(D) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g

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$\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}$

(D) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$, the change in moles of gas is $\Delta n = n_p - n_r = 2 - (1 + 3) = -2$.The temperature $T = 500 + 273 = 773 \text{ K}$.The gas constant $R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$.Substituting these values into the formula: $K_p = K_c(RT)^{-2}$.Rearranging for $K_c$, we get $K_c = \frac{K_p}{(RT)^{-2}}$.Substituting the given $K_p$ value: $K_c = \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}$.

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

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

KCET 2026 All KCET

A
$\frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}}$
B
$\frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}}$
C
$\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{2}}$
D
$\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}$

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1 Equilibrium: Chemical Equilibrium

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Kp and Kc Relationship

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In a chemical equilibrium,the rate constant of the backward reaction is $7.5 \times 10^{-4}$ and the equilibrium constant is $1.5$. The rate constant of the forward reaction is:

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For the reaction $A_{(g)} \rightleftharpoons 2 B_{(g)}$,the backward reaction rate constant is higher than the forward reaction rate constant by a factor of $2500$,at $1000 \ K$. [Given : $R = 0.0831 \ L \ atm \ mol^{-1} \ K^{-1}$] $K_p$ for the reaction at $1000 \ K$ is

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Given $K_p$ for the reaction $\frac{1}{2} C_{(g)} \rightleftharpoons \frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)}$ at a fixed temperature is $0.25 \ atm$. Find the $K_p$ for the reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$ at the same temperature.

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One mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:

EasyAP EAMCET 2023

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$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$
$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$
Then the equilibrium constant for the following reaction is ....
$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$

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

Similar Questions

In a chemical equilibrium,the rate constant of the backward reaction is $7.5 \times 10^{-4}$ and the equilibrium constant is $1.5$. The rate constant of the forward reaction is:

For the reaction $A_{(g)} \rightleftharpoons 2 B_{(g)}$,the backward reaction rate constant is higher than the forward reaction rate constant by a factor of $2500$,at $1000 \ K$. [Given : $R = 0.0831 \ L \ atm \ mol^{-1} \ K^{-1}$] $K_p$ for the reaction at $1000 \ K$ is

Given $K_p$ for the reaction $\frac{1}{2} C_{(g)} \rightleftharpoons \frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)}$ at a fixed temperature is $0.25 \ atm$. Find the $K_p$ for the reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$ at the same temperature.

$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$Then the equilibrium constant for the following reaction is ....$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$

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$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$
$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$
Then the equilibrium constant for the following reaction is ....
$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$