(A) Given reactions:$(i) \, CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \quad K_1 = 67$$(ii) \, CoO_{2(g)} + CO_{(g)} \rightleftha

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(A) Given reactions:$(i) \, CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \quad K_1 = 67$$(ii) \, CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \quad K_2 = 490$Target reaction:$(iii) \, CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \quad K = ?$To obtain reaction $(iii)$,subtract reaction $(ii)$ from reaction $(i)$:$(iii) = (i) - (ii)$Therefore,the equilibrium constant $K$ is given by:$K = \frac{K_1}{K_2} = \frac{67}{490} \approx 0.137$

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

Medium

$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$
$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$
Then the equilibrium constant for the following reaction is ....
$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$

A
$0.137$
B
$423$
C
$7.31$
D
$557$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

If $K_C$ for the equilibrium reaction $2 ABC_{(g)} \rightleftharpoons 2 AB_{(g)} + C_{2(g)}$ is $X$ at $T \ K$,its $K_P$ at the same temperature is:

EasyAP EAMCET 2017

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For the reactions:
$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$
$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$
Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to:

Difficult

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For the chemical equilibrium,$CaCO_{3_{(s)}} \rightleftharpoons CaO_{(s)} + CO_{2_{(g)}}$,$\Delta H_{r}^{\circ}$ can be determined from which one of the following plots?

DifficultAIIMS 2005

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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 $atm$?

Medium

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When $4.5 \, \text{mol}$ each of hydrogen and iodine are heated in a $10 \, \text{L}$ closed vessel,$3 \, \text{mol}$ of $HI$ are formed at equilibrium. The equilibrium constant $K_c$ for the reaction ${H_2} + {I_2} \rightleftharpoons 2HI$ is .........

Medium

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$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$Then the equilibrium constant for the following reaction is ....$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$

Similar Questions

If $K_C$ for the equilibrium reaction $2 ABC_{(g)} \rightleftharpoons 2 AB_{(g)} + C_{2(g)}$ is $X$ at $T \ K$,its $K_P$ at the same temperature is:

For the reactions:$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to:

For the chemical equilibrium,$CaCO_{3_{(s)}} \rightleftharpoons CaO_{(s)} + CO_{2_{(g)}}$,$\Delta H_{r}^{\circ}$ can be determined from which one of the following plots?

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 $atm$?

When $4.5 \, \text{mol}$ each of hydrogen and iodine are heated in a $10 \, \text{L}$ closed vessel,$3 \, \text{mol}$ of $HI$ are formed at equilibrium. The equilibrium constant $K_c$ for the reaction ${H_2} + {I_2} \rightleftharpoons 2HI$ is .........

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$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$
$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$
Then the equilibrium constant for the following reaction is ....
$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$

For the reactions:
$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$
$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$
Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to: