(A) The equilibrium constant $K_P$ is given by: $K_P = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = \frac{4 \times 4}{2} = 8 \ atm$. The relationship

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(A) The equilibrium constant $K_P$ is given by: $K_P = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = \frac{4 \times 4}{2} = 8 \ atm$. The relationship between $K_P$ and $K_C$ is $K_P = K_C(RT)^{\Delta n_g}$. Here,$\Delta n_g = (1 + 1) - 1 = 1$. Temperature $T = 27 + 273 = 300 \ K$. $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$. $K_C = \frac{K_P}{(RT)^{\Delta n_g}} = \frac{8}{0.0821 \times 300} \approx \frac{8}{24.63} \approx 0.325 \approx \frac{1}{3} \ mol \ L^{-1}$.

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

Easy

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at equilibrium,the partial pressures of $PCl_5, PCl_3$ and $Cl_2$ are $2 \ atm, 4 \ atm$ and $4 \ atm$ respectively. The value of $K_C$ at $27^{\circ}C$ is (approx):-

A
$\frac{1}{3} \ mol \ L^{-1}$
B
$3 \ mol \ L^{-1}$
C
$\frac{2}{3} \ mol \ L^{-1}$
D
$\frac{1}{2} \ mol \ L^{-1}$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

For a given exothermic reaction,$K_p$ and $K'_p$ are the equilibrium constants at temperatures $T_1$ and $T_2,$ respectively. Assuming that the heat of reaction is constant in the temperature range between $T_1$ and $T_2,$ where $T_2 > T_1,$ it is readily observed that:

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For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. If the equilibrium pressure is $3X \ bar$,then $\Delta _r G^o$ would be:

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For the equilibrium reaction $CO + 2H_2 \rightleftharpoons CH_3OH$ at $427 \, ^\circ C$,the partial pressures of $CH_3OH$,$CO$,and $H_2$ are $2.0 \, atm$,$1.0 \, atm$,and $0.1 \, atm$ respectively. What is the value of $K_P$ for the decomposition of $CH_3OH$?

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The reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is started by taking $0.8 \ mol$ of $N_2O_4$ in a $1 \ L$ flask. If the equilibrium constant at $298 \ K$ is $0.00466 \ M$,the equilibrium concentration of $NO_2$ will be ........... $M$.

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The following equilibrium is established at $STP$. $B_{2(g)} \rightleftharpoons 2B_{(g)}$. Atoms of $B$ occupy $20 \%$ of total volume at $STP$. The total pressure of the system is $1 \ bar$. What is its $K_p$?

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For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at equilibrium,the partial pressures of $PCl_5, PCl_3$ and $Cl_2$ are $2 \ atm, 4 \ atm$ and $4 \ atm$ respectively. The value of $K_C$ at $27^{\circ}C$ is (approx):-

Similar Questions

For a given exothermic reaction,$K_p$ and $K'_p$ are the equilibrium constants at temperatures $T_1$ and $T_2,$ respectively. Assuming that the heat of reaction is constant in the temperature range between $T_1$ and $T_2,$ where $T_2 > T_1,$ it is readily observed that:

For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. If the equilibrium pressure is $3X \ bar$,then $\Delta _r G^o$ would be:

For the equilibrium reaction $CO + 2H_2 \rightleftharpoons CH_3OH$ at $427 \, ^\circ C$,the partial pressures of $CH_3OH$,$CO$,and $H_2$ are $2.0 \, atm$,$1.0 \, atm$,and $0.1 \, atm$ respectively. What is the value of $K_P$ for the decomposition of $CH_3OH$?

The reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is started by taking $0.8 \ mol$ of $N_2O_4$ in a $1 \ L$ flask. If the equilibrium constant at $298 \ K$ is $0.00466 \ M$,the equilibrium concentration of $NO_2$ will be ........... $M$.

The following equilibrium is established at $STP$. $B_{2(g)} \rightleftharpoons 2B_{(g)}$. Atoms of $B$ occupy $20 \%$ of total volume at $STP$. The total pressure of the system is $1 \ bar$. What is its $K_p$?

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