(B) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$For the reaction $CO_{2(g)} + C_{(s)} \rightlefthar

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(B) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$For the reaction $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - 1 = 1$.Given $K_p = 3.0$,$T = 1000 \ K$,and $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$.Substituting these values into the equation: $3.0 = K_c(0.083 \times 1000)^1$$3.0 = K_c(83)$$K_c = \frac{3.0}{83} \approx 3.6 \times 10^{-2}$

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

Medium

$K_p$ for the following reaction is $3.0$ at $1000 \ K$.
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$
What will be the value of $K_c$ for the reaction at the same temperature?
(Given: $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)

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A
$0.36$
B
$3.6 \times 10^{-2}$
C
$3.6 \times 10^{-3}$
D
$3.6$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

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For which of the following reactions is $K_p = K_c$?

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For which of the following equilibria are $K_P$ and $K_C$ different?

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$K_p$ for the following reaction is $3.0$ at $1000 \ K$.$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$What will be the value of $K_c$ for the reaction at the same temperature?(Given: $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)

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For which of the following reactions is $K_p = K_c$?

For the reaction,$PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the value of $K_c$ at $250 \ ^oC$ is $26$. The value of $K_p$ at this temperature will be:

At $300 \ K$,$K_C$ for the reaction $A_2B_{2(g)} \rightleftharpoons A_{2(g)} + B_{2(g)}$ is $100 \ mol \ L^{-1}$. What is its $K_p$ (in $atm$) at the same temperature? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

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$K_p$ for the following reaction is $3.0$ at $1000 \ K$.
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$
What will be the value of $K_c$ for the reaction at the same temperature?
(Given: $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)