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

(N/A) If the volume is kept constant and an inert gas such as argon is added,which does not take part in the reaction,the equilibrium remains undisturbed.
This is because the addition of an inert gas at constant volume does not change the partial pressure or the molar concentrations of the substances involved in the reaction.
The reaction quotient $(Q_c)$ changes only if the added gas is a reactant or product involved in the reaction.

When a change in temperature of an equilibrium reaction occurs,the value of the equilibrium constant $K_{c}$ changes. In general,the temperature dependence of the equilibrium constant depends on the sign of $\Delta H$ for the reaction.
$1$. Exothermic reaction: The equilibrium constant for an exothermic reaction (negative $\Delta H$) decreases as the temperature increases. According to Le-Chatelier's principle,raising the temperature shifts the equilibrium to the left. For example,the production of ammonia is an exothermic reaction:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} ; \Delta H = -92.38 \ kJ \ mol^{-1}$
Raising the temperature decreases the yield of ammonia. Thus,low temperature is favorable for a high yield,but a catalyst is used to maintain a practical reaction rate.
$2$. Endothermic reaction: The equilibrium constant for an endothermic reaction (positive $\Delta H$) increases as the temperature increases. The forward reaction is favored. For example:
$[Co(H_{2}O)_{6}]^{2+}_{(aq)} + 4Cl^{-}_{(aq)} \rightleftharpoons [CoCl_{4}]^{2-}_{(aq)} + 6H_{2}O_{(l)}$
(Pink) (Blue)
At room temperature,the mixture is blue. When cooled,the equilibrium shifts to the left,and the color turns pink due to the formation of $[Co(H_{2}O)_{6}]^{2+}$.

(N/A) The effect of temperature on equilibrium can be demonstrated by taking $NO_{2}$ gas (brown in colour) which dimerises into $N_{2}O_{4}$ gas (colourless).
$2NO_{2(g)} \rightleftharpoons N_{2}O_{4(g)}$; $\Delta H = -57.2 \ kJ \ mol^{-1}$
(Brown colour gas) $\quad$ (Colourless gas)
Method of experiment: $NO_{2}$ gas,prepared by the addition of $Cu$ turnings to concentrated $HNO_{3}$,is collected in two $5 \ mL$ test tubes (ensuring the same intensity of colour of gas in each tube) and the stoppers are sealed with araldite. Take three $250 \ mL$ beakers labeled $1$,$2$,and $3$.
Beaker $1$ contains a freezing mixture $(270 \ K)$,beaker $2$ contains water at room temperature $(298 \ K)$,and beaker $3$ contains hot water $(363 \ K)$.
Both test tubes are placed in beaker $2$ for $8-10$ minutes to reach equilibrium. After this,one tube is placed in beaker $1$ and the other in beaker $3$. The effect of temperature on the direction of the reaction is observed.
Observation:
$(i)$ At low temperature in beaker $1$ $(270 \ K)$,the forward reaction (formation of $N_{2}O_{4}$) is favoured because the reaction is exothermic. Thus,the intensity of the brown colour due to $NO_{2}$ decreases.
$(ii)$ In beaker $3$ $(363 \ K)$,the high temperature favours the reverse reaction (formation of $NO_{2}$),and thus,the brown colour intensifies.
Deduction: In an exothermic reaction,a decrease in temperature shifts the equilibrium in the forward direction,while an increase in temperature shifts it in the reverse direction.

(N/A) catalyst is used to increase or decrease the rate of a chemical reaction. It does not affect the equilibrium position or the value of the equilibrium constant $K$. The catalyst does not appear in the balanced chemical equation or the expression for the equilibrium constant.
Effect of catalyst:
$(i)$ $A$ catalyst increases the rate of the chemical reaction by providing a new,lower-energy pathway for the conversion of reactants to products.
$(ii)$ It increases the rates of both forward and reverse reactions equally,thus not shifting the equilibrium.
$(iii)$ $A$ catalyst lowers the activation energy for both forward and reverse reactions by the same amount.
Example $1$: The Haber process for the synthesis of $NH_{3}$ from $N_{2}$ and $H_{2}$:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
This reaction is exothermic. At low temperatures,the reaction rate is very slow. Fritz Haber discovered that an iron catalyst allows the reaction to occur at a satisfactory rate at $500 \ ^\circ C$,where the equilibrium concentration of $NH_{3}$ is favorable.
Example $2$: The contact process for the manufacture of sulphuric acid:
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,$K_{c} = 1.7 \times 10^{26}$
Although the large $K_{c}$ value suggests the reaction goes to completion,the oxidation of $SO_{2}$ to $SO_{3}$ is very slow. Therefore,platinum or vanadium$(V)$ oxide $(V_{2}O_{5})$ is used as a catalyst to increase the reaction rate.

(N/A) According to Le Chatelier's principle,when the pressure of a system at equilibrium is decreased (by increasing the volume),the equilibrium shifts in the direction that increases the total number of moles of gaseous species.
$(a)$ $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$: The number of gaseous moles increases from $1$ (reactant) to $2$ (products). Thus,the equilibrium shifts forward,and the number of moles of products increases.
$(b)$ $CaO_{(s)} + CO_{2(g)} \rightleftharpoons CaCO_{3(s)}$: The number of gaseous moles is $1$ on the reactant side and $0$ on the product side. To increase gaseous moles,the equilibrium shifts in the reverse direction,so the number of moles of products decreases.
$(c)$ $3Fe_{(s)} + 4H_{2}O_{(g)} \rightleftharpoons Fe_{3}O_{4(s)} + 4H_{2(g)}$: The number of gaseous moles is $4$ on both sides. Therefore,a change in pressure has no effect on the equilibrium position,and the number of moles of products remains the same.

Reaction	Change in Moles of Gas (Products vs Reactants)
$(a)$	Products: $2$,Reactants: $1$. Equilibrium shifts forward.
$(b)$	Products: $0$,Reactants: $1$. Equilibrium shifts reverse.
$(c)$	Products: $4$,Reactants: $4$. No change in equilibrium.

(A) In chemical equilibrium,the effect of pressure is observed only if the change in the number of gaseous moles,$\Delta n_g \neq 0$.
According to Le Chatelier's principle,increasing the pressure shifts the equilibrium in the direction where the number of gaseous moles decreases.

$Reaction$	$Effect$ $of$ $increasing$ $pressure$
$(i).$ $\Delta n_g = +1$	Backward $direction$
$(ii).$ $\Delta n_g = 0$	No $effect$
$(iii).$ $\Delta n_g = +1$	Backward $direction$
$(iv).$ $\Delta n_g = -2$	Forward $direction$
$(v).$ $\Delta n_g = +1$	Backward $direction$
$(vi).$ $\Delta n_g = +1$	Backward $direction$

(N/A) The expression for $K_p$ is: $K_p = \frac{(p_{CO})(p_{H_2})^3}{(p_{CH_4})(p_{H_2O})}$
$(b)$ $(i)$ Increasing the pressure: According to Le Chatelier's principle,the equilibrium will shift in the direction that produces fewer moles of gas (the left side),so the amount of $H_2$ will decrease. $K_p$ remains constant.
$(ii)$ Increasing the temperature: Since the reaction is endothermic,increasing the temperature will shift the equilibrium in the forward direction,increasing the yield of $H_2$ and increasing the value of $K_p$.
$(iii)$ Using a catalyst: $A$ catalyst does not affect the value of $K_p$ or the equilibrium composition; it only helps the system attain equilibrium more quickly.

(N/A) The given reaction is: $2H_{2(g)} + CO_{(g)} \rightleftharpoons CH_3OH_{(g)}$
$(a)$ Addition of $H_2$: $H_2$ is a reactant. According to Le Chatelier's principle,adding a reactant shifts the equilibrium in the forward direction to consume the added $H_2$,resulting in an increase in the yield of $CH_3OH$.
$(b)$ Addition of $CH_3OH$: $CH_3OH$ is a product. Adding a product shifts the equilibrium in the reverse direction to consume the added $CH_3OH$.
$(c)$ Removal of $CO$: $CO$ is a reactant. Removing a reactant shifts the equilibrium in the reverse direction to produce more $CO$,resulting in a decrease in the amount of $CH_3OH$.
$(d)$ Removal of $CH_3OH$: $CH_3OH$ is a product. Removing a product shifts the equilibrium in the forward direction to produce more $CH_3OH$.

(B) According to Le Chatelier's Principle,for an exothermic reaction,increasing the concentration of reactants will shift the equilibrium in the forward direction.
Since the reaction is $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}$,adding $F_2$ (a reactant) will increase the yield of the product $ClF_3$.

(N/A) The orange-colored $Cr_{2}O_{7}^{2-}$ ion and the yellow-colored $CrO_{4}^{2-}$ ion are interconvertible depending on the $pH$ of the medium.
In an alkaline medium,the equilibrium shifts towards the formation of chromate ions:
$Cr_{2}O_{7}^{2-} + 2OH^{-} \rightleftharpoons 2CrO_{4}^{2-} + H_{2}O$
(Orange) $\quad$ (Yellow)
In an acidic medium,the equilibrium shifts towards the formation of dichromate ions:
$2CrO_{4}^{2-} + 2H^{+} \rightleftharpoons Cr_{2}O_{7}^{2-} + H_{2}O$
(Yellow) $\quad$ (Orange)

(A) According to Le Chatelier's principle,a catalyst provides an alternative pathway for the reaction by lowering the activation energy.
It increases the rate of both the forward and backward reactions equally.
Therefore,the addition of a catalyst does not change the position of the equilibrium or the equilibrium constant.

(N/A) $1$. Effect of temperature: Since the reaction is exothermic $(\Delta H < 0)$,according to the $Le \ Chatelier$ principle,a low temperature favors the forward reaction to increase the yield of $NH_3$. However,very low temperatures make the reaction rate too slow. Thus,an optimum temperature of approximately $700 \ K$ is maintained.
$2$. Effect of pressure: The forward reaction involves a decrease in the number of moles ($4 \ mol$ of reactants to $2 \ mol$ of products). According to the $Le \ Chatelier$ principle,high pressure favors the forward reaction. Therefore,a high pressure of about $200 \ atm$ is used to increase the yield.
$3$. Effect of addition of argon: At constant volume,the addition of an inert gas like argon does not change the partial pressures or concentrations of the reacting species. Consequently,the equilibrium remains unaffected.

(A) The given reaction is $2CO_{(g)} \rightleftharpoons O_{2(g)} + 2CO_{2(g)}$.
According to Le Chatelier's principle,if the volume of the system is decreased,the pressure increases.
The system will shift in the direction that reduces the number of moles of gas.
On the reactant side,there are $2$ moles of gas,and on the product side,there are $1 + 2 = 3$ moles of gas.
Since the reactant side has fewer moles of gas $(2 < 3)$,the equilibrium will shift towards the left (reactant side) to decrease the pressure.
Therefore,the concentration of $CO$ will increase.

(N/A) The given equilibrium is: $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)$.
$(i)$ Addition of $H_2C_2O_4$: Oxalic acid provides $C_2O_4^{2-}$ ions which react with $Fe^{3+}$ to form a stable complex $[Fe(C_2O_4)_3]^{3-}$. This decreases the concentration of free $Fe^{3+}$ ions. According to Le Chatelier's principle,the equilibrium shifts to the left,causing the deep red color to fade.
$(ii)$ Addition of $HgCl_2$: $Hg^{2+}$ ions react with $SCN^-$ to form a very stable complex $[Hg(SCN)_4]^{2-}$. This decreases the concentration of $SCN^-$ ions. The equilibrium shifts to the left,and the deep red color fades.
$(iii)$ Addition of $KSCN$: This increases the concentration of $SCN^-$ ions. According to Le Chatelier's principle,the equilibrium shifts to the right to consume the excess $SCN^-$,resulting in an increase in the intensity of the deep red color.

(A) The given reaction is endothermic $(\Delta H^{\circ} > 0)$.
$(a)$ According to Le Chatelier's principle,decreasing the temperature favors the exothermic direction to release heat. Since the forward reaction is endothermic,the reverse reaction is exothermic. Therefore,the equilibrium shifts towards the reactant side.
$(b)$ Adding an inert gas like $N_{2}$ at constant temperature and constant volume does not change the partial pressures of the reacting species. Therefore,there is no effect on the equilibrium position.

(A) The given reaction is $A_{2} \rightleftharpoons 2A$ with $\Delta H > 0$ (endothermic).
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium in the forward direction,thereby increasing the yield of the product $A$.
Additionally,increasing the concentration of the reactant $A_{2}$ shifts the equilibrium in the forward direction to counteract the change,further increasing the yield of $A$.
Therefore,an increase in both temperature and concentration of the reactant will increase the yield of the monomer $A$.

(D) According to the $Le \ Chatelier$ principle,a catalyst provides an alternative pathway with lower activation energy for both the forward and backward reactions. It increases the rate of both reactions equally,thus reaching the equilibrium state faster without changing the position of the equilibrium or the equilibrium constant. Therefore,the presence of a catalyst does not affect the equilibrium.

(D)
$C_2H_{6(g)} \rightleftharpoons C_2H_{4(g)} + H_{2(g)}$,$\Delta H = 137.0 \ kJ \ mol^{-1}$
Increasing the volume of the closed reaction vessel would shift the equilibrium to the right.
As the volume increases,pressure decreases,so the system shifts to increase the number of gaseous molecules.
In the given reaction,the number of moles of gaseous products $(2 \ mol)$ is greater than the number of moles of gaseous reactants $(1 \ mol)$.
Thus,according to Le Chatelier's principle,the equilibrium will shift in the forward direction to increase the number of gaseous molecules.

(D) The given reaction is $2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$ with $\Delta H = -190 \ kJ$.
$1$. Increasing temperature: Since the reaction is exothermic $(\Delta H < 0)$,increasing the temperature shifts the equilibrium to the left,decreasing the yield of $SO_3$.
$2$. Increasing pressure: The reaction involves a decrease in the number of moles of gas $(3 \text{ moles of reactants} \rightarrow 2 \text{ moles of products})$. According to Le Chatelier's principle,increasing pressure shifts the equilibrium to the right,increasing the yield of $SO_3$.
$3$. Adding more $SO_2$ or $O_2$: Increasing the concentration of reactants shifts the equilibrium to the right,increasing the yield of $SO_3$.
$4$. Addition of catalyst: $A$ catalyst only increases the rate of reaction and does not change the position of equilibrium or the yield of products.
Therefore,the factors that increase the yield are $B, C,$ and $D$. The total number of such factors is $3$.

(D) For the reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
$1$. At constant pressure: The addition of an inert gas like helium increases the volume of the system. According to Le Chatelier's principle,the equilibrium shifts towards the side with a greater number of moles of gas,which is the forward direction. Thus,more $PCl_3$ and $Cl_2$ are produced.
$2$. At constant volume: The addition of an inert gas does not change the partial pressures or concentrations of the reacting species. Therefore,the equilibrium remains unaffected.
Since the condition (constant pressure or constant volume) is not specified,in standard textbook problems,if not mentioned,it is often assumed to be at constant volume,or the question is considered ambiguous. However,based on the options provided,if the pressure is constant,$A$ is correct; if the volume is constant,$D$ is correct. In many competitive contexts,if the condition is not specified,the addition of an inert gas at constant volume is the standard assumption,leading to $D$.

(A) According to Le Chatelier's principle,the equilibrium position is affected by changes in the concentration of gaseous or aqueous species.
In the given reaction,$Fe_2O_{3(s)}$ is a solid.
The active mass of a pure solid or pure liquid is taken as unity $(1)$ and remains constant regardless of the amount present.
Therefore,the addition or removal of $Fe_2O_{3(s)}$ does not change the concentration of the reactants or products involved in the equilibrium expression,and thus it will not disturb the equilibrium.

(B) The given equilibrium is $Cr_2O_7^{2-} + H_2O \rightleftharpoons 2CrO_4^{2-} + 2H^{+}$.
According to Le Chatelier's principle,if we add a base ($OH^-$ ions) to the system,the $OH^-$ ions will react with the $H^+$ ions produced in the reaction to form water $(H^+ + OH^- \rightarrow H_2O)$.
This removal of $H^+$ ions from the product side decreases the concentration of $H^+$,which causes the equilibrium to shift to the right to produce more $H^+$ ions.
Therefore,the equilibrium shifts to the right in a basic medium.

(A) The equilibrium $2 Cu^{+} \rightleftharpoons Cu + Cu^{2+}$ shifts to the left if the concentration of $Cu^{+}$ is increased or the concentration of $Cu^{2+}$ is decreased,or if $Cu^{+}$ is removed from the solution by precipitation or complexation.
$Cu(I)$ forms insoluble compounds such as $CuCl$,$CuCN$,and $CuSCN$.
When $Cl^{-}$,$CN^{-}$,or $SCN^{-}$ are added,they react with $Cu^{+}$ to form these insoluble precipitates,effectively removing $Cu^{+}$ from the equilibrium mixture.
According to Le Chatelier's principle,removing a product (or in this case,a reactant $Cu^{+}$) shifts the equilibrium to the left (backward direction) to counteract the change.
Therefore,the presence of $Cl^{-}$,$SCN^{-}$,and $CN^{-}$ shifts the equilibrium to the left.

(C) The reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is exothermic $(\Delta H < 0)$.
According to Le Chatelier's principle,increasing the temperature $(T_2 > T_1)$ will shift the equilibrium in the backward direction,resulting in a lower equilibrium yield of ammonia at $T_2$ compared to $T_1$.
However,increasing the temperature increases the rate of reaction,so the system reaches equilibrium faster at $T_2$ than at $T_1$.
Therefore,the curve for $T_2$ will rise more steeply initially but will level off at a lower equilibrium yield value than the curve for $T_1$.

(A) $1$. According to Le Chatelier's Principle,when pressure is increased,the equilibrium shifts towards the side with fewer moles of gaseous species.
$2$. In the given reaction,the number of moles of gaseous reactants is $1 + 3 = 4$,and the number of moles of gaseous products is $1 + 1 = 2$.
$3$. Since the product side has fewer moles $(2 < 4)$,increasing the pressure shifts the equilibrium in the forward direction ($B$ is correct).
$4$. Increasing the pressure increases the concentration of all gaseous species present in the system ($A$ is correct).
$5$. The equilibrium constant ($K_c$ or $K_p$) depends only on temperature. Since the temperature is constant,the equilibrium constant remains unchanged ($C$ is incorrect,$D$ is incorrect because the concentrations do not remain the same).
$6$. Therefore,statements $(A)$ and $(B)$ are correct.

(D) When an inert gas like xenon is added to a gaseous equilibrium system at constant $T$ and $P$,the total volume of the system increases.
This leads to a decrease in the partial pressure of each reacting species.
According to Le Chatelier's principle,the system shifts in the direction that increases the total number of moles of gas.
In the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the number of moles of products $(2)$ is greater than the number of moles of reactants $(1)$.
Therefore,the equilibrium shifts in the forward direction.
As a result,the concentration of $PCl_5$ decreases,while the concentrations of $PCl_3$ and $Cl_2$ increase.

(A) The given reaction is $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ with $\Delta H = +180.7 \ kJ \ mol^{-1}$.
Since the reaction is endothermic $(\Delta H > 0)$,according to Le Chatelier's principle,an increase in temperature favors the forward reaction,leading to a higher yield of $NO$.
Increasing the concentration of reactants ($N_2$ or $O_2$) also shifts the equilibrium in the forward direction to consume the added reactants,thereby increasing the yield of $NO$.
Therefore,higher temperature $(A)$,higher concentration of $N_2$ $(C)$,and higher concentration of $O_2$ $(D)$ all favor the formation of $NO$.

(C) $1$. At $273 \ K$ and $1 \ atm$, the system $H_2O_{(s)} \rightleftharpoons H_2O_{(\ell)}$ is in equilibrium, so $\Delta G = 0$. Since melting is an endothermic process involving an increase in disorder, $\Delta S > 0$. Thus, statement $(A)$ is correct.
$2$. According to Le Chatelier's principle, for the process $H_2O_{(s)} \rightleftharpoons H_2O_{(\ell)}$, the density of liquid water is greater than that of ice. Increasing pressure favors the side with higher density (liquid phase). Therefore, more ice will melt. Statement $(B)$ is correct.
$3$. Since increasing pressure shifts the equilibrium towards the liquid phase, the melting point of ice decreases. Statement $(C)$ is correct and $(D)$ is incorrect.

(A) According to $Le \ Chatelier's \ Principle$,when the volume of a system at equilibrium is reduced,the pressure increases.
To counteract this increase in pressure,the system shifts in the direction that produces fewer moles of gas.
In the reaction $2 C_{(s)} + O_{2(g)} \rightleftharpoons 2 CO_{(g)}$,the number of moles of gaseous reactants is $1$ $(O_{2})$ and the number of moles of gaseous products is $2$ $(CO)$.
Since the number of moles of gaseous products $(2)$ is greater than the number of moles of gaseous reactants $(1)$,an increase in pressure (due to volume reduction) will shift the equilibrium to the left (towards the reactants) to decrease the total number of gas moles.

(C) According to Le Chatelier's Principle:
$1$. For an exothermic reaction $(\Delta H < 0)$,low temperature favours the formation of products.
$2$. For a reaction where the number of gaseous moles increases $(\Delta n_g > 0)$,low pressure favours the formation of products.
$3$. In option $C$,$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$,$\Delta H^{\circ} = -285 \ kJ$ (exothermic) and $\Delta n_g = 3 - 2 = 1$ (increase in gaseous moles).
$4$. Therefore,low temperature and low pressure favour the forward reaction.

(A) According to Le Chatelier's principle,the equilibrium shifts in the direction that counteracts the change applied to the system.
For the reaction $CO_{(g)} + 3H_{2(g)} \rightleftharpoons CH_{4(g)} + H_2O_{(g)}$,the formation of methane $(CH_4)$ is favoured by:
$1$. Increasing the concentration of reactants ($CO$ or $H_2$): This shifts the equilibrium to the right.
$2$. Decreasing the concentration of products ($CH_4$ or $H_2O$): This also shifts the equilibrium to the right to produce more product.
Evaluating the options:
$(a)$ Increasing the concentration of $CO$ favours the forward reaction.
$(b)$ Increasing the concentration of $H_2O$ favours the backward reaction.
$(c)$ Decreasing the concentration of $CH_4$ favours the forward reaction.
$(d)$ Decreasing the concentration of $H_2$ favours the backward reaction.
Therefore,the formation of methane is favoured by $a$ and $c$.

(C) For the given reaction,$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = (1 + 1) - (1 + 1) = 0$.
Since $\Delta n_g = 0$,the pressure change does not affect the equilibrium position or the equilibrium constant.
The equilibrium constant $K_c$ is a function of temperature only.
Therefore,increasing the pressure will not disturb the equilibrium constant.

(C) The given reaction is $A_{2(g)} + 2B_{(g)} \rightleftharpoons C_{(g)} + Q \ kJ$.
Since the reaction releases heat $(+Q \ kJ)$,it is an exothermic reaction.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction,thereby increasing the yield of the products.
Furthermore,the number of moles of gaseous reactants is $1 + 2 = 3$,while the number of moles of gaseous products is $1$.
Since the forward reaction proceeds with a decrease in the number of gaseous molecules,an increase in pressure will shift the equilibrium toward the product side.
Therefore,low temperature and high pressure favor the formation of products.

(A) $N_{2} + 3H_{2} \rightleftharpoons 2NH_{3} + \text{heat}$
According to Le Chatelier's principle,for an exothermic reaction,an increase in temperature shifts the equilibrium in the direction that absorbs heat.
Since the forward reaction is exothermic,increasing the temperature will favour the backward reaction.
Therefore,the equilibrium shifts to the left.

(D) According to Le Chatelier's principle,adding an inert gas like $He$ at constant volume does not change the partial pressures of the reacting species,and thus the equilibrium position remains unaffected.
Furthermore,for this reaction,the number of moles of gaseous reactants $(1 + 1 = 2)$ is equal to the number of moles of gaseous products $(2)$.
Therefore,even if the total pressure is changed,the equilibrium will not shift because the reaction quotient $Q_c$ remains equal to the equilibrium constant $K_c$.

(C) Addition of oxalic acid $(H_2C_2O_4)$ provides oxalate ions $(C_2O_4^{2-})$,which react with $Fe^{3+}$ ions to form a stable complex $[Fe(C_2O_4)_3]^{3-}$.
This reaction significantly decreases the concentration of free $Fe^{3+}$ ions in the solution.
According to Le Chatelier's principle,the equilibrium will shift to the left to compensate for the loss of $Fe^{3+}$ ions.
As the equilibrium shifts to the left,the concentration of the deep red complex $[Fe(SCN)]^{2+}$ decreases,leading to a decrease in the intensity of the deep red color.

(D) Consider the reaction for the formation of ammonia from nitrogen and hydrogen by Haber's process:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}; \Delta H = -92.4 \ kJ$
The forward reaction is exothermic $(\Delta H < 0)$,while the reverse reaction is endothermic $(\Delta H > 0)$.
According to Le Chatelier's principle,if the temperature of a system at equilibrium is increased,the system will shift in the direction that absorbs heat to counteract the change.
Therefore,an increase in temperature favors the endothermic (backward) reaction.
As a result,the backward reaction is favored,leading to a decrease in the yield of ammonia.

(B) The reaction is $A_{(g)} + \frac{1}{2} B_{(g)} \rightleftharpoons C_{(g)} + \text{heat}$.
Since the reaction releases heat,it is an exothermic reaction.
According to Le Chatelier's principle,low temperature favors the forward direction for exothermic reactions.
Calculate the change in the number of gaseous moles: $\Delta n_g = 1 - (1 + 0.5) = -0.5$.
Since $\Delta n_g < 0$,the number of gas molecules decreases in the forward reaction.
Therefore,high pressure is favorable for the forward reaction.

(B) The given reaction is $SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)}$.
In this reaction,the number of moles of gaseous products is $1$ and the number of moles of gaseous reactants is $1 + 0.5 = 1.5$.
According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with fewer moles of gas.
Since the product side has fewer moles $(1 < 1.5)$,increasing the pressure will increase the yield of $SO_3$.
From the graph,for a given temperature,the yield follows the order $P_3 > P_2 > P_1$.
Therefore,the correct relationship is $P_3 > P_2 > P_1$,which is equivalent to $P_1 < P_2 < P_3$.

(B) $Le-Chatelier$'s principle is not applicable to solid-solid equilibrium because the activities of pure solids and liquids are taken as $1$.
Pure solids and liquids are excluded from the equilibrium constant expression.
This is because they do not affect the concentration of reactants or products in the equilibrium state.
Therefore,adding or removing a pure solid from a system at equilibrium has no effect on the position of equilibrium.
Among the given options,the reaction $Fe_{(s)} + S_{(s)} \rightleftharpoons FeS_{(s)}$ involves only solid phases,making the principle inapplicable.

(C) According to Le Chatelier's principle,adding $H_{2(g)}$ (a reactant) shifts the equilibrium in the forward direction to consume the added $H_{2(g)}$.
Since the forward reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is exothermic (releases heat),the forward shift results in the release of additional heat.
Because the container is closed and the volume is constant,this released heat causes the temperature of the system to increase.

(A) $Le \ Chatelier's$ principle explains the transport of gases in the blood. As blood reaches the tissues,the partial pressure of $O_2$ is low,causing $O_2$ to dissociate from hemoglobin. Conversely,in the lungs,the high partial pressure of $O_2$ drives the formation of oxyhemoglobin. The equilibrium $Hb + 4O_2 \rightleftharpoons Hb(O_2)_4$ shifts according to the concentration of reactants and products,which is a direct application of $Le \ Chatelier's$ principle. Thus,the correct answer is option $(A)$.

(B) According to the van't Hoff equation,the effect of temperature on the equilibrium constant $K_{eq}$ depends on the enthalpy change $\Delta H$ of the reaction.
For an exothermic reaction,$\Delta H < 0$,so increasing the temperature decreases the equilibrium constant $K_{eq}$.
For an endothermic reaction,$\Delta H > 0$,so increasing the temperature increases the equilibrium constant $K_{eq}$.

(C) According to Le Chatelier's principle,when the total pressure of a system at equilibrium is increased,the equilibrium shifts in the direction that results in a decrease in the total number of gaseous moles.
For reaction $(I)$: $X_{2(g)} + 3Y_{2(g)} \rightleftharpoons 2XY_{3(g)}$. The number of gaseous moles on the reactant side is $1 + 3 = 4$,and on the product side is $2$. Since $2 < 4$,increasing the pressure shifts the equilibrium towards the products.
For reactions $(II)$,$(III)$,and $(IV)$: The number of gaseous moles on the reactant side is $1 + 1 = 2$,and on the product side is $2$. Since the number of moles is equal on both sides,a change in pressure does not affect the position of the equilibrium.

(A) The formation of $NH_3$ follows Le Chatelier's principle for the reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + \text{Heat}$.
Number of moles of reactants $= 1 + 3 = 4$.
Number of moles of products $= 2$.
Since the number of moles of products is less than the number of moles of reactants,an increase in pressure will shift the equilibrium in the forward direction.
As the reaction is exothermic (heat is released),a decrease in temperature will shift the equilibrium in the forward direction.
Therefore,high pressure and low temperature are the favourable conditions for the formation of $NH_3$.

(A) The formation of $NH_{3(g)}$ is given by the reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$; $\Delta H < 0$ (exothermic reaction).
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction,leading to a higher yield of the product.
Given $T_1 < T_2$,the yield of $NH_{3(g)}$ will be higher at $T_1$ than at $T_2$ for any given pressure.
Therefore,the curve for $T_1$ must lie above the curve for $T_2$.
This corresponds to the graph where the $T_1$ curve is higher than the $T_2$ curve.

(A) The reaction between ferric ions and thiocyanate ions is given by: $Fe^{3+} (aq) + SCN^- (aq) \rightleftharpoons [Fe(SCN)]^{2+} (aq)$ (red colour).
When oxalic acid $(C_2H_2O_4)$ is added,it reacts with $Fe^{3+}$ ions to form a stable complex $[Fe(C_2O_4)_3]^{3-}$,which decreases the concentration of $Fe^{3+}$,shifting the equilibrium to the left and decreasing the red colour intensity.
When mercuric chloride $(HgCl_2)$ is added,it reacts with $SCN^-$ ions to form a stable complex $[Hg(SCN)_4]^{2-}$,which decreases the concentration of $SCN^-$,shifting the equilibrium to the left and decreasing the red colour intensity.
Therefore,both $(I)$ and $(II)$ will decrease the intensity of the red colour.

(C) For a chemical equilibrium,the addition of an inert gas at constant volume does not change the partial pressures of the reacting species.
Since the total volume of the container remains constant,the concentration of each reactant and product remains unchanged.
According to Le Chatelier's principle,if the concentrations of the reacting species do not change,the position of the equilibrium remains unaffected.
Therefore,the addition of inert $Ar$ gas at constant volume will not shift the equilibrium for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.