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

(D) According to Le Chatelier's principle,if a product is removed from the reaction mixture,the equilibrium shifts in the forward direction to compensate for the loss.
In a lime kiln,$CO_2$ gas is allowed to escape continuously.
This constant removal of $CO_2$ drives the reaction to completion in the forward direction.

(C) The reaction is $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$.
According to Le Chatelier's principle,increasing the concentration of a reactant shifts the equilibrium in the forward direction to consume the added reactant.
Adding $Cl_2$ (a reactant) at constant volume increases its concentration,thereby shifting the equilibrium to the right.

(A) According to Le Chatelier's principle,a change in the volume of the system affects the equilibrium position only if there is a change in the total number of moles of gaseous species $(\Delta n_g \neq 0)$.
For the reaction $N_{2(g)} O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the change in the number of gaseous moles is $\Delta n_g = 2 - (1 1) = 0$.
Since $\Delta n_g = 0$,a change in volume will not shift the equilibrium,and the number of moles remains unchanged.

(D) For the given reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$
The change in the number of moles of gaseous products and reactants is $\Delta n_g = n_{p} - n_{r} = 2 - (1 + 1) = 0$.
According to Le Chatelier's principle,if $\Delta n_g = 0$,a change in pressure or volume has no effect on the position of the equilibrium.
Since the pressure is halved and volume is doubled,the total number of moles of gas on both sides of the equation is equal,so the equilibrium position remains unchanged.

(D) The chemical reaction is: $2NaNO_3(s) \rightleftharpoons 2NaNO_2(s) + O_2(g)$.
According to Le Chatelier's principle:
$1$. Adding more reactant $(NaNO_3)$ shifts the equilibrium in the forward direction to consume the excess reactant.
$2$. Adding more product $(NaNO_2)$ shifts the equilibrium in the reverse direction to consume the excess product.
$3$. Increasing the pressure in a system with gaseous products shifts the equilibrium toward the side with fewer moles of gas. Here,the product side has $1 \text{ mole}$ of $O_2(g)$ and the reactant side has $0 \text{ moles}$ of gas. Thus,increasing pressure shifts the equilibrium to the left (reverse reaction).
Therefore,all the given statements are correct.

(C) For the number of moles at equilibrium to remain unaffected by a change in volume at constant temperature,the reaction must have $\Delta n_g = 0$ (where $\Delta n_g$ is the change in the number of moles of gaseous species).
Calculating $\Delta n_g$ for the given reactions:
$A$) $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$; $\Delta n_g = (1 + 3) - 2 = 2 \neq 0$.
$B$) $C_{(s)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{(g)}$; $\Delta n_g = 1 - 0.5 = 0.5 \neq 0$.
$C$) $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$; $\Delta n_g = 2 - (1 + 1) = 0$.
Since reaction $C$ has $\Delta n_g = 0$,the equilibrium composition remains unaffected by a change in volume. Thus,the correct option is $C$.

(A) The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + 92.3 \ kJ$.
Since the reaction is exothermic $(\Delta H < 0)$,according to Le Chatelier's principle,a decrease in temperature favors the forward reaction.
Therefore,increasing the temperature is unfavourable for the production of ammonia.
Increasing the pressure favors the side with fewer moles of gas (product side),and removing ammonia also shifts the equilibrium to the right.
Thus,the correct option is $A$.

(B) The correct answer is $(B)$.
Chemical equilibrium of a reversible reaction is not influenced by a catalyst.
$A$ catalyst only increases the rate of both the forward and backward reactions equally,thereby helping the system reach equilibrium faster without changing the position of the equilibrium.
Pressure,temperature,and the concentration of reactants do affect the position of the equilibrium according to Le Chatelier's Principle.

(C) The given reaction is $I_{2(g)} \rightleftharpoons 2I_{(g)}$ with $\Delta H_r^o = +150 \ kJ$.
Since $\Delta H_r^o > 0$,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 (towards the product side) to absorb the excess heat.
Therefore,increasing the temperature will shift the reaction towards the product.

(C) The reaction is endothermic $(\Delta H > 0)$,so according to Le Chatelier's principle,increasing the temperature will shift the equilibrium in the forward direction.
The change in the number of moles of gaseous products is $\Delta n_g = (2+1) - 2 = +1$.
Since $\Delta n_g > 0$,decreasing the pressure will shift the equilibrium in the direction of more moles of gas,which is the forward direction.
Therefore,increasing the temperature and decreasing the pressure will lead to the maximum decomposition of $H_2O_{(g)}$.

(A) According to Le Chatelier's principle,for an exothermic reaction $(\Delta H < 0)$,a decrease in temperature shifts the equilibrium in the forward direction to release heat.
In the given reaction,the number of moles of gaseous reactants is $1 + 4 = 5$,and the number of moles of gaseous products is $2$.
Since the number of moles decreases in the forward direction $(5 \to 2)$,an increase in pressure shifts the equilibrium towards the side with fewer moles,which is the forward direction.
Therefore,the formation of $AB_{4}$ is favoured at low temperature and high pressure.

(D) According to Le Chatelier's principle,for the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$,$\Delta H = -45.2 \ kcal$.
Since the reaction is exothermic $(\Delta H < 0)$,a decrease in temperature favors the forward reaction.
For the change in the number of moles of gas,$\Delta n_g = n_p - n_r = 2 - (2 + 1) = -1$.
Since $\Delta n_g < 0$,an increase in pressure favors the forward reaction (the side with fewer moles of gas).
Therefore,the formation of $SO_3$ is favored by increasing the pressure.

(A) The given reaction is $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.
For this reaction,the change in the number of moles of gaseous products and reactants is $\Delta n = (1 + 1) - 1 = 1$.
According to Le Chatelier's principle,an increase in pressure shifts the equilibrium in the direction where the number of moles of gas decreases.
Since $\Delta n > 0$,the forward reaction involves an increase in the number of moles of gas.
Therefore,increasing the pressure will shift the equilibrium in the backward direction,which decreases the dissociation of $PCl_5$.

(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 to form more $cis-2-pentene$.

(B) When in any system at equilibrium state,pressure,temperature,and concentration are changed,the equilibrium shifts in a direction that neutralizes the effect of the change. This is known as $Le \ Chatelier's \ principle$.

(A) The given reaction is $N_2 + O_2 \rightleftharpoons 2NO - Q \, cal$,which can be rewritten as $N_2 + O_2 + Q \, cal \rightleftharpoons 2NO$.
Since the reaction absorbs heat $(-Q)$,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.
Therefore,a high temperature is the essential condition for the higher production of $NO$.

(C) According to Le Chatelier's Principle,if a system at equilibrium is subjected to a change in concentration,temperature,or pressure,the system will shift its equilibrium position to counteract the effect of the disturbance. Therefore,the rate of the reaction that opposes the change will increase to restore equilibrium.

(C) According to Le Chatelier's principle,for a reaction at equilibrium,a decrease in pressure favors the direction in which the number of moles of gas increases.
In reaction $(C)$,$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$,the number of moles of gas increases from $1$ to $2$.
Therefore,low pressure favors the forward reaction in this case.

(A) Le Chatelier's principle is applicable to all systems,whether chemical or physical,that are in a state of equilibrium.
According to this principle,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 or nullify the effect of the applied change.

(C) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + Q \ \text{cal}$.
Since the reaction is exothermic (indicated by $+ Q \ \text{cal}$),according to Le Chatelier's principle,a low temperature will shift the equilibrium in the forward direction to produce more $SO_3$.
Also,the number of moles of gaseous reactants is $3$ ($2$ moles of $SO_2$ and $1$ mole of $O_2$) and the number of moles of gaseous products is $2$ $(SO_3)$.
Since the volume decreases from left to right,a high pressure will shift the equilibrium in the forward direction to produce more $SO_3$.
Therefore,the most suitable conditions are low temperature and high pressure.

(A) The equilibrium is represented as: $\text{Ice (more volume)} \rightleftharpoons \text{Water (less volume)}$.
According to Le Chatelier's principle,increasing the pressure on a system at equilibrium shifts the equilibrium in the direction that results in a decrease in volume.
Since liquid water has a lower volume than ice at the same temperature,increasing the pressure shifts the equilibrium towards the formation of more water.

(C) The given reaction is $A + B \rightleftharpoons C + D + \text{heat}$.
Since heat is released as a product,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,decreasing the temperature shifts the equilibrium in the forward direction to produce more heat.

(C) The addition of a catalyst has no effect on the value of the equilibrium constant $(K_{eq})$.
$A$ catalyst increases the rate of both the forward and backward reactions by the same proportion by lowering the activation energy.
Since both rates are increased equally,the position of equilibrium and the ratio of the mixture at equilibrium remain unchanged.
Catalysts can exist in solid,liquid,or gaseous states,not just in solution.

(A) According to Le $Chatelier$ principle,for an endothermic process like melting $(Solid + Heat \rightleftharpoons Liquid)$,providing heat shifts the equilibrium in the forward direction.
As the equilibrium shifts forward,the solid converts into liquid.
Therefore,the quantity of solid will reduce.

(C) The given reaction is $A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)} + Q \ kJ$.
Since heat $(Q)$ is released,the reaction is exothermic,meaning $\Delta H < 0$.
According to Le Chatelier's principle,for an exothermic reaction,a low temperature favors the forward direction.
For the pressure effect,we look at the change in the number of moles of gaseous species: $\Delta n_g = n_p - n_r = 1 - (1 + 2) = -2$.
Since $\Delta n_g < 0$,an increase in pressure favors the side with fewer moles,which is the product side.
Therefore,high pressure and low temperature favor the forward reaction.

(B) The given reaction is $C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g)$ with $\Delta H = -32.7 \ kcal$.
Since $\Delta H$ is negative,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature shifts the equilibrium in the forward direction.
Therefore,reducing the temperature will increase the yield and equilibrium concentration of the product $C_2H_6$.

(C) For the given equilibrium reaction: $2A + 3B \rightleftharpoons 3A + 2B$
The change in the number of moles of gaseous products and reactants is calculated as: $\Delta n_g = (3 + 2) - (2 + 3) = 5 - 5 = 0$.
According to Le Chatelier's principle,if $\Delta n_g = 0$,the change in pressure has no effect on the position of the equilibrium.
Therefore,the correct option is $(c)$.

(B) The given reaction is $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)} + 14.6 \ kcal$.
Since heat is released in the forward direction,the forward reaction is exothermic and the reverse reaction is endothermic.
According to Le Chatelier's principle,an increase in temperature shifts the equilibrium in the direction that absorbs heat (endothermic direction).
Therefore,an increase in temperature will favour the reverse reaction,which is the decomposition of $N_2O_4$ into $NO_2$.

(C) According to $Le \ Chatelier's \ principle$,if the concentration of one or more products is increased in a system at equilibrium,the system will shift in the reverse or backward direction to counteract the change and re-establish equilibrium.
Therefore,option $(C)$ is correct.

(A) The reaction for the formation of $SO_3$ is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + \text{Heat}$.
According to Le Chatelier's principle,to shift the equilibrium in the forward direction (to produce more $SO_3$):
$1$. $A$ catalyst $(V_2O_5)$ is used to increase the rate of reaction.
$2$. An optimum temperature (around $673-723 K$) is maintained.
$3$. $A$ higher concentration of reactants ($SO_2$ and $O_2$) is used to favor the product formation.
Therefore,the correct conditions are catalyst,optimum temperature,and higher concentration of reactants.

(C) The given reaction is $2 X_{(g)} Y_{(g)} \rightleftharpoons 2 Z_{(g)} 80 \ kcal$.
Since the reaction is exothermic $(\Delta H < 0)$,according to Le Chatelier's principle,a low temperature will shift the equilibrium in the forward direction to produce more $Z$.
There are $3$ moles of gaseous reactants $(2 1)$ and $2$ moles of gaseous products. Since the number of moles decreases in the forward reaction,high pressure will shift the equilibrium in the forward direction to produce more $Z$.
Therefore,the combination of high pressure $(1000 \ atm)$ and low temperature $(100 \ ^oC)$ gives the highest yield of $Z$.

(B) According to Le Chatelier's principle,when a product $(CN_{(aq)}^{-})$ is added to a system at equilibrium,the reaction shifts in the backward direction to counteract the change.
This shift consumes $H_{(aq)}^{+}$ ions and $CN_{(aq)}^{-}$ ions to form more $HCN_{(aq)}$.
Therefore,the concentration of $H_{(aq)}^{+}$ ions decreases.

(B) The given reaction is $H_2X_2 + \text{heat} \rightleftharpoons 2HX$.
$1$. The reaction is endothermic because heat is absorbed in the forward direction.
$2$. According to Le Chatelier's principle,an increase in temperature favors the endothermic reaction.
$3$. The number of moles of gaseous products ($2$ moles of $HX$) is greater than the number of moles of gaseous reactants ($1$ mole of $H_2X_2$).
$4$. According to Le Chatelier's principle,a decrease in pressure favors the direction with more moles of gas.
$5$. Therefore,high temperature and low pressure favor the formation of $HX$.

(B) 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.
Since endothermic reactions absorb heat $(\Delta H > 0)$,an increase in temperature favours the endothermic reaction.
Therefore,the correct option is $(B)$.

(C) The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ with $\Delta H = -93.6 \ kJ$.
Since $\Delta H < 0$,the reaction is exothermic. According to Le Chatelier's principle,lowering the temperature shifts the equilibrium to the right,increasing the yield of $NH_3$.
The total number of moles of gaseous reactants is $1 + 3 = 4$,and the number of moles of gaseous products is $2$.
Increasing the pressure or decreasing the volume of the reaction vessel shifts the equilibrium toward the side with fewer moles,which is the product side,thereby increasing the yield of $NH_3$.
Conversely,lowering the pressure shifts the equilibrium toward the side with more moles (the reactant side),which decreases the yield of $NH_3$.

(A) According to Le Chatelier's principle,a change in pressure affects the equilibrium position only if there is a change in the total number of moles of gaseous species between reactants and products (i.e.,$\Delta n_g \neq 0$).
If $\Delta n_g = 0$,the equilibrium remains unaffected by a change in pressure.
For option $A$: $\Delta n_g = 2 - (1 + 1) = 0$.
For option $B$: $\Delta n_g = 2 - (2 + 1) = -1$.
For option $C$: $\Delta n_g = 3 - 2 = 1$.
For option $D$: $\Delta n_g = 1 - 2 = -1$.
Since $\Delta n_g = 0$ for reaction $A$,it is unaffected by pressure changes.

(A) According to Le Chatelier's principle,for an endothermic reaction,the forward reaction is favored by an increase in temperature.
Since the reaction $(M + N \rightleftharpoons P)$ is endothermic $(\Delta H > 0)$,raising the temperature will shift the equilibrium to the right,thereby increasing the formation of product $P$.

(A) The given reaction is $N_2 + O_2 \rightleftharpoons 2NO - 43,200 \ kcal$.
This can be rewritten as $N_2 + O_2 + 43,200 \ kcal \rightleftharpoons 2NO$.
Since the reaction absorbs heat (enthalpy change $\Delta H > 0$),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.
Therefore,an increase in temperature will increase the yield of $NO$.

(C) The reaction is $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) + Q \ kcal$.
Since the reaction is exothermic $(+Q \ kcal)$,according to Le Chatelier's principle,a low temperature favors the forward reaction to increase the yield of $NH_3$.
Since the number of moles of gaseous products $(2)$ is less than the number of moles of gaseous reactants $(1+3=4)$,high pressure favors the forward reaction to increase the yield of $NH_3$.
Therefore,low temperature and high pressure are the ideal conditions for maximum yield.

(C) The reaction is $PCl_{5(s)} \rightleftharpoons PCl_{3(s)} + Cl_{2(g)}$.
Since $PCl_{5(s)}$ and $PCl_{3(s)}$ are in the solid state,their active masses are taken as $1$.
The equilibrium constant expression is $K_c = [Cl_{2(g)}]$.
Because the equilibrium constant $K_c$ depends only on temperature,and the temperature is constant,the concentration of $Cl_{2(g)}$ at equilibrium remains unchanged regardless of the amount of solid $PCl_5$ added,provided some solid is present.

(A) According to Le Chatelier’s principle,when the total pressure of a system at equilibrium is increased,the equilibrium shifts in the direction that produces a smaller number of moles of gaseous species to counteract the pressure increase.
For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$,the number of moles of gaseous reactants is $1 + 3 = 4$,and the number of moles of gaseous products is $2$.
Since the product side has fewer moles $(2 < 4)$,increasing the total pressure will shift the equilibrium to the right.
For options $B$ and $C$,the number of moles of gaseous reactants equals the number of moles of gaseous products $(2 = 2)$,so pressure change has no effect.
For option $D$,the product side has more moles $(2 > 1)$,so increasing pressure shifts the equilibrium to the left.

(B) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with a smaller number of gaseous moles.
For option $A$: $\Delta n_g = (1+1) - 2 = 0$. No effect.
For option $B$: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$. $\Delta n_g = 2 - (2+1) = -1$. Since the product side has fewer moles,increasing pressure increases the yield of $SO_3$.
For option $C$: $\Delta n_g = 2 - (1+1) = 0$. No effect.
For option $D$: $\Delta n_g = (1+1) - (1+1) = 0$. No effect.
Therefore,the correct option is $B$.

(A) According to Le Chatelier's principle,for a gaseous reaction,an increase in pressure shifts the equilibrium towards the side with fewer moles of gas.
In the given reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$,the reactant side has $2$ moles of gas ($1$ mole of $A$ + $1$ mole of $B$) and the product side has $1$ mole of gas $(C)$.
Since the product side has fewer moles,an increase in pressure will favour the forward reaction.
Conversely,a decrease in pressure will favour the side with more moles of gas,which is the reactant side.
Therefore,the backward reaction is favoured by a decrease in pressure.

(D) According to Le Chatelier's principle,for the reaction $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g) + \text{heat}$:
$1$. The reaction is exothermic (releases heat). Therefore,decreasing the temperature shifts the equilibrium to the right to produce more heat,favoring the formation of $NO_2$.
$2$. The total number of moles of gaseous reactants is $3$ ($2$ moles of $NO$ and $1$ mole of $O_2$),while the number of moles of gaseous products is $2$ ($2$ moles of $NO_2$). Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas,which is the product side,thus favoring the formation of $NO_2$.
Therefore,both high pressure and low temperature favor the formation of $NO_2$.

(C) For the reaction,$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} Cl_{2(g)}$,the total number of moles of gaseous products is $2$ and reactants is $1$.
According to Le Chatelier's principle,for a reaction where the number of moles of gaseous products is greater than the number of moles of gaseous reactants $(\Delta n_g > 0)$:
$1$. Introducing an inert gas at constant volume does not change the partial pressures of the reacting species,so the equilibrium remains unaffected.
$2$. Introducing chlorine gas $(Cl_2)$ increases the concentration of products,which shifts the equilibrium in the backward direction.
$3$. Introducing an inert gas at constant pressure increases the volume of the container. Since the number of moles of products is greater than reactants,the system shifts towards the side with more moles to counteract the increase in volume,thus favouring the forward reaction.
$4$. Decreasing the volume of the container increases the total pressure,which shifts the equilibrium towards the side with fewer moles (backward direction).

(D) The production of ammonia by Haber's process is given by the reaction: $N_2(g) + 3H_2(g) leftharpoons 2NH_3(g); \Delta H = -92.4 \ kJ \ mol^{-1}$.
$1.$ According to Le Chatelier's principle, increasing the concentration of reactants ($N_2$ and $H_2$) shifts the equilibrium to the , favoring the formation of $NH_3$.
$2.$ Since the reaction is exothermic, low temperature favors the forward reaction. High pressure is favorable because the number of moles of gaseous products $(2)$ is less than the number of moles of gaseous reactants $(4)$.
$3.$ Continuous removal of $NH_3$ from the reaction mixture shifts the equilibrium to the , promoting further production of $NH_3$.
Therefore, all the given conditions are favorable.

(D) Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions,such as concentration,pressure,or temperature,the position of equilibrium shifts to counteract the change to re-establish a new equilibrium.
Therefore,all three factors—concentration,pressure,and temperature—influence a chemical system at equilibrium.

(B) The given equilibrium reaction is $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$.
According to Le Chatelier's principle,when pressure is increased,the equilibrium shifts in the direction that decreases the total number of moles of gaseous species.
In this reaction,the number of moles of gaseous reactants is $1$ $(H_2O_{(g)})$ and the number of moles of gaseous products is $2$ $(CO_{(g)} + H_{2(g)})$.
Since the number of gaseous moles on the product side $(2)$ is greater than on the reactant side $(1)$,an increase in pressure will shift the equilibrium towards the side with fewer gaseous moles,which is the reactant side.
Therefore,the equilibrium will shift in the backward direction.

(D) According to Le Chatelier's principle,for an exothermic reaction $(\Delta H < 0)$,increasing the concentration of reactants shifts the equilibrium to the right to produce more products.
Since the reaction is $Cl_{2(g)} + 3F_{2(g)} \rightleftharpoons 2ClF_{3(g)}$,adding more $F_2$ (a reactant) will shift the equilibrium towards the product side,thereby increasing the quantity of $ClF_3$.

(D) When an inert gas is added to a system at equilibrium at constant pressure,the total volume of the system increases.
As a result,the concentration (moles per unit volume) of all gaseous reactants and products decreases.
According to Le Chatelier's principle,the equilibrium will shift in the direction that increases the total number of moles of gaseous species to counteract the decrease in concentration.
If the number of moles of gaseous products is greater than the number of moles of gaseous reactants,the reaction shifts towards the product side.
If the number of moles of gaseous reactants is greater than the number of moles of gaseous products,the reaction shifts towards the reactant side.
If the total number of moles of gaseous reactants equals the total number of moles of gaseous products,the equilibrium remains unaffected.