CBSE Class 11 Chemistry Notes For Le Chateuer’s Principle

Le Chateuer’s Principle

Equilibrium of a chemical reaction is established under some conditions such as pressure, temperature, Concentration, etc. Le Chatelier, a celebrated French chemist studied the effect of such conditions on a large number of chemical equilibria.

He summed up his observations regarding the effect of these factors on equilibrium in the form of generalization which is commonly known as Le Chatelier’s principle.

Le Chateuer’s Principle:

If a system under equilibrium Is subjected to a change In pressure, temperature, or concentration then the equilibrium will shift Itself in such a way as to reduce the effect of that change.

The effect of the change in the various conditions of the chemical reactions at equilibrium is discussed below based on Le Chatelier’s principle.

Effect of change in concentration of reactant or product at equilibrium of a reaction

According to Le Chatelier’s principle, at a constant temperature, keeping the volume fixed, if the concentration of reactant or product at equilibrium is changed, the equilibrium will shift in the direction in which the effect of change in concentration is reduced as far as possible.

With the help of the following general reaction, let us discuss how the change in concentration of reactant or product affects the equilibrium of a chemical reaction: A + B⇌ C+D

Read and Learn More CBSE Class 11 Chemistry Notes

Effect of addition of reactant to the reaction system at equilibrium at constant volume and temperature

Reaction: A+B⇌C+D

1. At constant temperature, keeping the volume un¬ changed, if a certain amount of reactant (A or J3) is added to the system at equilibrium, the concentration of that reactant will increase.
2. As a result, the reaction will no longer remain in the state of equilibrium.
3. According to Le Chatelier’s principle, the system will arrange itself in such a manner so that the effect of increased concentration of that reactant is reduced as far as possible. Naturally, the equilibrium will tend to shift in a direction that causes a decrease in the concentration of the added reactant.
4. Since the concentrations of the reactants reduce in the forward direction, the net reaction will occur in this direction until a new equilibrium is established when the rates of both the forward and reverse reactions become the same.
5. Therefore, the addition of reactant to the reaction system at equilibrium causes the equilibrium to shift to the right. As a result, the yields of the products (C andD) increase.

Conclusion:

At constant volume and temperature, when some quantity of reactant is added to a reaction system at equilibrium, the equilibrium shifts to the right, and the yield of product (s) increases.

Example: Reaction: N₂(g) + 3H₂(g) 2NH₃(g)

At constant temperature, keeping the volume fixed, if some quantity of H₂(g) is added to the above system at equilibrium, the net reaction will occur In the forward direction until a new equilibrium is established.

This means that the addition of some H₂(g) to the reaction system at equilibrium causes the equilibrium to shift to the tire right. As a result, the yield of NH₃(g) will increase.

Effect of addition of the product to the reaction system at equilibrium at constant volume and temperature:

Reaction: A+B ⇌C+D

At constant temperature, keeping the volume fixed, when some quantity of the product (C or D) is added to the system at equilibrium, the concentration of that product increases.

• As a result, the reaction will no longer remain in the state of equilibrium.
• According to Le Chatelier’s principle, the system will adjust itself in such a way that the effect of an increase in concentration of that product is reduced as far as possible. Naturally, the equilibrium of the system will try to shift in a direction that reduces the concentration of the added product.
• Since the concentrations of the product decrease in the reverse direction, the net reaction will occur in this direction until a new equilibrium is established when the rates of both the forward and reverse reactions become identical.
• Therefore, the addition of aproduct to the reaction system at equilibrium makes the equilibrium shift to the left. As a result, the concentrations of the reactants (4 and B) increase.

Conclusion:

At constant volume and temperature, if some quantity of product is added to a reaction system remaining at equilibrium, then the equilibrium will shift to the left. As a result, the concentration of product(s) decreases, whereas that of the reactant(s) increases.

Effect of removal of reactant from the reaction system at equilibrium at constant volume and temperature:

Reaction: A+B⇌C+D

At constant temperature without changing the volume If some amount of reactant (4 orB) is removed from the system at equilibrium, the concentration of that reactant decreases.

• As a result, the reaction will no longer exist in the state of equilibrium.
• According to Le Chatelier’s principle, the system will adjust itself in such a manner so that the effect of a decrease in concentration of that reactant is reduced as far as possible. Naturally, the equilibrium of the system will shift in a direction that increases the concentration of that reactant.
• Since the concentration of the reactants increases in the reverse direction, the net reaction will occur in this direction until a new equilibrium is established when both the forward and reverse reactions take place at equal rates.
• Therefore, the removal of a reactant from the reaction system results in shifting the equilibrium position to the left. As a result, the yield of the products (C and D) will decrease and that of the reactants (A and B) will increase.

Conclusion:

At constant volume and temperature, if some quantity of reactant(s) is removed from the reaction system at equilibrium, the equilibrium will shift to the left. As a result, the yield of product(s) decreases, whereas that of the reactant(s) increases

Effect of removal of the product from the reaction system at equilibrium at constant volume and temperature:

Reaction: A+B⇌C+D

At fixed temperature, keeping the volume unaltered, when some quantity of product (C or D) is removed from the system at equilibrium, the concentration of that product will decrease.

• As a result, the reaction will no longer remain in the equilibrium state.
• According to Le Chatelier’s principle, the system will adjust itself in such a manner, so that the effect of a decrease in the concentration ofthatproduct is reduced as far as possible. Hence, the equilibrium will shift in a direction that increases the concentrations of that removed product.
• Since the concentration of the products increases in the forward direction, the net reaction will occur in this direction until a new equilibrium is established.
• Therefore, the removal of a product from the reaction system results in shifting the equilibrium position to the right. As a result, the yields of products( C or D) increase.

Conclusion:

At constant volume and temperature, if some quantity of a product is removed from the reaction system at equilibrium, the equilibrium will shift to the right. As a result, the yield of product(s) increases and that of the reactant(s)

Explanation of the effect of addition or removal of reactant or product on equilibrium in terms of reaction quotient: Let us suppose, at a given temperature, the following reaction Is at equilibrium: A+B ⇌ C+D

Equilibrium constant:

⇒ \(K_c=\frac{[C]_{e q} \times[D]_{e q}}{[A]_{e q} \times[B]_{e q}}\) ……………………………..(1)

Where (A)_eq,(B)_eq, (C)_eq and [D]eq are equilibrium molar concentrations of A, B, C, and D respectively.

Effect of addition of the reactant:

Suppose, keeping the temperature and volume fixed, some amount of A is added to the reaction system at equilibrium. Consequently, the concentration of A in the mixture will increase.

Let the concentration of A increase from [A]eq to [A]. At this condition, the reaction quotient will be—

⇒ \(Q_c=\frac{[C]_{e q} \times[D]_{e q}}{[A] \times[B]_{e q}}\) …………………………………(2)

Since,[(A)>(A)_eq , Q_c<K_c. Thus, the reaction is not in equilibrium now (because at equilibrium, Q_c = K_c). Due to the increase in concentration of A, the forward reaction will take place to a greater extent compared to the reverse reaction. As a result, the value of the numerator in equation (2) increases and that of the denominator decreases, leading to a net increase in the value of Q_c.

A time comes when Q_c = K_c and the equilibrium is re-established. Therefore, the addition of reactant (A) to the above reaction system at equilibrium will result in a shift of the equilibrium to the right. As a result, the yields of the products (C and D) increase.

Effect of addition of the product:

At constant temperature and volume, if some amount of C is added to the reaction at equilibrium, then the concentration of C in the reaction mixture will increase. Suppose, the concentration of C increases from (c). at this condition, the reaction quotient Will Be-

⇒ \(Q_c=\frac{[C] \times[D]_{e q}}{[A]_{e q} \times[B]_{e q}}\)

Since,(C)>[C]_eq ,Q_c>K_c. As Q_c≠K_c, the reaction is no longer at equilibrium. Re-establishing of the equilibrium occurs when Q_c = K_c

This is possible if the shifting of equilibrium occurs to the left because this will cause the numerator to decrease and the denominator to increase in the equation (3). As a result of the shifting of equilibrium to the left, the yields of the products decrease.

Effect Of Removal Of The Reactant:

Keeping the temperature and volume fixed, let some amount of A be removed from the reaction system at equilibrium.

Consequently, the concentration of A in the reaction mixture will be reduced and eventually, the equilibrium will be disturbed. At this condition, Q_c will be greater than Kc i.e., Q_c > K_c. This will cause the equilibrium to shift to (lie left. As u result, the yields of the products (C and I) decrease.

Effect of removal of the product:

At constant temperature and volume, If some amount of product C is removed from the reaction system, the equilibrium will be disturbed because of a decrease in the concentration of C in the reaction mixture.

At this condition, Q_c < K_c. As a result, equilibrium will shift to the right. Consequently, the yields of the products ( C and D) increase.

Some examples from everyday life:

Drying of clothes:

Clothes dry quicker when there is a breeze or we keep on shaking them. This is because water vapor of the nearby air is removed and cloth loses more water vapor to re-establish equilibrium with the surrounding air.

Transport of O₂ by hemoglobin in blood:

Oxygen breathed in combines with the hemoglobin in the lungs according to equilibrium, Hb(s) + O₂(g) HbO₂(s). In the tissue the pressure of oxygen is low. To re-establish equilibrium oxyhemoglobin gives up oxygen. But in the lungs, more oxyhemoglobin is formed due to the high pressure of oxygen.

Removal of CO₂ from tissues by blood: This equilibrium

⇒ \(\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3(a q)\) ⇌ \(\mathrm{H}^{+}(a q)+\mathrm{HCO}_3^{-}(a q)\)

In tissue, partial pressure of CO₂ is high thus, CO₂ dissolves in the blood. In the lungs, the partial pressure of CO₂ is low, it is released from the blood.

Tooth decay by sweets:

Our teeth are coated with an enamel of insoluble substance known as hydroxylapatite,

If we do not brush our teeth after eating sweets, the sugar gets fermented on the teeth and produces H ions which combine with the OH ions shifting the above equilibrium in the forward direction thereby causing tooth decay.

Effect of pressure on equilibrium at constant temperature

The effect of change in pressure at equilibrium is observed only for those chemical reactions whose reactants and products are in a gaseous state and have different numbers of moles.

The effect of change in pressure is not significant for the chemical reactions occurring in a solid or liquid state because the volume of a liquid or a solid does not undergo any appreciable change with the variation of pressure.

Effect of increase in pressure:

• At constant temperature, a reaction exists at equilibrium under a definite pressure. Keeping the temperature constant, if the pressure on the system at equilibrium is increased, then the reaction will no longer exist at equilibrium.
• According to Le Chatelier’s principle, the system will tend to adjust itself in such a way as to minimize the effect of the increased pressure as far as possible.
• At constant temperature, the only way to counteract the effect of the increase in pressure is to decrease the volume or to reduce the number of moles (or molecules).
• Hence, at a constant temperature, if the pressure on the system at equilibrium is increased, then the net reaction will take place in a direction that is accompanied by a decrease in volume or several moles (or molecules).

Effect Of decrease in pressure:

• At constant temperature, if the pressure of a reaction system at equilibrium is decreased, then the reaction will no longer remain at equilibrium.
• According to Le Chatelier’s principle, the net reaction will tend to occur in a direction that is associated with an increase in the volume or number of molecules.
• So, at a constant temperature, the decrease in pressure at the equilibrium of a reaction will result in a shifting of equilibrium in a direction that is accompanied by an increase in volume or an increase in the number of molecules (or moles).

Example:

Let, at a constant temperature, the following reaction is at equilibrium: N₂(g) + 3H₂(g) 2NH₃(g). In the reaction, the number of moles of the product is fewer than that of the reactants. So, the forward reaction is accompanied by a decrease in volume.

Effect of increase in pressure at equilibrium:

At constant temperature, if the pressure of the reaction system at equilibrium is increased, then according to Le Chatelier’s principle, the equilibrium of the reaction will shift to the right i.e., the forward reaction will occur to a greater extent compared to the reverse reaction.

So the yield of NH₃(g) will increase. Effect of decrease in pressure at equilibrium: Keeping the temperature constant, if the pressure of the reaction system at equilibrium is decreased, then according to Le Chatelier’s principle, the equilibrium will shift to the left i.e., the backward reaction will occur to a greater extent, leading to a reduction in the yield of NH₃.

For gaseous reactions in which the total number of mole of reactants is equal to that of the products (i.e.,Δn = 0), equilibrium is unaffected by the change in pressure.

This is because these types of reactions are not accompanied by any volume change. Some examples are:

⇒ \(\mathrm{H}_2(g)+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(g) ; \quad \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\)

⇒ \( \text { and } \mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HCl}(\mathrm{g}) \text {. }\)

Effect of pressure on the equilibrium of some chemical reactions at constant temperature:

Effect of temperature on equilibrium

Chemical reactions are usually associated with the evolution or absorption of heat. A reaction in which heat is evolved is called an exothermic reaction, while a reaction in which heat is absorbed is called an endothermic reaction.

In a reversible reaction, if the reaction in any one direction is endothermic, then the reaction in the reverse direction will be exothermic. The temperature of a system at equilibrium can be increased by supplying heat from an external source while the temperature of the system can be lowered by cooling.

Effect of increase in temperature:

At equilibrium, if the temperature of a system is increased, then according to Le Chatelier’s principle, the system will try to offset the effect of the increase in temperature as far as possible. As a result, when the temperature of a reaction system at equilibrium is raised, equilibrium will shift in a direction in which heat is absorbed because it is possible to neutralize the effect of an increase in temperature through the absorption of heat.

Effect of increase In temperature in case of endothermic reactions:

If the temperature is increased at the equilibrium of an endothermic reaction, the forward reaction will take place to a greater extent compared to the reverse reaction until a new equilibrium is established when both the forward and backward reaction occur at equal rates. As a result, equilibrium shills to the right, and the yields of products Increase.

Effect of Increase In Temperature In the case of exothermic reactions:

The temperature Is Increased at the equilibrium of an exothermic reaction, and then the reverse reaction will occur to a greater extent than the forward reaction until a new equilibrium is established when both the forward and backward reactions take place at equal rates. This makes the equilibrium of the reaction shift to the left, resulting In decreased yields of the products.

Effect of decrease in temperature:

According to I.C. Chntolicr’s principle, when the temperature of any system at equilibrium is decreased, the system will try to offset the effect of a decrease in temperature as far as possible.

Therefore, when the temperature is decreased at equilibrium, the equilibrium will shift in a direction in which heat is evolved because it is possible to neutralize the effect of a decrease in temperature through the evolution of heat.

Effect of decrease in temperature In case of endothermic reactions:

As heat is absorbed in an endothermic reaction, a decrease in temperature at equilibrium of such a reaction causes the backward reaction to take place to a greater extent compared to the forward reaction until a new equilibrium is established. As a result, the equilibrium shifts to the left, and the yields of products decrease.

Effect of decrease in temperature in case of exothermic reactions:

If the temperature is decreased at the equilibrium of an exothermic reaction, the forward reaction will take place to a greater extent compared to the reverse reaction until a new equilibrium is established. As a result, shifts to the right, and the yields of products increase.

Example: Manufacturing of ammonia (NH₃) by Haber’s process is an example of an exothermic reaction:

N₂(g) + 3H₂(g) ?=± 2NH₃(g); AH = -22kcl. In this case, the forward reaction is exothermic.

Hence, the backward reaction is endothermic. Effect of increase in temperature at equilibrium:

If the temperature is increased at the equilibrium of the reaction, then according to Le Chatelier’s principle, the backward reaction will take place to a greater extent compared to the forward reaction’ until a new equilibrium is formed. Hence, the equilibrium will shift to the left, causing a decrease in the yield of NH₃.

Effect of decrease In temperature at equilibrium:

If the temperature  Is decreased at the equilibrium of the reaction, then according to i.e., Cliuteller’sprinciple, the equilibrium will be In (the direction In which heat Is generated ie., in ibis case, the forward reaction will be favored, and consequently the yield of N 1 L, will be higher.

Formation of NO(g) from N₂(g) and O₂(g) Is an endothermic reaction:

N₂(g) + O₂(g) ⇌ 2NO(g); ΔH = + 44 kcal Here the forward reaction is endothermic. Hence, the backward reaction Is exothermic.

Effect of increase In temperature on equilibrium:

If the temperature is Increased at the equilibrium of the reaction, then according to Lc Cliatelier’s principle, the forward reaction will take place to a greater extent compared to the backward reaction until a new equilibrium is formed. Hence, the equilibrium will shift to the right, leading to a higher NO.

Effect of decrease in temperature on equilibrium:

If the temperature is decreased at the equilibrium of the reaction, then according to Le Cliatelier’s principle, the equilibrium will shift in the direction in which heat is generated i.e., in this case, the backward reaction will be favored over the forward reaction. Consequently, the yield of NO will be reduced.

Effect Of Catalyst On Equilibrium

A catalyst has no role in the equilibrium of a reaction. At a given temperature, when a reaction is conducted separately in the presence and the absence of a catalyst, the composition of the equilibrium mixture formed in either case remains the same. This is because the catalyst increases the rates of the forward and backward reactions equally.

The catalyst functions to make the attainment of equilibrium faster by accelerating the rates of both the forward and reverse reactions to the same extent. The yield of product in a reaction cannot be increased with the use of a catalyst.

Effect of addition of inert gas on equilibrium

At constant temperature, adding an inert gas (He, Ne, Ar, etc.) to an equilibrium reaction system can be done at constant volume or pressure.

Effect of addition of inert gas at constant volume:

At constant temperature, keeping the volume fixed, when an inert gas is added to a reaction system at equilibrium, the total number of molecules (or moles) in the system increases. So, the total pressure of the system increases, but the partial pressure of the components does not change. Hence, the equilibrium of the system remains undisturbed.

Effect of addition of inert gas at constant pressure:

Keeping both temperature & pressure fixed, the addition of inert gas to a reaction system at equilibrium causes an increase in the volume of the system (because the total number of moles in the system increases) with a consequent decrease in partial pressures of the components.

So, the sum of the partial pressures of the reactants and the products also decreases. In this situation, equilibrium will shift in a direction that increases the volume of the reaction system i.e., the number of molecules in the reaction system.

Depending on An, three situations may arise—

Numerical Examples

Question 1. At 986°C, 3 mol of H₂O(g) and 1 mol of CO(g) react with each other according to the reaction, CO(g) + H₂O(g)⇌CO₂(g) + H₂(g). At equilibrium, the total pressure of the reaction mixture is found to be 2.0 atm. If Kc = 0.63 (at 986°C), then at equilibrium find O the number of moles of H₂(g), 0 the partial pressure of each of the gases.
Answer:

Let, a decrease in several moles of H₂O(g) be x after the reaction attains equilibrium. Consequently, number of moles of CO also decreases by x. According to the reaction, each of CO₂(g) and H₂(g) increases by x number of moles.

Therefore, number of moles of different substances will be as follows:

So, the total number of moles of different substances at equilibrium =1 – x + 3 – x + x + x = 4

Partial pressures of different substances at equilibrium:

⇒ \(p_{\mathrm{CO}}=\frac{(1-x)}{4} \times 2=\frac{1}{2}(1-x)\)

⇒ \(p_{\mathrm{H}_2 \mathrm{O}}=\left(\frac{3-x}{4}\right) \times 2=\frac{1}{2}(3-x)\)

⇒ \(p_{\mathrm{CO}_2}=\left(\frac{x}{4}\right) \times 2=\frac{x}{2}\)

⇒ \(p_{\mathrm{H}_2}=\left(\frac{x}{4}\right) \times 2=\frac{x}{2}\)

⇒ \(p_{\mathrm{CO}_2}=\left(\frac{x}{4}\right) \times 2=\frac{x}{2} ; p_{\mathrm{H}_2}=\left(\frac{x}{4}\right) \times 2=\frac{x}{2}\)

In the given reaction, Δn = 0.

Therefore, K_p – K_c = 0.63.

In the given, ,K_p=\(\frac{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}}\)

⇒ \(\text { So, } 0.63=\frac{\left(\frac{x}{2}\right) \times\left(\frac{x}{2}\right)}{\left(\frac{1-x}{2}\right) \times\left(\frac{3-x}{2}\right)}=\frac{x^2}{(1-x) \times(3-x)}\)

or, x²= 0.63x²- 2.52x+ 1.89

or, x² + 6.81x- 5.108 = 0

∴ x = 0.681

∴ At equilibrium, the number of moles of H2(g) = 0.68

∴ At equilibrium \(p_{\mathrm{CO}}=\frac{1}{2}(1-0.681)=0.1595 \mathrm{~atm}\)

⇒ \(p_{\mathrm{H}_2 \mathrm{O}}=\frac{1}{2}(3-0.681)=1.1595 \mathrm{~atm}\)

⇒ \(p_{\mathrm{CO}_2}=p_{\mathrm{H}_2}=\frac{0.681}{2}=0.3405 \mathrm{~atm}\)

Question 2. For the reaction, N₂O₄(g), and – 2NO₄(g) occurring in a closed vessel at 300K, the partial pressures of N₂O₄(g) and NO₂(g) at equilibrium are 0.28 atm and 1.1 atm respectively. What will be the partial pressures of these gases if the volume of the reaction system is doubled keeping the temperature constant?
Answer:

Equilibrium constant,

⇒  \(K_p=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{(1.1)^2}{0.28}=4.32\)

If the volume of the reaction system is doubled at constant temperature, then partial pressures of N₂O₄ and NO₂ will decrease to half of their initial values. Therefore, partial pressures of N₂O₄ and NO₂ will be 0.14 and 0.55 atm respectively.

So, the equilibrium of the reaction will be disturbed. Now, according to Le Chaterlier’s principle, a reaction will attain a new equilibrium by shifting to the right because in such a case the number of moles as well as the volume will increase.

Let, at the new equilibrium, the partial pressure of N₂O₄(g) decreases to p atm. According to the equation, the partial pressure of NO₂(g) will increase to 2p atm. Therefore, at a new equilibrium,

The partial pressure of each of the component gases will be:

⇒ \(K_p=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{(0.55+2 p)^2}{(0.14-p)}=4.32\)

Or, 4p² + 2.2p + 0.3025 = 4.32(0.14-p) = 0.6048-4.32p

or, p² + 1.63p- 0.0755 = 0

∴ p = 0.045 atm

So, at new equilibrium, partial pressure of N₂O₄(g) = (0.14-0.045) atm = 0.095 atm and partial pressure of

NO₂ = (0.55 + 2 × 0.045) atm = 0.64 atm

Question 3. PCl₅(g)⇌PCl₃(g) + Cl₂(g); K_p =1.8 At 250°C if 50% of PCl₅ dissociates at equilibrium then what should be the pressure of the reaction system?
Answer:

Let the initial number of moles of PCl₅g) be a. At equilibrium, 50% dissociation of PCl₅(g) will occur if the pressure of the reaction system = P atm. After 50% dissociation of PCl₅(g), the number of moles of PCl₅(g) decreases by an amount of 0.5a, and for PCl₃(g) and Cl₂(g) it increases by 0.5a.

So at equilibrium:

At equilibrium, total no. moles = 0.5a + 0.5a + 0.5a = 1.5a

∴ At equilibrium, partial pressure of different components

⇒ \(\text { are, } p_{\mathrm{PCl}_5}=\frac{(0.5 a)}{(1.5 a)} P=\frac{P}{3} ; p_{\mathrm{PCl}_3}=\frac{P}{3} \text { and } p_{\mathrm{Cl}_2}=\frac{P}{3}\)

Equilibrium constant of the reaction \(K_p=\frac{p_{\mathrm{PCl}_3} \times p_{\mathrm{Cl}_2}}{p_{\mathrm{PCl}_5}}\)

So at 5.4 atm pressure, 50% of PCl₅(g) will be dissociated at 250°C.

Question 4. At a particular temperature and 0.50 aim pressure, NH) Ami some amount of .solid NH₄HS arc present In a rinsed container. Solid NH₃(g) dissociates to give NH₄(g) and H₂S(g). At equilibrium, the total pressure of (ho reaction-mixture Is found to be 0.8 1 atm. Hud the value of the equilibrium constant of this reaction at that temperature.
Answer:

Reaction: NH₄H(s) ⇌ NH₃(g) + H₂S(g)

From the reaction, it is dear that, ) mol NH₄HS(s) dissociation produces 1 mol NH₃(g) and 1 mol H₂S(g). So, at a particular temperature and volume, the partial pressure of NHa(g) and I H₂S(g) will be the same and independent of the amount of NH₄HS(s).

Let, partial pressure of H₂S(g) at equilibrium =p atm. Therefore, partial pressure of NH₃(g) and H₂S(g) at Initial stage and at equilibrium arc as follows:

∴ 0.5 + 2p = 0.04 or, p = 0.17 atm

∴ At equilibrium, partial pressure of NH₃,

PNH₃ = (0.5 + 0.17) atm =0.67 atm and partial pressure of

H₂S,PH₂S = 0.17 atm

So, the equilibrium constant of the reaction, K_p = PNH₃ × PH₂O

= 0.67 ×  0.17 = 0.1139 atm²