(D) The given reaction is a heterogeneous equilibrium: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$.For heterogeneous equilibria involving s

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(D) The given reaction is a heterogeneous equilibrium: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$.For heterogeneous equilibria involving solids and gases,the equilibrium constant $K_p$ is defined only by the partial pressure of the gaseous species.Thus,$K_p = P_{CO_2}$.Given that $K_p = 1.16 \ atm$,the partial pressure of $CO_2$ at equilibrium is $P_{CO_2} = 1.16 \ atm$.

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

Medium

For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the value of $K_p$ at $800 \ ^oC$ is $1.16 \ atm$. If $1 \ mol$ of $CaCO_{3(s)}$ is taken in a $1 \ L$ container to start the reaction,what will be the partial pressure of $CO_2$ at equilibrium in $atm$?

A
$0.29$
B
$0.58$
C
$0.73$
D
$1.16$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

For the decomposition of the compound,represented as $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,the $K_p = 2.9 \times 10^{-5} \ atm^3$. If the reaction is started with $1 \ mol$ of the compound,the total pressure at equilibrium would be............ $\times 10^{-2} \ atm$.

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At $700 \, K$,the equilibrium constant $K_p$ for the reaction $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ is $1.80 \times 10^{-3}$. The numerical value in $mol \, L^{-1}$ of $K_c$ for this reaction at the same temperature will be $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.

Medium

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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:

Medium

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For the reaction $NH_4COONH_2(s) \rightleftharpoons 2NH_3(g) + CO_2(g)$,the equilibrium pressure is $3 \ atm$. Calculate $K_p$.

Medium

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At $400 \ K$,in a $1.0 \ L$ vessel,$N_2O_4$ is allowed to attain equilibrium,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,the total pressure is $600 \ mm \ Hg$,when $20 \%$ of $N_2O_4$ is dissociated. The value of $K_p$ for the reaction is

DifficultAP EAMCET 2016

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For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the value of $K_p$ at $800 \ ^oC$ is $1.16 \ atm$. If $1 \ mol$ of $CaCO_{3(s)}$ is taken in a $1 \ L$ container to start the reaction,what will be the partial pressure of $CO_2$ at equilibrium in $atm$?

Similar Questions

For the decomposition of the compound,represented as $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,the $K_p = 2.9 \times 10^{-5} \ atm^3$. If the reaction is started with $1 \ mol$ of the compound,the total pressure at equilibrium would be............ $\times 10^{-2} \ atm$.

At $700 \, K$,the equilibrium constant $K_p$ for the reaction $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ is $1.80 \times 10^{-3}$. The numerical value in $mol \, L^{-1}$ of $K_c$ for this reaction at the same temperature will be $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.

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:

For the reaction $NH_4COONH_2(s) \rightleftharpoons 2NH_3(g) + CO_2(g)$,the equilibrium pressure is $3 \ atm$. Calculate $K_p$.

At $400 \ K$,in a $1.0 \ L$ vessel,$N_2O_4$ is allowed to attain equilibrium,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,the total pressure is $600 \ mm \ Hg$,when $20 \%$ of $N_2O_4$ is dissociated. The value of $K_p$ for the reaction is

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