(D) The given reaction is $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$.For this reaction,the equilibrium constant $K_{P1}$ is given by:$K_{P1} = \fr

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(D) The given reaction is $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$.For this reaction,the equilibrium constant $K_{P1}$ is given by:$K_{P1} = \frac{P_{CH_3OH}}{P_{CO} \cdot P_{H_2}^2} = \frac{2.0}{1.0 \times (0.1)^2} = \frac{2.0}{0.01} = 200 \ atm^{-2}$.The reaction for the decomposition of $CH_3OH$ is the reverse of the given reaction:$CH_3OH(g) \rightleftharpoons CO(g) + 2H_2(g)$.The equilibrium constant $K_P$ for this reverse reaction is the reciprocal of $K_{P1}$:$K_P = \frac{1}{K_{P1}} = \frac{1}{200} = 0.005 = 5 \times 10^{-3} \ atm^2$.

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

Medium

The partial pressures of $CH_3OH$,$CO$ and $H_2$ in the equilibrium mixture for the reaction $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$ at $427 \ ^oC$ are $2.0 \ atm$,$1.0 \ atm$ and $0.1 \ atm$ respectively. The value of $K_P$ for the decomposition of $CH_3OH$ to $CO$ and $H_2$ is:

A
$1 \times 10^2 \ atm^2$
B
$2 \times 10^2 \ atm^{-2}$
C
$50 \ atm^2$
D
$5 \times 10^{-3} \ atm^2$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_p = 41$ at $400 \ K$. Calculate $K_c$ for the following reactions at $400 \ K$:
$(a)$ $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$
$(b)$ $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$(c)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$

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$2X_{6(g)} \rightleftharpoons 4X_{3(g)}$
$A$ $20 \ L$ box contains $X_6$ and $X_3$ at equilibrium at $1500 \ K$. The equilibrium constant $K_p = 4 \times 10^{18} \ atm$. Assuming $P_{X_3} \gg P_{X_6}$ and the total pressure is $10 \ atm$,the partial pressure of $X_6$ will be:

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$A$ schematic plot of $\ln K_{eq}$ versus inverse of temperature $(1/T)$ for a reaction is shown below. The reaction must be

Easy

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For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is

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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?

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The partial pressures of $CH_3OH$,$CO$ and $H_2$ in the equilibrium mixture for the reaction $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$ at $427 \ ^oC$ are $2.0 \ atm$,$1.0 \ atm$ and $0.1 \ atm$ respectively. The value of $K_P$ for the decomposition of $CH_3OH$ to $CO$ and $H_2$ is:

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$2X_{6(g)} \rightleftharpoons 4X_{3(g)}$$A$ $20 \ L$ box contains $X_6$ and $X_3$ at equilibrium at $1500 \ K$. The equilibrium constant $K_p = 4 \times 10^{18} \ atm$. Assuming $P_{X_3} \gg P_{X_6}$ and the total pressure is $10 \ atm$,the partial pressure of $X_6$ will be:

$A$ schematic plot of $\ln K_{eq}$ versus inverse of temperature $(1/T)$ for a reaction is shown below. The reaction must be

For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is

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?

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$2X_{6(g)} \rightleftharpoons 4X_{3(g)}$
$A$ $20 \ L$ box contains $X_6$ and $X_3$ at equilibrium at $1500 \ K$. The equilibrium constant $K_p = 4 \times 10^{18} \ atm$. Assuming $P_{X_3} \gg P_{X_6}$ and the total pressure is $10 \ atm$,the partial pressure of $X_6$ will be: