(C) The reaction $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ is a reversible reaction occurring in a closed system. However,when the reaction is c

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(C) The reaction $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ is a reversible reaction occurring in a closed system. However,when the reaction is carried out in an open vessel,the gaseous product $CO_2$ escapes into the atmosphere. According to Le Chatelier's principle,the removal of a product from the system shifts the equilibrium in the forward direction to compensate for the loss. Consequently,the reaction proceeds to completion.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Le-Chaterlier principle and It’s application

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The reaction $CaCO_3 \rightleftharpoons CaO + CO_2$ goes to completion in an open vessel because:

A
$CaO$ does not react with $CO_2$ to give $CaCO_3$
B
The backward reaction is very slow
C
$CO_2$ formed escapes out
D
None of these

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6-1.Equilibrium (Chemical Equilibrium)

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Le-Chaterlier principle and It’s application

For the following reaction,which change will shift the equilibrium towards the product?
$I_{2(g)} \rightleftharpoons 2I_{(g)}; \Delta H_r^o(298 \ K) = +150 \ kJ$

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The equation for the reaction in the figure below is: $H_{2(g)} + I_{2(g)} + \text{heat} \rightleftharpoons 2HI_{(g)}$
At the instant $3 \text{ min}$,what change was imposed on the equilibrium?

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For the reaction $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}; \Delta H = -329 \, kJ$,which of the following will increase the amount of $ClF_{3(g)}$ in the equilibrium mixture?

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In the equilibrium $N_2 + 3H_2 \rightleftharpoons 2NH_3 + 22 \ kcal$,the formation of ammonia is favoured by

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For the chemical reaction: $BaO_{2(s)} \rightleftharpoons BaO_{(s)} + \frac{1}{2} O_{2(g)}$,$\Delta H = +ve$. At equilibrium,the pressure of $O_2$ depends upon:

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The reaction $CaCO_3 \rightleftharpoons CaO + CO_2$ goes to completion in an open vessel because:

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For the following reaction,which change will shift the equilibrium towards the product?$I_{2(g)} \rightleftharpoons 2I_{(g)}; \Delta H_r^o(298 \ K) = +150 \ kJ$

The equation for the reaction in the figure below is: $H_{2(g)} + I_{2(g)} + \text{heat} \rightleftharpoons 2HI_{(g)}$At the instant $3 \text{ min}$,what change was imposed on the equilibrium?

For the reaction $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}; \Delta H = -329 \, kJ$,which of the following will increase the amount of $ClF_{3(g)}$ in the equilibrium mixture?

In the equilibrium $N_2 + 3H_2 \rightleftharpoons 2NH_3 + 22 \ kcal$,the formation of ammonia is favoured by

For the chemical reaction: $BaO_{2(s)} \rightleftharpoons BaO_{(s)} + \frac{1}{2} O_{2(g)}$,$\Delta H = +ve$. At equilibrium,the pressure of $O_2$ depends upon:

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For the following reaction,which change will shift the equilibrium towards the product?
$I_{2(g)} \rightleftharpoons 2I_{(g)}; \Delta H_r^o(298 \ K) = +150 \ kJ$

The equation for the reaction in the figure below is: $H_{2(g)} + I_{2(g)} + \text{heat} \rightleftharpoons 2HI_{(g)}$
At the instant $3 \text{ min}$,what change was imposed on the equilibrium?