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Some inert gas is added at constant volume to the following reaction at equilibrium $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ Predict the effect of adding the inert gas: (a) the equilibrium shifts in the forward direction (b) the equilibrium shifts in the backward direction (c) the equilibrium remains unaffected (d) the value of \(K_{p}\) is increased

Short Answer

Expert verified
The equilibrium remains unaffected (option c), and the value of Kp is not changed.

Step by step solution

01

Understand the Effect of Adding an Inert Gas

Adding an inert gas to a reaction at constant volume does not change the partial pressures of the reacting gases since the total volume of the gas mixture remains constant. The inert gas does not react with the other substances, so it does not alter the equilibrium concentrations or partial pressures of the reactants or products.
02

Analyze the Equilibrium Shift

Since the addition of an inert gas does not affect the partial pressure of the reactants or the products, there is no change in the positions of the equilibrium. The reaction quotient Q remains unchanged, and as a result, the equilibrium position is maintained.
03

Consider the Effect on The Equilibrium Constant

The equilibrium constant K or Kp for a reaction at a given temperature is a constant value that is not affected by changes in pressure, volume, or the addition of an inert gas. Therefore, the value of Kp remains the same.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Imagine a dance floor where dancers are constantly switching partners - this dynamic situation is akin to what happens in a chemical reaction reaching equilibrium. Chemical equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, although they are not necessarily equal.

This balance is not static; it's a dynamic equilibrium because the reactions continue to occur, but without any net change in concentration. This concept is essential in understanding how reactions adapt to changes in conditions, such as temperature, pressure, and concentrations, governed by the principle known as Le Chatelier's principle.
Equilibrium Constant
The equilibrium constant, represented by the symbol K, is a number that describes the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their coefficients in the balanced equation. In gaseous systems, we often use the term Kp to reflect the equilibrium involving partial pressures.

Let's break it down: if we have a reaction where aA + bB ⇌ cC + dD, the equilibrium constant K would be defined by the equation \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \), where the square brackets indicate concentrations. The equilibrium constant is crucial because it provides insight into the reaction's favorability - a high K indicates a greater concentration of products, while a low K suggests the reactants are more prevalent.
Partial Pressure
Under the dome of a stadium, the cheering of fans creates pressure, much like particles in a gaseous state exert pressure within a container. Partial pressure is the pressure that a single gas in a mixture of gases would exert if it alone occupied the entire volume of the container. It is a measure of the impact of individual gas species in a system and plays a critical role in calculating the equilibrium constant Kp for reactions involving gases.

To calculate partial pressure, we use Dalton's Law, which states that the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual gas within the mixture. For example, if gas A and gas B are in a container, the total pressure P is \( P = P_A + P_B \) where \( P_A \) and \( P_B \) are the partial pressures of A and B, respectively.
Effect of Inert Gas on Equilibrium
Adding mystery to the equation, what happens when an inert gas is introduced to a reaction at equilibrium? Unlike reactive gases, inert gases do not participate in the chemical reaction but can still influence the system. When inert gas is added to a closed system at constant volume, the total pressure increases, but the partial pressures of the individual gases remain the same since the inert gas does not react.

This addition does not disturb the equilibrium position (the ratio of products to reactants) because the equilibrium constant Kp is dependent on the partial pressures of the reacting gases only. For the reaction \( \mathrm{NH}_4\mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_3(g) + \mathrm{H}_2\mathrm{~S}(g) \), the addition of an inert gas at constant volume would not affect the concentrations of \( \mathrm{NH}_3 \) or \( \mathrm{H}_2\mathrm{~S} \) — hence, option (c) the equilibrium remains unaffected is correct. The value of the equilibrium constant Kp also remains unchanged.

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Most popular questions from this chapter

In the presence of excess of anhydrous \(\mathrm{SrCl}_{2}\), the amount of water taken up is governed by \(K_{p}=10^{12} \mathrm{~atm}^{-4}\) for the following reaction at \(273 \mathrm{~K}\) $$ \mathrm{SrCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{SrCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s) $$ What is equilibrium vapour pressure (in torr) of water in a closed vessel that contains \(\mathrm{SrCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) ?\) (a) \(0.001\) torr (b) \(10^{3}\) torr (c) \(0.76\) torr (d) \(1.31\) torr

\(A B_{3}(g)\) is dissociates as \(A B_{3}(g) \rightleftharpoons A B_{2}(g)+\frac{1}{2} B_{2}(g)\) When the initial pressure of \(A B_{2}\) is 800 torr and the total pressure developed at equilibrium is 900 torr. What fraction of \(A B_{3}(g)\) is dissociated? (a) \(10 \%\) (b) \(20 \%\) (c) \(25 \%\) (d) \(30 \%\)

One mole of pure ethyl alcohol was treated with one mole of pure acetic acid at \(25^{\circ} \mathrm{C}\). One-third of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be : (a) \(\frac{1}{4}\) (b) 2 (c) 3 (d) 4

A nitrogen-hydrogen mixture initially in the molar ratio of \(1: 3\) reached equilibrium to form ammonia when \(25 \%\) of the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) had reacted. If the total pressure of the system was 21 atm, the partial pressure of ammonia at the equilibrium was: (a) \(4.5\) atm (b) \(3.0 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.5 \mathrm{~atm}\)

The equilibrium constant for the reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) is \(5 .\) How many moles of \(\mathrm{CO}_{2}\) must be added to 1 litre container already containing 3 moles each of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) to make \(2 \mathrm{M}\) equilibrium concentration of \(\mathrm{CO}\) ? (a) 15 (b) 19 (c) 5 (d) 20

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