(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

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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: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

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

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In the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,if at equilibrium there are $2 \ mol$ each of $PCl_5$,$PCl_3$,and $Cl_2$,and the total pressure is $3 \ atm$,then the value of the equilibrium constant $K_p$ is ....... $atm$.

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The plot of $log_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where $K$ is equilibrium constant).

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For the gaseous reaction $aA + bB \rightleftharpoons cC + dD$,the relation between $K_p$ and $K_c$ 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:

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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: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

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

In the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,if at equilibrium there are $2 \ mol$ each of $PCl_5$,$PCl_3$,and $Cl_2$,and the total pressure is $3 \ atm$,then the value of the equilibrium constant $K_p$ is ....... $atm$.

The plot of $log_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where $K$ is equilibrium constant).

For the gaseous reaction $aA + bB \rightleftharpoons cC + dD$,the relation between $K_p$ and $K_c$ is:

$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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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: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

$2X_{6(g)} \rightleftharpoons 4X_{3(g)}$
$A$ $20 \ L$ 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: