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

(A) The given equilibrium is $Ice \rightleftharpoons Water$.
According to Le Chatelier's principle,for an endothermic process (melting of ice),an increase in temperature favors the forward reaction,producing more water.
Since the density of water is greater than that of ice,the volume decreases during melting $(V_{water} < V_{ice})$.
According to Le Chatelier's principle,an increase in pressure favors the direction in which volume decreases.
Therefore,both an increase in temperature and an increase in pressure will shift the equilibrium to the right,resulting in a higher amount of water.

(A) The equilibrium constant $K$ decreases from $2$ to $1 \times 10^{-3}$ as the temperature increases from $273 \ K$ to $298 \ K$.
According to Le Chatelier's principle,for an exothermic reaction,the equilibrium constant decreases with an increase in temperature.
Since the value of $K$ decreases as the temperature increases,the forward reaction (formation of $C$ and $D$) is exothermic.

(D) According to Le Chatelier's principle,the addition of an inert gas at constant volume does not change the partial pressures of the reacting species.
Since the partial pressures remain constant,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 equilibrium position.

(A) The equilibrium reaction is $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$.
According to Le Chatelier's principle,when pressure is increased,the equilibrium shifts in the direction that reduces the number of moles of gas.
Since the gaseous phase occupies a larger volume than the liquid phase,increasing the pressure favors the liquid phase.
Therefore,a higher temperature is required to overcome the increased pressure and convert the liquid into vapor,which results in an increase in the boiling point of $H_2O$.

(B) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with fewer moles of gaseous species.
For the given reaction: $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$
Moles of gaseous reactants = $1$ $(H_2O)$
Moles of gaseous products = $2$ $(CO + H_2)$
Since the number of moles of gaseous products $(2)$ is greater than the number of moles of gaseous reactants $(1)$,an increase in pressure will shift the equilibrium in the backward direction to reduce the pressure.

(C) The reaction is $2B_{(g)} + 2C_{(g)} \rightleftharpoons 3A_{(g)} + \text{heat}$.
$1$. Effect of Temperature: The reaction is exothermic (heat is released). According to Le Chatelier's principle,for an exothermic reaction,a low temperature favors the forward direction to produce more product $A$.
$2$. Effect of Pressure: The change in the number of gaseous moles is $\Delta n_g = 3 - (2 + 2) = -1$. Since $\Delta n_g < 0$,an increase in pressure shifts the equilibrium toward the side with fewer moles,which is the product side. Thus,high pressure favors the formation of $A$.
Therefore,the necessary conditions are low temperature and high pressure.

(C) According to $Le \ Chatelier's \ principle$,if a product is continuously removed from the reaction mixture,the equilibrium shifts in the forward direction to compensate for the loss,eventually driving the reaction to completion.

(D) The correct answer is $D$.
According to Le Chatelier's principle,the equilibrium position and the equilibrium concentrations of reactants and products are affected by changes in temperature,pressure,or concentration.
However,a catalyst provides an alternative pathway with a lower activation energy,increasing the rate of both the forward and backward reactions equally.
Therefore,a catalyst helps the system reach equilibrium faster but does not change the position of the equilibrium or the equilibrium concentrations of the reactants and products.

(A) According to Le Chatelier's principle:
$1$. For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + 22 \ kcal$,the forward reaction is exothermic. Therefore,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 $1 + 3 = 4$.
$3$. According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with fewer moles of gas.
$4$. Thus,an increase in pressure favors the formation of $NH_3$.

(C) According to Le Chatelier's principle,an increase in pressure favors the direction that has a smaller number of moles of gaseous species.
For the given reaction: $2A_{(g)} + 3B_{(g)} \rightleftharpoons 3C_{(g)} + 2D_{(g)}$
Total moles of gaseous reactants = $2 + 3 = 5$
Total moles of gaseous products = $3 + 2 = 5$
Since the number of moles of gaseous reactants is equal to the number of moles of gaseous products $(\Delta n_g = 0)$,a change in pressure will have no effect on the equilibrium position.

(B) The given reaction is $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)} + 21.9 \ kcal$.
Since the heat is released on the product side,the reaction is exothermic $(\Delta H < 0)$.
According to Le Chatelier's principle,for an exothermic reaction,an increase in temperature shifts the equilibrium in the backward direction to absorb the excess heat.
Therefore,the backward reaction is favored,which means $N_2O_4$ will decompose into $NO_2$.
Thus,an increase in temperature favors the decomposition of $N_2O_4$.

(B) According to Le Chatelier's principle,for a gaseous reaction,a decrease in pressure favours the direction in which the number of moles of gaseous products is greater than the number of moles of gaseous reactants (i.e.,$\Delta n_g > 0$).
Let us analyze the change in the number of moles $(\Delta n_g)$ for each reaction:
$(A)$ $H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \implies \Delta n_g = 2 - (1 + 1) = 0$
$(B)$ $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \implies \Delta n_g = (1 + 1) - 1 = +1$
$(C)$ $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \implies \Delta n_g = 2 - (1 + 3) = -2$
$(D)$ $N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \implies \Delta n_g = 2 - (1 + 1) = 0$
Since reaction $(B)$ has $\Delta n_g > 0$,it is favoured by low pressure.

(B) The given reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} + q$.
Here,$q$ represents the heat released,meaning the reaction is exothermic $(\Delta H < 0)$.
According to Le Chatelier's principle,for an exothermic reaction,a low 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 $(3 \rightarrow 2)$,high pressure favors the forward reaction.
Therefore,low temperature and high pressure favor the formation of $SO_3$.

(A) According to Le Chatelier's principle,for a gaseous reaction,increasing the pressure shifts the equilibrium towards the side with a smaller number of moles of gaseous species.
If the number of moles of gaseous reactants equals the number of moles of gaseous products (i.e.,$\Delta n_g = 0$),then the pressure change has no effect on the equilibrium position.
In reaction $(A)$: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the number of moles of gaseous reactants is $1 + 1 = 2$ and the number of moles of gaseous products is $2$. Thus,$\Delta n_g = 2 - 2 = 0$.
Therefore,increasing the pressure will not shift the equilibrium and the amount of products will not increase.

(C) According to $Le \ Chatelier's$ principle,when pressure is increased on a system at equilibrium,the system shifts in the direction that reduces the volume.
Since the density of water is greater than that of ice,the melting of ice $(Ice \rightarrow Water)$ is accompanied by a decrease in volume.
Therefore,increasing the pressure favors the forward reaction,resulting in the formation of more water.

(B) $1$. The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + 21.9 \ kcal$.
$2$. Since the reaction is exothermic $(\Delta H < 0)$,according to Le Chatelier's principle,low temperature favors the forward reaction.
$3$. The number of moles of gaseous products $(2)$ is less than the number of moles of gaseous reactants $(1+3=4)$. Therefore,high pressure favors the forward reaction (shift towards fewer moles).
$4$. $A$ catalyst is used to increase the rate of the reaction to reach equilibrium faster.
Thus,the favorable conditions are low temperature,high pressure,and the presence of a catalyst.

(D) According to Le Chatelier's principle,if the volume of the container is increased (doubled),the pressure decreases. The system will shift in the direction that increases the number of moles of gas to counteract this change.
For a reaction to shift to the right (forward direction) upon increasing the volume,the number of moles of gaseous products must be greater than the number of moles of gaseous reactants $(\Delta n_g > 0)$.
Checking the options:
$A$: $\Delta n_g = 2 - (2 + 1) = -1$
$B$: $\Delta n_g = 2 - (1 + 1) = 0$
$C$: $\Delta n_g = 2 - (1 + 3) = -2$
$D$: $\Delta n_g = (1 + 1) - 1 = +1$
Since $\Delta n_g > 0$ only for option $D$,the equilibrium will shift to the right.

(B) The given reaction is: ${N_2}_{(g)} + {O_2}_{(g)} \rightleftharpoons 2NO - \text{heat}$.
This can be rewritten as: ${N_2}_{(g)} + {O_2}_{(g)} + \text{heat} \rightleftharpoons 2NO$.
Since the reaction involves the absorption of heat,it is an endothermic reaction $(\Delta H > 0)$.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium in the forward direction to absorb the extra heat.
Therefore,high temperature favors the production of $NO$.

(A) According to Le Chatelier's principle,if the concentration of a product $(trans-2-pentene)$ is increased in a system at equilibrium,the system will shift in the direction that consumes the added substance to counteract the change.
Therefore,the equilibrium will shift in the backward direction (towards the reactants) to produce more $cis-2-pentene$.

(C) The dissociation of molecular hydrogen is represented by the equation: $H_{2(g)} \rightleftharpoons 2H_{(g)}$.
This reaction is endothermic $(\Delta H > 0)$ because energy is required to break the $H-H$ bond.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature favors the forward reaction.
Regarding pressure,the number of moles of gas increases from $1$ mole on the reactant side to $2$ moles on the product side.
According to Le Chatelier's principle,a decrease in pressure favors the side with more moles of gas,which is the product side.
Therefore,high temperature and low pressure favor the dissociation of molecular hydrogen.

(B) The given reaction is $2X_{(g)} + Y_{(g)} \rightleftharpoons 2Z_{(g)} + 80 \ kcal$.
Since the reaction is exothermic $(\Delta H < 0)$,according to Le Chatelier's principle,a lower temperature will shift the equilibrium in the forward direction to produce more $Z$.
For 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,increasing the pressure will shift the equilibrium in the forward direction to produce more $Z$.
Therefore,high pressure and low temperature are the ideal conditions.
Comparing the given options,$1000 \ atm$ (high pressure) and $100 \ ^\circ C$ (lowest temperature among options) will yield the maximum amount of $Z$.

(C) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.
Given $K_p > K_c$,it implies that $(RT)^{\Delta n_g} > 1$.
Since $T > 15 \ K$ and $R$ is a positive constant,this condition is satisfied when $\Delta n_g > 0$.
$\Delta n_g$ is the change in the number of moles of gaseous products and reactants,defined as $\Delta n_g = n_p - n_r$.
When $\Delta n_g > 0$,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 with $\Delta n_g > 0$,decreasing the pressure favors the forward reaction (the side with more moles of gas).
Therefore,the forward reaction is favored by low pressure.

(C) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with a smaller number of gaseous moles.
For the given reaction: $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$.
Number of gaseous moles on the reactant side = $1$ $(H_2O)$.
Number of gaseous moles on the product side = $1 + 1 = 2$ ($CO$ and $H_2$).
Since the product side has more gaseous moles,an increase in pressure will shift the equilibrium to the side with fewer gaseous moles,which is the reactant side (backward direction).

(B) The given reaction is $SO_2(g) + Cl_2(g) \rightleftharpoons SO_2Cl_2(g) + \text{Heat}$.
According to Le Chatelier's principle,adding $SO_2$ (a reactant) will shift the equilibrium in the forward direction to consume the added $SO_2$.
Since the forward reaction is exothermic,it releases heat into the system.
Therefore,the temperature of the system will increase.

(C) According to Le Chatelier's principle:
$1$. For an endothermic reaction $(\Delta H > 0)$,high temperature favors the forward reaction (higher yield).
$2$. For a reaction where the number of moles of gaseous products is less than the number of moles of gaseous reactants $(\Delta n_g < 0)$,high pressure favors the forward reaction.
Analyzing the options:
- Option $A$: Endothermic $(\Delta H > 0)$,$\Delta n_g = (1+3) - 1 = +3$. High temperature helps,but high pressure does not.
- Option $B$: Exothermic $(\Delta H < 0)$,$\Delta n_g = 2 - (1+3) = -2$. High pressure helps,but high temperature does not.
- Option $C$: Endothermic $(\Delta H > 0)$,$\Delta n_g = 2 - (1+2) = -1$. High temperature helps,and high pressure helps.
- Option $D$: Exothermic $(\Delta H < 0)$,$\Delta n_g = (2+7) - 2 = +7$. Neither high temperature nor high pressure helps.
Therefore,option $C$ is the correct answer.

(C) $1$. According to Le Chatelier's principle,adding $CO$ to the second equilibrium $COCl_{2(g)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$ will shift the equilibrium to the left to consume the added $CO$.
$2$. This shift causes the concentration of $Cl_2$ to decrease.
$3$. Now,consider the first equilibrium $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. Since the concentration of $Cl_2$ has decreased,the system will shift to the right to produce more $Cl_2$ to restore equilibrium.
$4$. As the first equilibrium shifts to the right,$PCl_5$ dissociates to form more $PCl_3$ and $Cl_2$.
$5$. Consequently,the concentration of $PCl_5$ decreases.

(B) $1$. According to Le Chatelier's principle,since the reaction is endothermic $(\Delta H = +69 \ kcal)$,an increase in temperature will shift the equilibrium in the forward direction.
$2$. The number of moles of gas on the reactant side is $3$,and on the product side is $2$. According to the principle,an increase in pressure will shift the equilibrium towards the side with fewer moles of gas,which is the forward direction.
$3$. Therefore,the reaction is favoured by high temperature and high pressure.

(A) The melting of ice is represented by the equilibrium: $H_2O(s) \rightleftharpoons H_2O(l)$.
According to Le Chatelier's principle,since the density of liquid water is higher than that of ice,increasing the pressure favors the formation of the denser phase (liquid water).
Additionally,melting is an endothermic process,so increasing the temperature favors the forward reaction.
Therefore,high temperature and high pressure are favorable for the melting of ice.

(B) According to Le Chatelier's principle,when an inert gas is added to a system at constant pressure,the volume of the system increases to maintain the pressure.
This results in a decrease in the partial pressure of the reacting species.
For a decomposition reaction (e.g.,$A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$),the total number of moles of gaseous products is greater than the number of moles of gaseous reactants $(\Delta n_g > 0)$.
Increasing the volume shifts the equilibrium in the direction of the greater number of moles,which is the product side.
Therefore,more products will be produced.

(C) The reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ is exothermic $(\Delta H < 0)$ and involves a decrease in the number of moles of gaseous species (from $4$ moles to $2$ moles).
According to Le Chatelier's principle,high pressure favors the forward reaction.
Although low temperature favors the forward reaction for an exothermic process,the reaction has a high activation energy $(E_a)$.
Therefore,in a continuous flow process,an elevated optimum temperature is required to ensure a sufficient rate of reaction to produce $NH_3$ efficiently.

(C) According to Le Chatelier's principle,an increase in temperature favors the endothermic direction of a reaction.
Reaction $(a)$ is the water-gas shift reaction,which is exothermic $(\Delta H < 0)$.
Reaction $(b)$ is the contact process step,which is exothermic $(\Delta H < 0)$.
Reaction $(c)$ represents the thermal decomposition of water,which is highly endothermic $(\Delta H > 0)$.
Reaction $(d)$ is the Deacon process,which is exothermic $(\Delta H < 0)$.
Therefore,only reaction $(c)$ will shift to the right side upon increasing the temperature.

(A) catalyst provides an alternate reaction pathway with a lower activation energy $(E_a)$.
It speeds up both the forward and backward reactions to the same extent.
Since the rate of both reactions increases equally,the equilibrium position remains unchanged,and the equilibrium constant $(K_{eq})$ is not altered.

(A) The given reaction is $A_{2(g)} + B_{2(g)} \rightleftharpoons X_{2(g)}$.
$1$. The enthalpy change $\Delta_{r} H = -X \ kJ$ indicates that the reaction is exothermic $(\Delta H < 0)$. According to Le Chatelier's principle,for an exothermic reaction,a low temperature favours the formation of the product.
$2$. The change in the number of gaseous moles is $\Delta n_{g} = n_{p} - n_{r} = 1 - (1 + 1) = -1$. Since $\Delta n_{g} < 0$,an increase in pressure will shift the equilibrium towards the side with fewer moles,which is the product side.
Therefore,low temperature and high pressure favour the maximum formation of the product.

(N/A) Ammonia is prepared using the Haber's process. The reaction is given by: $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g); \Delta H = -92.4 \, kJ \, mol^{-1}$.
The yield of ammonia can be maximized under the following conditions:
$(i)$ High pressure: $A$ pressure of $\sim 200 \, atm$ is used to shift the equilibrium towards the product side as there is a decrease in the number of moles.
$(ii)$ Optimum temperature: $A$ temperature of $\sim 700 \, K$ is used. Although the reaction is exothermic,this temperature is a compromise to ensure a reasonable rate of reaction.
$(iii)$ Catalyst: Use of a catalyst such as iron oxide mixed with small amounts of $K_{2}O$ and $Al_{2}O_{3}$ to increase the rate of attainment of equilibrium.

(N/A) Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in concentration,temperature,or pressure,the system will adjust itself to counteract the effect of the change and establish a new equilibrium state.

(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) According to Le Chatelier's principle,increasing the pressure shifts the equilibrium toward the side with fewer moles of gaseous species.
$(i)$ $\Delta n_g = (1+1) - 1 = 1$. Pressure increase shifts it backward.
$(ii)$ $\Delta n_g = (1+2) - (1+2) = 0$. No effect.
$(iii)$ $\Delta n_g = 2 - 1 = 1$. Pressure increase shifts it backward.
$(iv)$ $\Delta n_g = 1 - (2+1) = -2$. Pressure increase shifts it forward.
$(v)$ $\Delta n_g = 1 - 0 = 1$. Pressure increase shifts it backward.
$(vi)$ $\Delta n_g = (4+6) - (4+5) = 1$. Pressure increase shifts it backward.

(N/A) According to Le Chatelier's principle,on addition of $H_2$,the concentration of reactants increases,so the equilibrium will shift in the forward direction to consume the added $H_2$.
$(b)$ On addition of $CH_3OH$,the concentration of the product increases,so the equilibrium will shift in the backward direction to counteract the increase.
$(c)$ On removing $CO$,the concentration of a reactant decreases,so the equilibrium will shift in the backward direction to produce more $CO$.
$(d)$ On removing $CH_3OH$,the concentration of the product decreases,so the equilibrium will shift in the forward direction to produce more $CH_3OH$.

(A) The dissolution of $CO_2$ in water is an exothermic process,represented by the equilibrium: $CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq) + \text{heat}$.
According to Le Chatelier's Principle,increasing the temperature of an exothermic reaction shifts the equilibrium in the backward direction to absorb the excess heat.
Therefore,heating a saturated solution of $CO_2$ causes the equilibrium to shift to the left,resulting in the evolution of $CO_2$ gas from the solution.

(B) The Haber process for the industrial production of ammonia $(NH_3)$ is an exothermic reaction represented by: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$; $\Delta H = -92.4 \ kJ \ mol^{-1}$.
According to Le Chatelier's principle,low temperature favors the formation of ammonia,but a very low temperature makes the reaction rate extremely slow.
Therefore,an optimum temperature of approximately $700 \ K$ is maintained to balance the rate of reaction and the yield of ammonia.

(A) The industrial production of ammonia $(NH_3)$ is carried out by the Haber process according to the reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$; $\Delta H = -92.4 \ kJ \ mol^{-1}$.
According to Le Chatelier's principle,since the reaction is exothermic and involves a decrease in the number of moles,high pressure and low temperature favor the formation of ammonia.
However,at very low temperatures,the rate of reaction is very slow.
Therefore,an optimum temperature of about $700 \ K$ and a high pressure of about $200 \ atm$ are maintained to achieve a balance between yield and rate of reaction.

(N/A) One of the principal goals of chemical synthesis is $(i)$ to maximize the products and $(ii)$ to minimize the expenditure of energy.
This implies achieving the maximum yield of products under mild temperature and pressure conditions. If this is not achieved,the experimental conditions must be adjusted. For example,in the Haber process for the synthesis of ammonia from $N_2$ and $H_2$,the choice of experimental conditions is of significant economic importance.
The equilibrium constant,$K_c$,is independent of initial concentrations. However,if a system at equilibrium is subjected to a change in the concentration of one or more of the reacting substances,the system is no longer at equilibrium,and a net reaction occurs in a direction that allows the system to return to equilibrium.
Changes in temperature or pressure of the system may also alter the equilibrium state.
To determine the direction of the reaction and make a qualitative prediction about the effect of changing conditions on equilibrium,Le Chatelier's principle is used.
Le Chatelier's principle: "$A$ change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner as to reduce or counteract the effect of the change." This principle is applicable to all physical and chemical equilibria.

(N/A) The principal goals of chemical synthesis are $(i)$ to maximize the yield of products and $(ii)$ to minimize the expenditure of energy.
This implies achieving maximum yield under mild temperature and pressure conditions. If this is not achieved,experimental conditions must be adjusted. For example,in the Haber process for the synthesis of ammonia from $N_2$ and $H_2$,the choice of experimental conditions is of significant economic importance.
The equilibrium constant,$K_c$,is independent of initial concentrations. However,if a system at equilibrium is subjected to a change in the concentration of one or more reacting substances,the system is no longer at equilibrium. $A$ net reaction occurs in a direction that allows the system to return to equilibrium.
Changes in temperature or pressure can also alter the equilibrium state.
To predict the effect of changing conditions on equilibrium,$Le \ Chatelier's$ principle is used.
$Le \ Chatelier's$ principle states: "$A$ change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner as to reduce or counteract the effect of the change." This principle is applicable to all physical and chemical equilibria.

(N/A) According to Le Chatelier's principle,when the concentration of any of the reactants or products in a reaction at equilibrium is changed,the composition of the equilibrium mixture changes so as to minimize the effect of the concentration change.
$(i)$ If the concentration of a reactant is increased,the equilibrium shifts in the forward direction to consume the added reactant.
$(ii)$ If the concentration of a product is increased,the equilibrium shifts in the backward direction to consume the added product.
Example: Consider the synthesis of $HI$ at a constant temperature:
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$
If $H_{2(g)}$ is added to the reaction mixture at equilibrium,the system is disturbed. To restore equilibrium,the reaction proceeds in the forward direction,consuming $H_{2}$ and $I_{2}$ to form more $HI$. Consequently,the concentration of $HI$ increases,and the concentration of $I_{2}$ decreases until a new equilibrium state is reached.

(B) The equilibrium reaction is: $Fe^{3+}_{(aq)} + SCN^{-}_{(aq)} \rightleftharpoons [Fe(SCN)]^{2+}_{(aq)}$.
Adding $KSCN$ increases the concentration of $SCN^{-}$ ions.
According to Le Chatelier's principle,the equilibrium shifts in the direction that consumes the added reactant,which is the forward direction (right).
This results in an increase in the concentration of the blood-red complex $[Fe(SCN)]^{2+}$,thereby increasing the intensity of the red color.

(C) According to Le Chatelier's principle,an increase in the concentration of any of the reactants ($Fe^{3+}$ or $SCN^{-}$) will shift the equilibrium in the forward direction to consume the added species,thereby increasing the concentration of $[Fe(SCN)]^{2+}$ and the intensity of the deep red color.
Conversely,adding reagents like $HgCl_2$ that react with $SCN^{-}$ will decrease its concentration,shifting the equilibrium to the left.
Therefore,both statements $(A)$ and $(B)$ are correct.

(N/A) The effect of pressure change on equilibrium is only observed in reactions involving gaseous components.
$1$. If the total number of moles of gaseous products and reactants are the same,pressure change has no effect on the equilibrium position.
$2$. If the total number of moles of gaseous products and reactants are different,a change in pressure will shift the equilibrium to establish a new state.
$3$. According to Le Chatelier's principle,if the pressure is increased (volume decreased),the equilibrium shifts in the direction that produces fewer moles of gas to counteract the pressure increase.
Example: $CO_{(g)} + 3H_{2(g)} \rightleftharpoons CH_{4(g)} + H_{2}O_{(g)}$
In this reaction,the reactant side has $1 + 3 = 4$ moles of gas,while the product side has $1 + 1 = 2$ moles of gas.
If the pressure is increased (volume decreased),the system shifts towards the side with fewer moles,which is the forward direction (towards products).
Using the reaction quotient $Q_c$:
$Q_c = \frac{[CH_{4}][H_{2}O]}{[CO][H_{2}]^3}$
If the volume is halved,the concentration of each species doubles. The new $Q_c$ becomes smaller than $K_c$ $(Q_c < K_c)$,causing the reaction to proceed in the forward direction until a new equilibrium is established.