(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

In which of the following equilibria,the value of $K_p$ is less than $K_c$?

Medium

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Find out the value of $K_{c}$ for each of the following equilibria from the value of $K_{p}$:
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$; $K_{p} = 1.8 \times 10^{-2}$ at $500 \ K$
$(ii)$ $CaCO_{3(s)} \longleftrightarrow CaO_{(s)} + CO_{2(g)}$; $K_{p} = 167$ at $1073 \ K$

Medium

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$x A_{(s)} \rightleftharpoons y B_{(g)} + z C_{(g)}$. If $\frac{K_c}{K_p} = (RT)^{-2}$,then which is correct?

Medium

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Two gaseous equilibria $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$ and $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ have equilibrium constants $K_1$ and $K_2$ respectively at $298 \ K$. Which of the following relationships between $K_1$ and $K_2$ is correct?

MediumAIPMT 1989

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For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the value of $K_p$ at $800 \ ^oC$ is $1.16 \ atm$. If $1 \ mol$ of $CaCO_{3(s)}$ is taken in a $1 \ L$ container to start the reaction,what will be the partial pressure of $CO_2$ at equilibrium in $atm$?

Medium

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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

In which of the following equilibria,the value of $K_p$ is less than $K_c$?

Find out the value of $K_{c}$ for each of the following equilibria from the value of $K_{p}$:$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$; $K_{p} = 1.8 \times 10^{-2}$ at $500 \ K$$(ii)$ $CaCO_{3(s)} \longleftrightarrow CaO_{(s)} + CO_{2(g)}$; $K_{p} = 167$ at $1073 \ K$

$x A_{(s)} \rightleftharpoons y B_{(g)} + z C_{(g)}$. If $\frac{K_c}{K_p} = (RT)^{-2}$,then which is correct?

Two gaseous equilibria $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$ and $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ have equilibrium constants $K_1$ and $K_2$ respectively at $298 \ K$. Which of the following relationships between $K_1$ and $K_2$ is correct?

For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the value of $K_p$ at $800 \ ^oC$ is $1.16 \ atm$. If $1 \ mol$ of $CaCO_{3(s)}$ is taken in a $1 \ L$ container to start the reaction,what will be the partial pressure of $CO_2$ at equilibrium in $atm$?

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Find out the value of $K_{c}$ for each of the following equilibria from the value of $K_{p}$:
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$; $K_{p} = 1.8 \times 10^{-2}$ at $500 \ K$
$(ii)$ $CaCO_{3(s)} \longleftrightarrow CaO_{(s)} + CO_{2(g)}$; $K_{p} = 167$ at $1073 \ K$