If the equilibrium concentration of $D$ is twice the equilibrium concentration of $A$,then the equilibrium constant $K_c$ is:

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For the following given equilibrium reaction,$\frac{K_{c}}{K_{p}}$ is equal to $1076$ at $T \ K$. What is the value of $T$ (in $K$)? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

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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:

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For the reaction $CO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons CO_{2(g)}$,the value of $K_P/K_C$ is

The value of $K_P / K_C$ for the reaction at $T(K)$ is:$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$

For the reaction $A + B \rightleftharpoons C + D$,the initial concentrations of $A$ and $B$ are equal. If the equilibrium concentration of $D$ is twice the equilibrium concentration of $A$,then the equilibrium constant $K_c$ is:

For the following given equilibrium reaction,$\frac{K_{c}}{K_{p}}$ is equal to $1076$ at $T \ K$. What is the value of $T$ (in $K$)? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

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:

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The value of $K_P / K_C$ for the reaction at $T(K)$ is:
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$

For the following given equilibrium reaction,$\frac{K_{c}}{K_{p}}$ is equal to $1076$ at $T \ K$. What is the value of $T$ (in $K$)? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$