Medium
Does the number of moles of reaction products increase,decrease,or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
$(a) \quad PCl_{5(g)} \longleftrightarrow PCl_{3(g)} + Cl_{2(g)}$
$(b) \quad CaO_{(s)} + CO_{2(g)} \longleftrightarrow CaCO_{3(s)}$
$(c) \quad 3Fe_{(s)} + 4H_2O_{(g)} \longleftrightarrow Fe_3O_{4(s)} + 4H_{2(g)}$
$(a) \quad PCl_{5(g)} \longleftrightarrow PCl_{3(g)} + Cl_{2(g)}$
$(b) \quad CaO_{(s)} + CO_{2(g)} \longleftrightarrow CaCO_{3(s)}$
$(c) \quad 3Fe_{(s)} + 4H_2O_{(g)} \longleftrightarrow Fe_3O_{4(s)} + 4H_{2(g)}$
(N/A) According to Le Chatelier's principle,decreasing the pressure (by increasing the volume) shifts the equilibrium toward the side with a greater number of moles of gaseous species.
$(a)$ $PCl_{5(g)} \longleftrightarrow PCl_{3(g)} + Cl_{2(g)}$: There are $1$ mole of gas on the reactant side and $2$ moles of gas on the product side. Since the product side has more moles,the equilibrium shifts forward,and the number of moles of products increases.
$(b)$ $CaO_{(s)} + CO_{2(g)} \longleftrightarrow CaCO_{3(s)}$: There is $1$ mole of gas on the reactant side and $0$ moles of gas on the product side. Since the reactant side has more moles,the equilibrium shifts backward,and the number of moles of products decreases.
$(c)$ $3Fe_{(s)} + 4H_2O_{(g)} \longleftrightarrow Fe_3O_{4(s)} + 4H_{2(g)}$: There are $4$ moles of gas on the reactant side and $4$ moles of gas on the product side. Since the number of moles of gas is equal on both sides,the equilibrium position remains unchanged,and the number of moles of products remains the same.
$(a)$ $PCl_{5(g)} \longleftrightarrow PCl_{3(g)} + Cl_{2(g)}$: There are $1$ mole of gas on the reactant side and $2$ moles of gas on the product side. Since the product side has more moles,the equilibrium shifts forward,and the number of moles of products increases.
$(b)$ $CaO_{(s)} + CO_{2(g)} \longleftrightarrow CaCO_{3(s)}$: There is $1$ mole of gas on the reactant side and $0$ moles of gas on the product side. Since the reactant side has more moles,the equilibrium shifts backward,and the number of moles of products decreases.
$(c)$ $3Fe_{(s)} + 4H_2O_{(g)} \longleftrightarrow Fe_3O_{4(s)} + 4H_{2(g)}$: There are $4$ moles of gas on the reactant side and $4$ moles of gas on the product side. Since the number of moles of gas is equal on both sides,the equilibrium position remains unchanged,and the number of moles of products remains the same.
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Similar Questions
Suppose the reaction $PCl_{5(s)} \rightleftharpoons PCl_{3(s)} + Cl_{2(g)}$ is in a closed vessel at equilibrium. What is the effect on the equilibrium concentration of $Cl_{2(g)}$ by adding $PCl_{5(s)}$ at constant temperature?
Easy
View SolutionOn addition of an inert gas at constant volume to the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ at equilibrium,
Medium
View SolutionThe reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is exothermic and reversible. $A$ mixture of $N_{2(g)}$,$H_{2(g)}$,and $NH_{3(g)}$ is at equilibrium in a closed container. When a certain quantity of extra $H_{2(g)}$ is introduced into the container,while keeping the volume constant,then which statement among the following is true?
Match List-$I$ (Equilibrium) with List-$II$ (Conditions for the process) and select the correct answer from the options given below.
List-$I$ (Equilibrium) List-$II$ (Conditions for the process) $P. A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ (Endothermic) $1. \text{High temperature}$ $Q. 2AB_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ (Exothermic) $2. \text{Low temperature}$ $R. 2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$ (Endothermic) $3. \text{High temperature}$ $4. \text{Low temperature}$ $5. \text{Independent of pressure}$
| List-$I$ (Equilibrium) | List-$II$ (Conditions for the process) |
|---|---|
| $P. A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ (Endothermic) | $1. \text{High temperature}$ |
| $Q. 2AB_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ (Exothermic) | $2. \text{Low temperature}$ |
| $R. 2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$ (Endothermic) | $3. \text{High temperature}$ |
| $4. \text{Low temperature}$ | |
| $5. \text{Independent of pressure}$ |
Medium
View SolutionWhich of the following reactions will be affected by increasing the pressure? Also,mention whether the change will cause the reaction to proceed in the forward or backward direction.
$(i)$ $CaCl_{2(s)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$
$(ii)$ $CH_{4(g)} + 2S_{2(g)} \rightleftharpoons CS_{2(g)} + 2H_{2}S_{(g)}$
$(iii)$ $CO_{2(g)} \rightleftharpoons C_{(s)} + 2CO_{(g)}$
$(iv)$ $2H_{2(g)} + CO_{(g)} \rightleftharpoons CH_{3}OH_{(g)}$
$(v)$ $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
$(vi)$ $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_{2}O_{(g)}$
$(i)$ $CaCl_{2(s)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$
$(ii)$ $CH_{4(g)} + 2S_{2(g)} \rightleftharpoons CS_{2(g)} + 2H_{2}S_{(g)}$
$(iii)$ $CO_{2(g)} \rightleftharpoons C_{(s)} + 2CO_{(g)}$
$(iv)$ $2H_{2(g)} + CO_{(g)} \rightleftharpoons CH_{3}OH_{(g)}$
$(v)$ $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
$(vi)$ $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_{2}O_{(g)}$
Difficult
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