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In the equilibrium mixture: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\) \(\rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), the mole ratio of gases are \(\sqrt{2}: \sqrt{2}: 10\), respectively. What would be the effect on the mole ratio on adding 5 mole of He gas at constant pressure? (a) No change (b) The new molar ratio becomes \(1: 1: 5\) (c) The new molar ratio becomes \(2: 2: 5\) (d) The new molar ratio becomes \(2: 2: 5 \sqrt{2}\)

Short Answer

Expert verified
(a) No change

Step by step solution

01

Understand the properties of inert gases

Adding an inert gas to a reaction mixture at constant pressure does not affect the equilibrium position of the reaction because inert gases do not react with the substances in the mixture. Therefore, the mole ratio of the reacting substances remains unchanged.
02

Assess the effect of adding helium

The addition of helium (He) gas to the system will increase the total number of moles of gas in the container, but it will not change the mole ratios of the reacting gases themselves. Helium is an inert gas, and at constant pressure, it will not affect the equilibrium concentrations of the reacting gases.
03

Determine the new mole ratio

Since helium does not participate in the reaction and the moles of each substance in the reaction remain the same, the mole ratio of \text{H}_2, \text{I}_2, and \text{HI} gases will not change. Therefore, the mole ratio after adding helium gas will remain the same as the initial ratio, which is \(\sqrt{2} : \sqrt{2} : 10\).
04

Select the correct option

Based on the understanding that the addition of helium does not change the mole ratio of the reacting gases, the correct option is that there is no change in the mole ratio. Hence, the correct answer is (a) No change.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle plays a pivotal role in understanding chemical reactions at equilibrium. It states that if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the system will adjust to counteract that change and a new equilibrium will be established. For example, increasing the concentration of reactants will push the equilibrium to produce more products, while increasing the temperature of an exothermic reaction will shift the equilibrium to favor the reactants.

In the given exercise, if the pressure or concentration of either reactants or products is altered, Le Chatelier's Principle gives us a qualitative prediction of how the system will respond in order to re-establish equilibrium. Since adding helium gas at constant pressure does not affect the concentrations of the reacting gases, the equilibrium position of the reaction remains unchanged.
Mole Ratio
The mole ratio in a chemical reaction indicates the proportion of reactants to products. It is determined by the coefficients in the balanced chemical equation. In the context of the given equilibrium, \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), the stoichiometric coefficients represent the mole ratio of hydrogen (H2) to iodine (I2) to hydrogen iodide (HI), which is 1:1:2 under ideal conditions.

The mole ratio is important in predicting the amounts of products formed and reactants needed. When an inert gas is added at constant pressure, the total volume and number of moles in the container increase, but the mole ratio of the substances involved in the equilibrium remains constant because the inert gas does not participate in the reaction.
Inert Gas Effect
The inert gas effect describes how an inert gas, like helium, affects a system at equilibrium when it is added to the reaction mixture. Inert gases are non-reactive and do not change the concentrations of the reactants or products. They can change the total pressure of the system if the system is closed and the volume is held constant.

In this exercise, helium is added at constant pressure which means the volume of the system had to increase to accommodate the additional gas. This does not affect the partial pressures of the reacting gases or the mole ratios. Understanding the inert gas effect is crucial to correctly predict that, under these conditions, the equilibrium position will not change.
Equilibrium Constant
The equilibrium constant, represented as K, is a value that expresses the ratio of the concentrations of the products to the reactants at equilibrium, each raised to the power of their respective stoichiometric coefficients. It is a way to quantify the position of equilibrium and is determined by the specific reaction at a given temperature.

In the exercise's reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), the equilibrium constant would be expressed as \(K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\). Because the addition of an inert gas at constant pressure does not alter the concentration of any species in the reaction, it has no effect on the value of the equilibrium constant. Thus, the equilibrium constant remains a reliable measure of the system's equilibrium state regardless of the inert gas addition.

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Most popular questions from this chapter

What is the approximate value of \(\log K_{\mathrm{p}}\) for the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\) The standard enthalpy of formation of \(\mathrm{NH}_{3}(\mathrm{~g})\) is \(-40.0 \mathrm{~kJ} / \mathrm{mol}\) and standard entropies of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{NH}_{3}(\mathrm{~g})\) are 191,130 and \(192 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. (a) \(0.04\) (b) \(7.05\) (c) \(8.6\) (d) \(3.73\)

For a reversible reaction: \(\mathrm{A} \underset{K_{2}}{\stackrel{K_{1}}{K}} \mathrm{~B}\), the initial molar concentration of \(\mathrm{A}\) and \(\mathrm{B}\) are \(a \mathrm{M}\) and \(b \mathrm{M}\), respectively. If \(x \mathrm{M}\) of \(\mathrm{A}\) is reacted till the achievement of equilibrium, then \(x\) is (a) \(\frac{K_{1} a-K_{2} b}{K_{1}+K_{2}}\) (b) \(\frac{K_{1} a-K_{2} b}{K_{1}-K_{2}}\) (c) \(\frac{K_{1} a-K_{2} b}{K_{1} K_{2}}\) (d) \(\frac{K_{1} a+K_{2} b}{K_{1}+K_{2}}\)

\(\Delta_{t} G^{\circ}\) for the formation of \(\mathrm{HI}(\mathrm{g})\) from its gaseous elements is \(-2.303 \mathrm{kcal} / \mathrm{mol}\) at \(500 \mathrm{~K}\). When the partial pressure of HI is \(10 \mathrm{~atm}\) and of \(\mathrm{I}_{2}(\mathrm{~g})\) is \(0.001 \mathrm{~atm}\), what must be the partial pressure of hydrogen be at this temperature to reduce the magnitude of \(\Delta G\) for the reaction to zero? (a) \(1000 \mathrm{~atm}\) (b) \(10000 \mathrm{~atm}\) (c) \(100 \mathrm{~atm}\) (d) \(31.63 \mathrm{~atm}\)

The complexion of \(\mathrm{Fe}^{2+}\) with the chelating agent dipyridyl has been studied kinetically in both the forward and reverse directions. \(\mathrm{Fe}^{2+}+3\) dipy \(\rightleftharpoons\left[\mathrm{Fe}(\mathrm{dipy})_{3}\right]^{2+}\) For this reaction, the rates of forward and reverse reactions are \(\left(1.45 \times 10^{13}\right.\) \(\left.\mathrm{M}^{-3} \mathrm{~s}^{-1}\right)\left[\mathrm{Fe}^{2+}\right][\mathrm{dipy}]^{3}\) and \(\left(1.22 \times 10^{-4} \mathrm{~s}^{-1}\right)\) \(\left[\mathrm{Fe}(\mathrm{dipy})_{3}{ }^{2+}\right]\), at \(25^{\circ} \mathrm{C}\). What is the stability constant of the complex? (a) \(1.77 \times 10^{9}\) (b) \(8.4 \times 10^{-18}\) (c) \(1.18 \times 10^{17}\) (d) \(5.65 \times 10^{-10}\)

For the reaction: \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\), the relation between the degree of dissociation, \(\alpha\), of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at pressure, \(P\), with its equilibrium constant \(K_{\mathrm{P}}\) is (a) \(\alpha=\frac{K_{P} / P}{4+K_{P} / P}\) (b) \(\alpha=\frac{K_{P}}{4+K_{P}}\) (c) \(\alpha=\left[\frac{K_{P} / P}{4+K_{P} / P}\right]^{1 / 2}\) (d) \(\alpha=\left[\frac{K_{P}}{4+K_{P}}\right]^{1 / 2}\)

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