(D) For an elementary reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$,the equilibrium constant $K_c$ is the ratio of the forward rate cons

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(D) For an elementary reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$,the equilibrium constant $K_c$ is the ratio of the forward rate constant $k_f$ to the reverse rate constant $k_r$.Given $r_f = k_f [A_2][B_2] = 1.7 \times 10^{-18} [A_2][B_2]$,so $k_f = 1.7 \times 10^{-18}$.Given $r_r = k_r [AB]^2 = 2.4 \times 10^{-21} [AB]^2$,so $k_r = 2.4 \times 10^{-21}$.$K_c = \frac{k_f}{k_r} = \frac{1.7 \times 10^{-18}}{2.4 \times 10^{-21}} = \frac{1.7}{2.4} \times 10^3 \approx 0.708 \times 10^3 \approx 0.7 \times 10^3$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Kp and Kc Relationship

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For the elementary reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$,the rate of the forward reaction is given by $r_f = 1.7 \times 10^{-18} [A_2][B_2]$. If the rate of decomposition of gaseous $AB$ into $A_2$ and $B_2$ is given by $r_r = 2.4 \times 10^{-21} [AB]^2$,then the equilibrium constant for the formation of $AB$ from $A_2$ and $B_2$ will be ...

A
$1.4 \times 10^{-3}$
B
$2.8 \times 10^{-3}$
C
$1.4 \times 10^{2}$
D
$0.7 \times 10^{3}$

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1 Equilibrium: Chemical Equilibrium

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Kp and Kc Relationship

For the reactions $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ and $\frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{(g)}$,if the equilibrium constants are $K_1$ and $K_2$ respectively,then their relationship is:

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For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at equilibrium,the partial pressures of $PCl_5, PCl_3$ and $Cl_2$ are $2 \ atm, 4 \ atm$ and $4 \ atm$ respectively. The value of $K_C$ at $27^{\circ}C$ is (approx):-

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For the equilibrium,$2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$,the value of the equilibrium constant,$K_{c}$ is $3.75 \times 10^{-6}$ at $1069 \ K$. Calculate the $K_{p}$ for the reaction at this temperature?

Easy

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In the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the partial pressures of $PCl_3$,$Cl_2$,and $PCl_5$ are $0.3 \ atm$,$0.2 \ atm$,and $0.6 \ atm$ respectively. If the partial pressures of $PCl_3$ and $Cl_2$ are doubled,what will be the partial pressure of $PCl_5$ in $atm$?

Medium

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At $T(K)$,the following gaseous equilibrium is established: $W + X \rightleftharpoons Y + Z$. The initial concentration of $W$ is two times the initial concentration of $X$. The system is heated to $T(K)$ to establish equilibrium. At equilibrium,the concentration of $Y$ is four times the concentration of $X$. What is the value of $K_c$?

MediumAP EAMCET 2025

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For the reactions $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ and $\frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{(g)}$,if the equilibrium constants are $K_1$ and $K_2$ respectively,then their relationship is:

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at equilibrium,the partial pressures of $PCl_5, PCl_3$ and $Cl_2$ are $2 \ atm, 4 \ atm$ and $4 \ atm$ respectively. The value of $K_C$ at $27^{\circ}C$ is (approx):-

For the equilibrium,$2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$,the value of the equilibrium constant,$K_{c}$ is $3.75 \times 10^{-6}$ at $1069 \ K$. Calculate the $K_{p}$ for the reaction at this temperature?

In the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the partial pressures of $PCl_3$,$Cl_2$,and $PCl_5$ are $0.3 \ atm$,$0.2 \ atm$,and $0.6 \ atm$ respectively. If the partial pressures of $PCl_3$ and $Cl_2$ are doubled,what will be the partial pressure of $PCl_5$ in $atm$?

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