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

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(C) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the change in the number of moles of gas is $\Delta n_g = 2 - (1 + 3) = -2$.Given $T = 300 + 273 = 573 \, K$,$K_c = 0.65$,and $R = 0.082 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.Substituting these values: $K_p = 0.65 \times (0.082 \times 573)^{-2}$.$K_p = 0.65 \times (46.986)^{-2} = 0.65 \times (1 / 2207.68) \approx 2.94 \times 10^{-4}$.

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

Medium

For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at $300 \, ^\circ C$,the value of $K_c$ is $0.65$. If $R = 0.082 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,then the value of $K_p$ will be:

A
$0.29 \times 10^{-4}$
B
$29 \times 10^{-4}$
C
$2.9 \times 10^{-4}$
D
$0.029 \times 10^{-4}$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

Explain why the partial pressure $p$ of a gas is proportional to its concentration $c$.

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In the thermal dissociation of $PCl_5$,the total pressure in the gaseous equilibrium mixture is $1.0 \ atm$ when half of $PCl_5$ is found to dissociate. The equilibrium constant of the reaction $(K_p)$ in atmosphere is

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For the reaction $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$. If the initial concentration of $[H_2] = [CO_2] = 1 \ M$ and $x \ mol/L$ of hydrogen is consumed at equilibrium,then the correct expression of $K_p$ is

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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:

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For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant is $10$. If the rate constant for the forward reaction is $203$,what will be the rate constant for the backward reaction?

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For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at $300 \, ^\circ C$,the value of $K_c$ is $0.65$. If $R = 0.082 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,then the value of $K_p$ will be:

Similar Questions

Explain why the partial pressure $p$ of a gas is proportional to its concentration $c$.

In the thermal dissociation of $PCl_5$,the total pressure in the gaseous equilibrium mixture is $1.0 \ atm$ when half of $PCl_5$ is found to dissociate. The equilibrium constant of the reaction $(K_p)$ in atmosphere is

For the reaction $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$. If the initial concentration of $[H_2] = [CO_2] = 1 \ M$ and $x \ mol/L$ of hydrogen is consumed at equilibrium,then the correct expression of $K_p$ is

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 $A + B \rightleftharpoons C + D$,the equilibrium constant is $10$. If the rate constant for the forward reaction is $203$,what will be the rate constant for the backward reaction?

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