Le-Chaterlier principle and It’s application Questions in English

(B) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ with $\Delta H = -45.2 \, kcal$.
$1.$ The reaction is exothermic $(\Delta H < 0)$,so a decrease in temperature would favor the forward reaction.
$2.$ The number of moles of gaseous products is $2$,while the number of moles of gaseous reactants is $2 + 1 = 3$.
$3.$ According to Le Chatelier's principle,for a reaction where the number of moles of gas decreases,an increase in pressure shifts the equilibrium towards the side with fewer moles,i.e.,the product side $(SO_3)$.
Therefore,an increase in pressure favors the formation of $SO_3$.

(A) The reaction is $3X_{(g)} + Y_{(g)} \rightleftharpoons X_3Y_{(g)}$.
Since the number of moles of gaseous reactants $(4 \ mol)$ is different from the number of moles of gaseous products $(1 \ mol)$,the equilibrium position is affected by pressure.
Since chemical reactions involve enthalpy changes,the equilibrium constant and the position of equilibrium are affected by temperature.
$A$ catalyst only increases the rate of the forward and backward reactions equally and does not affect the equilibrium position or the amount of product at equilibrium.
Therefore,the amount of $X_3Y$ at equilibrium is affected by temperature and pressure.

(A) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with a smaller number of gaseous moles.
In the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$,the reactant side has $1 + 3 = 4$ moles of gas,while the product side has $2$ moles of gas.
Since $2 < 4$,an increase in pressure will shift the equilibrium towards the product side,which is the forward direction.

(B) The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
This is an exothermic reaction $(\Delta H < 0)$.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction.
Since the number of moles of gaseous products $(2)$ is less than the number of moles of gaseous reactants $(1+3=4)$,an increase in pressure shifts the equilibrium towards the side with fewer moles,which is the product side.
Therefore,the yield of $NH_3$ increases with high pressure and low temperature.
Among the given options,increasing pressure is the correct factor for increasing the yield.

(D) According to Le Chatelier's principle,increasing the volume of the container decreases the total pressure of the system.
To counteract this,the equilibrium shifts in the direction where the total number of gaseous moles increases (i.e.,$\Delta n_g > 0$).
For option $(A)$: $\Delta n_g = 2 - (1+1) = 0$.
For option $(B)$: $\Delta n_g = 2 - (2+1) = -1$.
For option $(C)$: $\Delta n_g = 2 - (1+3) = -2$.
For option $(D)$: $\Delta n_g = (1+1) - 1 = +1$.
Since $\Delta n_g$ is positive only for option $(D)$,the equilibrium will shift to the right when the volume is doubled.

(D) According to Le Chatelier's principle,whenever a constraint (such as change in concentration,pressure,or temperature) is applied to a system in equilibrium,the system tends to readjust so as to nullify the effect of the constraint by bringing a shift in the equilibrium position.
Therefore,it helps in predicting the direction of the shift in equilibrium.

(B) According to Le Chatelier's principle,if the concentration of a product is decreased,the equilibrium shifts in the forward direction to counteract the change.
$A$. $A$ catalyst only increases the rate of both forward and backward reactions equally,so it does not shift the equilibrium.
$B$. Removing $SO_3$ as soon as it is formed decreases the concentration of the product,which shifts the equilibrium forward to produce more $SO_3$.
$C$. The reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ involves $3$ moles of gaseous reactants and $2$ moles of gaseous products. According to Le Chatelier's principle,low pressure shifts the equilibrium towards the side with more moles of gas,which is the backward direction.
$D$. Removing reactants shifts the equilibrium in the backward direction.

(D) The correct answer is $D$. When an inert gas is added to an equilibrium mixture at constant volume,the total pressure of the system increases,but the partial pressures and concentrations of the reacting species remain unchanged. Therefore,the position of the equilibrium does not shift,and the concentrations of $SO_2Cl_2, SO_2$,and $Cl_2$ remain constant.

(B) The dissolution of sodium sulphate is an exothermic process,which can be represented as: $Na_2SO_4(s) + aq \rightleftharpoons Na_2SO_4(aq) + \text{Heat}$.
According to Le Chatelier's principle,if the temperature of an exothermic system at equilibrium is increased,the equilibrium will shift in the direction that absorbs the excess heat,which is the reverse direction.
Therefore,the reverse reaction occurs,causing some of the dissolved $Na_2SO_4$ to precipitate out of the solution.

(A) According to Le Chatelier's principle,for an exothermic reaction $(\Delta H < 0)$,a decrease in temperature favors the forward reaction to produce more product.
Since the number of moles of gaseous products $(2 \ mol)$ is less than the number of moles of gaseous reactants $(1 + 3 = 4 \ mol)$,an increase in pressure shifts the equilibrium towards the side with fewer moles,which is the product side.
Therefore,the maximum yield of ammonia is obtained by decreasing the temperature and increasing the pressure.

(A) Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration,temperature,or pressure,the equilibrium will shift in a direction that tends to counteract the effect of the change. Therefore,it is applicable only to a $System \text{ in equilibrium}$.

(C) The equilibrium reaction is $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$.
When an inert gas like helium is added to a system at equilibrium at constant volume,the total pressure of the system increases,but the partial pressures of the reacting species remain unchanged.
Since the partial pressures of the reactants and products do not change,the reaction quotient $Q_c$ remains equal to the equilibrium constant $K_c$.
Therefore,the addition of an inert gas at constant volume has no effect on the position of the equilibrium.
Thus,the dissociation of $SO_3$ remains unaltered.

(C) For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = n_p - n_r = 2 - (1 + 1) = 0$.
Since $\Delta n_g = 0$,the equilibrium position is unaffected by a change in pressure.
Given $\Delta H = +q \ cal$,the reaction is endothermic.
According to Le Chatelier's principle,increasing the temperature favours the forward (endothermic) reaction,while decreasing the temperature favours the backward reaction.
Therefore,the formation of $HI$ is unaffected by a change in pressure.

(A) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with a smaller number of gaseous moles.
Conversely,the backward reaction is favoured by an increase in pressure if the number of gaseous moles of products is greater than the number of gaseous moles of reactants (i.e.,$\Delta n_g > 0$).
For option $A$: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,$\Delta n_g = (1+1) - 1 = 1 > 0$.
Since the product side has more moles,increasing the pressure will shift the equilibrium to the left (backward reaction).

(A) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with a smaller number of moles of gaseous species.
If the total number of moles of gaseous reactants equals the total number of moles of gaseous products,then $\Delta n_g = 0$,and the equilibrium position remains unaffected by a change in pressure.
For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$:
$\Delta n_g = (2) - (1 + 1) = 0$.
Since $\Delta n_g = 0$,this equilibrium is not shifted by an increase in pressure.

(C) The equilibrium is represented as: $Solid + \text{Heat} \rightleftharpoons Liquid$ (for melting,which is endothermic,$\Delta H > 0$).
According to Le Chatelier's principle,if heat is added to a system at equilibrium,the system will shift in the direction that absorbs the added heat to counteract the change.
Since the forward reaction (melting) is endothermic,adding heat shifts the equilibrium to the right (forward direction).
Consequently,the amount of solid decreases as it converts into liquid.

(A) According to Le Chatelier's principle,the addition of an inert gas at constant volume does not change the partial pressures or the molar concentrations of the reacting species.
Therefore,the reaction quotient $Q_c$ remains equal to the equilibrium constant $K_c$.
Thus,the equilibrium position remains unaffected.

(B) Le Chatelier's principle describes how a system at equilibrium responds to changes in concentration,pressure,or temperature.
It is primarily applicable to systems involving gases or solutes in solution where changes in volume or pressure significantly affect the position of equilibrium.
In the reaction $Fe_{(s)} + S_{(s)} \rightleftharpoons FeS_{(s)}$,all reactants and products are in the solid state.
Since solids have negligible compressibility and their concentrations (density) remain constant,changes in pressure or volume do not affect the equilibrium position of this reaction.
Therefore,the principle is not applicable to this system.

(A) The given reaction is $A + B + Q \rightleftharpoons C + D$.
Since heat $(Q)$ is on the reactant side,the reaction is endothermic.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium in the forward direction.
Therefore,the concentration of the products ($C$ and $D$) will increase.

(A) For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the number of moles of gaseous reactants is $1 + 1 = 2$ and the number of moles of gaseous products is $2$.
Since the change in the number of moles of gas $(\Delta n_g = 2 - 2 = 0)$ is zero,the equilibrium position is not affected by a change in pressure or volume.
Therefore,the reaction direction does not change.

(C) According to Le Chatelier's principle,the effect of pressure on a gaseous reaction depends on the change in the number of moles of gaseous products and reactants,denoted by $\Delta n_g = n_p - n_r$.
If $\Delta n_g = 0$,the equilibrium position is not affected by pressure.
For option $(C)$,$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$,$\Delta n_g = 2 - (1 + 1) = 0$.
Therefore,the reaction is not affected by pressure.

(A) The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + 22 \ kcal$.
This is an exothermic reaction $(\Delta H < 0)$ where the number of moles of gaseous products $(2)$ is less than the number of moles of gaseous reactants $(1 + 3 = 4)$.
According to Le Chatelier's principle,for a reaction with a decrease in the number of moles,increasing the pressure will shift the equilibrium in the forward direction to reduce the pressure.
Therefore,the formation of ammonia is favoured by increasing the pressure.

(B) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$; $\Delta H = -ve$.
Since $\Delta H$ is negative,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,a low temperature favours the forward reaction.
Regarding pressure,the number of moles of gaseous reactants is $2 + 1 = 3$,and the number of moles of gaseous products is $2$.
Since the number of moles decreases in the forward direction,high pressure favours the forward reaction.
Therefore,the reaction is favoured by low temperature and high pressure.

(C) For a general reversible reaction: $A + B \rightleftharpoons C + D$
According to Le Chatelier's principle,if the concentration of any of the products is increased,the system will counteract this change by consuming the added product.
Therefore,the equilibrium shifts in the backward (reverse) direction to form more reactants.
Thus,increasing the concentration of one or more of the products favours the reverse reaction.

(C) The dissociation of molecular hydrogen into atomic hydrogen is represented by the equation: $H_2(g) \rightleftharpoons 2H(g)$.
This reaction is endothermic $( \Delta H > 0 )$ and involves an increase in the number of moles of gas $( \Delta n_g = 2 - 1 = 1 )$.
According to Le Chatelier's principle:
$1$. Since the reaction is endothermic,a high temperature will shift the equilibrium to the right (product side).
$2$. Since the reaction involves an increase in the number of moles of gas,a low pressure will shift the equilibrium to the right (product side).
Therefore,the formation of atomic hydrogen is favoured at high temperature and low pressure.

(D) According to Le Chatelier's principle:
$1$. Since the reaction is endothermic $(\Delta H > 0)$,a high temperature favors the forward reaction.
$2$. Since the number of moles of gaseous products $(2 \ mol)$ is equal to the number of moles of gaseous reactants $(1+1 = 2 \ mol)$,the pressure has no effect on the equilibrium position.
$3$. According to the law of mass action,increasing the concentration of reactants ($N_2$ and $O_2$) shifts the equilibrium in the forward direction.
Therefore,the correct conditions are high temperature and excess reactant concentration.

(C) The given reaction is $BaO_{2(s)} \rightleftharpoons BaO_{(s)} + \frac{1}{2} O_{2(g)}$.
Since $BaO_2$ and $BaO$ are in the solid state,their active masses are taken as unity $(1)$.
The equilibrium constant expression is $K_p = P_{O_2}^{1/2}$.
This shows that the equilibrium pressure of $O_2$ is independent of the amounts of solid reactants or products present.
However,since the reaction is endothermic $(\Delta H > 0)$,according to Le Chatelier's principle,an increase in temperature will shift the equilibrium in the forward direction to absorb the heat,thereby increasing the equilibrium pressure of $O_2$.

(C) The given reaction is $A_{2(g)} + 2B_{(g)} \rightleftharpoons C_{(g)} + Q \ kJ$.
Since the reaction is exothermic $(+Q \ kJ)$,according to Le Chatelier's principle,a low temperature favors the forward reaction to increase the yield of the product.
The total number of moles of gaseous reactants is $1 + 2 = 3$,and the number of moles of gaseous products is $1$.
Since the number of moles decreases in the forward direction,high pressure favors the forward reaction to increase the yield of the product.
Therefore,the yield of the product is high at low temperature and high pressure.

(A) According to Le Chatelier's principle,a change in pressure affects the equilibrium position of reactions where there is a change in the number of moles of gaseous species,i.e.,$\Delta n_g \neq 0$.
For option $A$: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$,$\Delta n_g = 2 - (1 + 1) = 0$.
Since $\Delta n_g = 0$,the equilibrium position is not affected by a change in pressure.

(D) According to Le Chatelier's principle:
$1$. For an exothermic reaction (indicated by $+ Q$),the reaction is favoured at low temperature.
$2$. For the pressure effect,we compare the number of moles of gaseous products and reactants.
Reactants: $1 + 1 = 2$ moles.
Products: $2 + 1 = 3$ moles.
Since the number of moles of products $(3)$ is greater than the number of moles of reactants $(2)$,the reaction is favoured at low pressure to shift the equilibrium towards the side with more moles.

(A) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.
Given that $K_p > K_c$,it implies that $(RT)^{\Delta n} > 1$,which means $\Delta n > 0$.
$\Delta n$ represents the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.
Since $\Delta n > 0$,the number of moles of gaseous products is greater than the number of moles of gaseous reactants.
According to Le Chatelier's principle,for a reaction where the number of moles of gas increases,decreasing the pressure will shift the equilibrium in the forward direction to increase the number of moles and counteract the pressure change.
Therefore,the forward reaction is favoured by low pressure.

(A) In the reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$,the number of moles of gaseous reactants is $1 + 1 = 2$ and the number of moles of gaseous products is $2$.
Since the total number of moles of gas on both sides of the equation is equal $(\Delta n_g = 2 - 2 = 0)$,the equilibrium position is unaffected by changes in pressure.

(C) The given reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ with $\Delta H^\circ = -198 \ kJ$.
Since $\Delta H^\circ < 0$,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction.
Regarding pressure,the number of moles of gaseous products is $2$,while the number of moles of gaseous reactants is $2 + 1 = 3$.
Since the number of moles decreases in the forward direction $(3 \rightarrow 2)$,an increase in pressure favors the forward reaction.
Therefore,lowering the temperature and increasing the pressure are the favorable conditions.

(D) According to Le Chatelier's principle,if the concentration of one of the reactants is increased,the equilibrium shifts in the forward direction to counteract the change.
Since $CO_{(g)}$ is a reactant,increasing its amount will shift the equilibrium to the right,thereby increasing the production of the products $CO_{2(g)}$ and $H_{2(g)}$.
Adding a catalyst does not change the equilibrium position.
Adding an inert gas at constant volume has no effect on the equilibrium.
Decreasing the volume of the container has no effect on this reaction because the total number of moles of gaseous reactants equals the total number of moles of gaseous products $(1 + 1 = 1 + 1)$.

(D) The dissociation of $H_2S$ is given by: $H_2S \rightleftharpoons 2H^{+} + S^{2-}$.
When $NH_4OH$ is added,it dissociates as: $NH_4OH \rightleftharpoons NH_4^{+} + OH^{-}$.
The $OH^{-}$ ions react with $H^{+}$ ions from $H_2S$ to form water: $H^{+} + OH^{-} \rightleftharpoons H_2O$.
According to Le Chatelier's principle,the removal of $H^{+}$ ions shifts the equilibrium of $H_2S$ dissociation to the right.
Therefore,the concentration of $S^{2-}$ increases.

(B) According to Le Chatelier's principle,if $CO_2$ escapes from the system,the concentration of $CO_2$ decreases.
To counteract this change,the equilibrium shifts in the backward direction to produce more $CO_2$.
As the reaction shifts backward,$H^{+}$ ions are consumed to form $H_2CO_3$,which then decomposes into $H_2O$ and $CO_2$.
Consequently,the concentration of $H^{+}$ ions decreases,which leads to an increase in $pH$.

(D) According to the van't Hoff equation,for an exothermic reaction,the equilibrium constant $K_c$ is inversely proportional to temperature.
When the temperature of an exothermic reaction is increased,the reaction proceeds in the backward direction to absorb the excess heat,according to Le Chatelier's principle.
Consequently,the value of the equilibrium constant $K_c$ decreases with an increase in temperature.

(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.

(C) The reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$ is a reversible reaction.
According to Le Chatelier's principle,if one of the products is continuously removed from the system,the equilibrium shifts in the forward direction to compensate for the loss.
In a lime kiln,$CO_{2(g)}$ is a gas that escapes into the atmosphere as it is produced.
Because $CO_2$ is continuously removed,the reaction proceeds to completion in the forward direction.

(C) The dissociation of molecular chlorine into atomic chlorine is an endothermic process:
$Cl_2(g) \to 2Cl(g), \Delta H > 0$.
According to Le Chatelier's principle,for an endothermic reaction with an increase in the number of moles of gas (from $1$ mole to $2$ moles),high temperature favors the forward reaction.
Additionally,low pressure favors the side with more moles of gas.
Therefore,high temperature and low pressure are the favorable conditions.

(C) The role of a catalyst in a reversible reaction is to allow the equilibrium to be achieved quickly.
$A$ catalyst increases the rate of both the forward and reverse reactions to an equal extent.
However,it does not affect the value of the equilibrium constant.

(D) catalyst provides an alternative reaction pathway with a lower activation energy for both the forward and backward reactions.
This increases the rate of both the forward and backward reactions to the same extent.
Consequently,the catalyst does not change the equilibrium constant or the position of equilibrium,but it allows the system to reach the state of equilibrium more quickly.

(C) catalyst provides an alternative pathway with lower activation energy for both the forward and backward reactions.
It increases the rate of both reactions to the same extent.
Therefore,the equilibrium position and the equilibrium concentrations remain unchanged.
Also,the equilibrium constant $(K_{eq})$ depends only on temperature and is not affected by the presence of a catalyst.

(C) In the presence of a catalyst,both the forward and reverse reaction rates increase equally.
This allows the system to reach the state of equilibrium in a shorter amount of time.
However,a catalyst does not alter the final equilibrium position or the equilibrium constant of the reaction.

(C) catalyst provides an alternative pathway with lower activation energy for both the forward and backward reactions.
It increases the rate of both reactions equally,allowing the system to reach equilibrium faster.
However,it does not change the equilibrium constant or the final composition of the equilibrium mixture.

(D) The decomposition of limestone is a reversible reaction: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$.
According to Le Chatelier's principle,to increase the yield of products in a reversible reaction,the concentration of the product should be decreased.
Therefore,by continuously pumping out $CO_2$ from the kiln,the equilibrium shifts in the forward direction to produce more $CO_2$.

(C) According to Le Chatelier's principle,a change in the volume of the reaction vessel affects the equilibrium position only if there is a change in the number of moles of gaseous species between the reactants and products (i.e.,$\Delta n_g \neq 0$).
If $\Delta n_g = 0$,the equilibrium is not affected by a change in volume or pressure.
For option $A$: $\Delta n_g = (1+1) - 1 = 1$.
For option $B$: $\Delta n_g = 2 - (1+3) = -2$.
For option $C$: $\Delta n_g = 2 - (1+1) = 0$.
For option $D$: $\Delta n_g = (1+1) - 1 = 1$.
Since $\Delta n_g = 0$ for option $C$,it is not affected by a change in volume.

(C) The effect of pressure on the rate of reaction depends on the change in the number of moles of gaseous products and reactants,denoted by $\Delta n_g$.
If $\Delta n_g = 0$,the change in pressure does not affect the equilibrium or the rate of reaction.
For the reaction $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$:
$\Delta n_g = (n_{products}) - (n_{reactants}) = 2 - (1 + 1) = 0$.
Therefore,the rate of this reaction is independent of pressure changes.

(C) The given reaction is $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ with $\Delta H = +ve$,which means it is an endothermic reaction.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium in the forward direction to absorb the excess heat.
Since the number of moles of gaseous reactants $(1+1=2)$ is equal to the number of moles of gaseous products $(2)$,the change in pressure will have no effect on the equilibrium position.
Therefore,the production of $NO$ is favored by increasing the temperature.

(C) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ with $\Delta H = -198.2 \, kJ/mol$.
$1$. Effect of Pressure: The reaction involves a decrease in the number of moles of gas ($3$ moles of reactants to $2$ moles of product). According to Le Chatelier's principle,high pressure favors the side with fewer moles,so high pressure increases the yield of $SO_3$.
$2$. Effect of Temperature: The reaction is exothermic $(\Delta H < 0)$. According to Le Chatelier's principle,low temperature favors the forward (exothermic) reaction.
$3$. Effect of Concentration: Increasing the concentration of reactants ($SO_2$ and $O_2$) shifts the equilibrium to the right,increasing the yield of $SO_3$.
Therefore,high pressure,low temperature,and high concentration of reactants favor the maximum yield of $SO_3$.