(A) The reaction is: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$The equilibrium constant expression for this reaction is $K_p = p_{CO_2}$.G

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(A) The reaction is: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$The equilibrium constant expression for this reaction is $K_p = p_{CO_2}$.Given that the pressure of $CO_{2(g)}$ is $2.0 \times 10^5 \ Pa$.Since $1 \ bar = 10^5 \ Pa$,the pressure in bar is $2.0 \ bar$.Therefore,$K_p = 2.0$.

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

Medium

At $1100 \ K$ temperature,$CaCO_{3(s)}$ and $CaO_{(s)}$ are in equilibrium. The pressure of $CO_{2(g)}$ is $2.0 \times 10^5 \ Pa$.
Find the equilibrium constant $(K_p)$ for the reaction: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

A
$2.0$
B
$2.0 \times 10^5$
C
$1.0$
D
$0.5$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant at temperature $T$ is $4 \times 10^{-4}$. Find the value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$.

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For the reaction $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$,the relationship is given by $K_P = K_C(RT)^x$. Find the value of $x$.

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For the reaction $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the partial pressures of $CO$ and $CO_2$ at equilibrium are $2.0 \ atm$ and $4.0 \ atm$ respectively. The value of $K_p$ for the reaction is:

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One mole of $N_2O_4$ in a $1 \ L$ flask decomposes to attain the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At the equilibrium the mole fraction of $NO_2$ is $1/2$. Hence $K_C$ will be:

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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:

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At $1100 \ K$ temperature,$CaCO_{3(s)}$ and $CaO_{(s)}$ are in equilibrium. The pressure of $CO_{2(g)}$ is $2.0 \times 10^5 \ Pa$.Find the equilibrium constant $(K_p)$ for the reaction: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

Similar Questions

For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant at temperature $T$ is $4 \times 10^{-4}$. Find the value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$.

For the reaction $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$,the relationship is given by $K_P = K_C(RT)^x$. Find the value of $x$.

For the reaction $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the partial pressures of $CO$ and $CO_2$ at equilibrium are $2.0 \ atm$ and $4.0 \ atm$ respectively. The value of $K_p$ for the reaction is:

One mole of $N_2O_4$ in a $1 \ L$ flask decomposes to attain the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At the equilibrium the mole fraction of $NO_2$ is $1/2$. Hence $K_C$ will be:

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:

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At $1100 \ K$ temperature,$CaCO_{3(s)}$ and $CaO_{(s)}$ are in equilibrium. The pressure of $CO_{2(g)}$ is $2.0 \times 10^5 \ Pa$.
Find the equilibrium constant $(K_p)$ for the reaction: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$