(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

In the reversible reaction $A + B \rightleftharpoons C + D$,the concentration of each $C$ and $D$ at equilibrium was $0.8 \ mol/L$. If the initial concentration of $A$ and $B$ was $1 \ mol/L$ each,then the equilibrium constant $K_c$ will be:

Medium

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For the reversible reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at $500 \ ^oC$,the value of $K_P$ is $1.44 \times 10^{-5}$ when partial pressure is measured in atmospheres. The corresponding value of $K_c$ with concentration in $\text{mol L}^{-1}$ is:

MediumIIT 2000

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In a $1 \ L$ closed container,$4 \ moles$ of $PCl_5$ are heated. If $80\%$ of $PCl_5$ remains undissociated at equilibrium,what is the value of the equilibrium constant $(K_c)$?

Medium

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At a definite temperature and $3 \ atm$ pressure,$75\%$ of $PCl_5$ decomposes into $PCl_3$ and $Cl_2$. Find $K_p$ for the reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ (in $atm$)

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For the reaction $N_2O_{4_{(g)}} \rightleftharpoons 2NO_{2_{(g)}}$ at $298 \ K$,the equilibrium constant $K_p$ is $0.14 \ atm$. Calculate the value of $K_c$. $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

Medium

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In the reversible reaction $A + B \rightleftharpoons C + D$,the concentration of each $C$ and $D$ at equilibrium was $0.8 \ mol/L$. If the initial concentration of $A$ and $B$ was $1 \ mol/L$ each,then the equilibrium constant $K_c$ will be:

For the reversible reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at $500 \ ^oC$,the value of $K_P$ is $1.44 \times 10^{-5}$ when partial pressure is measured in atmospheres. The corresponding value of $K_c$ with concentration in $\text{mol L}^{-1}$ is:

In a $1 \ L$ closed container,$4 \ moles$ of $PCl_5$ are heated. If $80\%$ of $PCl_5$ remains undissociated at equilibrium,what is the value of the equilibrium constant $(K_c)$?

At a definite temperature and $3 \ atm$ pressure,$75\%$ of $PCl_5$ decomposes into $PCl_3$ and $Cl_2$. Find $K_p$ for the reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ (in $atm$)

For the reaction $N_2O_{4_{(g)}} \rightleftharpoons 2NO_{2_{(g)}}$ at $298 \ K$,the equilibrium constant $K_p$ is $0.14 \ atm$. Calculate the value of $K_c$. $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

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