(B) The relation between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n_g}$.Here,$\Delta n_g = (2+1) - 0 = 3$.Given $K_p = 3.2 \times 10^2\ bar^3$. Since

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(B) The relation between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n_g}$.Here,$\Delta n_g = (2+1) - 0 = 3$.Given $K_p = 3.2 \times 10^2\ bar^3$. Since $1\ bar \approx 0.987\ atm$,$K_p$ in $atm^3$ is $3.2 \times 10^2 \times (0.987)^3 \approx 307.5\ atm^3$.$T = 600 + 273 = 873\ K$.$K_c = \frac{K_p}{(RT)^{\Delta n_g}} = \frac{307.5}{(0.082 \times 873)^3}$.$K_c = \frac{307.5}{(71.586)^3} \approx \frac{307.5}{367000} \approx 8.37 \times 10^{-4}$.

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

Medium

At $600\ ^{\circ}C$,$NH_4COONH_2\ (s) \rightleftharpoons 2NH_3\ (g) + CO_2\ (g)$ has an equilibrium constant $K_p = 3.2 \times 10^2\ bar^3$. Calculate $K_c$. $\left( R = 0.082\ L\ atm\ K^{-1}\ mol^{-1} \right)$

A
$1.2 \times 10^{-5}$
B
$1.5 \times 10^{-4}$
C
$2.1 \times 10^{-6}$
D
$3.4 \times 10^{-5}$

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Class 11

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Chapter

1 Equilibrium: Chemical Equilibrium

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Kp and Kc Relationship

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,$1 mol$ of $PCl_5$ is taken at $5 atm$ pressure. If $50\%$ of $PCl_5$ dissociates at equilibrium,calculate $K_p$.

Medium

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In a closed container of $1000\, cm^3$,$2\, mol$ of $PCl_5$,$2\, mol$ of $PCl_3$,and $3\, mol$ of $Cl_2$ are found to be at equilibrium at $27\, ^oC$. Then $K_P$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at $27\, ^oC$ is $.....$ $atm$.

Medium

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For the reaction $AB_{(g)} \rightleftharpoons A_{(g)} + B_{(g)}$,$AB$ is $33.3\%$ dissociated at a total equilibrium pressure of $P$. Therefore,$P$ is correctly related to $K_p$ by which of the following options?

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The value of $K_{p}$ for the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is $3.0$ at $1000 \ K$. If initially $P_{CO_{2}} = 0.48 \ bar$ and $P_{CO} = 0 \ bar$ and pure graphite is present,calculate the equilibrium partial pressures of $CO$ and $CO_{2}$.

Medium

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In a $500 \,mL$ flask,the degree of dissociation of $PCl_5$ at equilibrium is $40 \%$ and the initial amount is $5 \,moles$. The value of equilibrium constant in $mol \,L^{-1}$ for the decomposition of $PCl_5$ is

DifficultTS EAMCET 2008

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At $600\ ^{\circ}C$,$NH_4COONH_2\ (s) \rightleftharpoons 2NH_3\ (g) + CO_2\ (g)$ has an equilibrium constant $K_p = 3.2 \times 10^2\ bar^3$. Calculate $K_c$. $\left( R = 0.082\ L\ atm\ K^{-1}\ mol^{-1} \right)$

Similar Questions

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,$1 mol$ of $PCl_5$ is taken at $5 atm$ pressure. If $50\%$ of $PCl_5$ dissociates at equilibrium,calculate $K_p$.

In a closed container of $1000\, cm^3$,$2\, mol$ of $PCl_5$,$2\, mol$ of $PCl_3$,and $3\, mol$ of $Cl_2$ are found to be at equilibrium at $27\, ^oC$. Then $K_P$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at $27\, ^oC$ is $.....$ $atm$.

For the reaction $AB_{(g)} \rightleftharpoons A_{(g)} + B_{(g)}$,$AB$ is $33.3\%$ dissociated at a total equilibrium pressure of $P$. Therefore,$P$ is correctly related to $K_p$ by which of the following options?

The value of $K_{p}$ for the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is $3.0$ at $1000 \ K$. If initially $P_{CO_{2}} = 0.48 \ bar$ and $P_{CO} = 0 \ bar$ and pure graphite is present,calculate the equilibrium partial pressures of $CO$ and $CO_{2}$.

In a $500 \,mL$ flask,the degree of dissociation of $PCl_5$ at equilibrium is $40 \%$ and the initial amount is $5 \,moles$. The value of equilibrium constant in $mol \,L^{-1}$ for the decomposition of $PCl_5$ is

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