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For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \quad \Delta H=-93.6 \mathrm{~kJ} \mathrm{~mol}^{-1} $$ the number of moles of \(\mathrm{H}_{2}\) at equilibrium will increase if : (a) volume is increased (b) volume is decreased (c) argon gas is added at constant volume (d) \(\mathrm{NH}_{3}\) is removed

Short Answer

Expert verified
(a) The number of moles of H2 at equilibrium will increase if the volume is increased.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. Examine the options based on this principle.
02

Assess the Effect of Increasing Volume

Increasing the volume of the reaction vessel decreases the pressure. According to Le Chatelier's Principle, the system will shift to increase the pressure. The side with more moles of gas (the reactants: 1 mole of N2 + 3 moles of H2) will be favored.
03

Assess the Effect of Decreasing Volume

Decreasing the volume increases the pressure. The system will shift to decrease the pressure, favoring the side with fewer moles of gas (the products: 2 moles of NH3).
04

Evaluate the Effect of Adding Argon Gas at Constant Volume

Adding an inert gas like argon at constant volume does not change the partial pressures of the reacting gases. Therefore, there is no shift in equilibrium.
05

Determine the Effect of Removing NH3

Removing NH3 from the system will disturb the equilibrium. According to Le Chatelier's Principle, the system will shift to produce more NH3, thus consuming more N2 and H2. This does not result in an increase in the number of moles of H2 at equilibrium.

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Key Concepts

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

Chemical Equilibrium
Imagine a busy city intersection where cars are continuously flowing in and out, yet the overall picture appears stable; this is akin to chemical equilibrium in reactions. When a reaction reaches equilibrium, it means the forward and reverse reactions occur at the same rate, leading to no net change in the concentrations of reactants and products over time. Despite this apparent static condition, particles are constantly moving, colliding, and reacting.

In the context of the provided exercise, where nitrogen and hydrogen react to form ammonia, reaching equilibrium means the rate at which ammonia is formed from nitrogen and hydrogen becomes equal to the rate at which ammonia decomposes back into nitrogen and hydrogen. Please note that even though rates are equal, the actual amounts of reactants and products don't have to be.
Equilibrium Shift
Equilibrium is all about balance, but what happens when that balance is disturbed? This is where Le Chatelier's Principle comes into play. According to this principle, if an external condition is changed, the equilibrium will adjust itself to minimize that change. This is known as an equilibrium shift.

For example, if the pressure of the system in our exercise is increased by decreasing volume, the equilibrium will shift towards the side with fewer gas molecules to reduce pressure. Conversely, if the volume is increased and pressure is lowered, the system will favor the side with more gas molecules to restore the pressure.
Reaction Volume Effects
The size of the dance floor can determine how freely people can dance—it's not much different in a reaction vessel with gases. The reaction volume has a direct impact on pressure when temperature and the amount of gas are constant. Increasing the volume lowers the pressure because molecules are more spread out and hit the walls of the container less frequently.

In the reaction given in the exercise, increasing the reaction volume would cause a shift toward the reactants (nitrogen and hydrogen) because there are more gas molecules on that side, thus increasing pressure. Similarly, decreasing volume would increase pressure and shift the equilibrium towards the side with fewer gas molecules (ammonia), according to Le Chatelier’s Principle.
Effect of Inert Gas on Equilibrium
Adding an inert gas, like argon, to a reaction at constant volume is like adding more dancers to a full dance floor without actually giving them more space to move. Since argon does not react with the other gases, it doesn't change the partial pressures of the reacting gases. Therefore, it has no effect on the equilibrium position of the reaction.

When argon or any other inert gas is added at constant volume, the total pressure of the system increases but it does not affect the equilibrium concentrations of the reactants and products because it does not change the rate at which the substances react. This is critical to understand for accurately predicting how adding an inert gas influences the reaction's behavior.

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Most popular questions from this chapter

\(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) ; K_{p}=4 \times 10^{-4} \mathrm{~atm}^{2}\). If the vapour pressure of water is 38 torr then percentage of relative humidity is : (Assume all data at constant temperature) (a) 4 (b) 10 (c) 40 (d) None of these

At a certain temperature, only \(50 \% \mathrm{HI}\) is dissociated at equilibrium in the following reaction: $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ The equilibrium constant for this reaction is : (a) \(0.25\) (b) \(1.0\) (c) \(3.0\) (d) \(0.5\)

When sulphur (in the form of \(S_{8}\) ) is heated at temperature \(T\), at equilibrium, the pressure of \(S_{8}\) falls by \(30 \%\) from \(1.0 \mathrm{~atm}\), because \(\mathrm{S}_{8}(g)\) is partially converted into \(\mathrm{S}_{2}(g)\). Find the value of \(K_{p}\) for this reaction. (a) \(2.96\) (b) \(6.14\) (c) \(204.8\) (d) None of these

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). If the initial concentration of \(\left[\mathrm{H}_{2}\right]=\left[\mathrm{CO}_{2}\right]\) and \(x\) moles/litre of hydrogen is consumed at equilibrium, the correct expression of \(K_{p}\) is : (a) \(\frac{x^{2}}{(1-x)^{2}}\) (b) \(\frac{(1+x)^{2}}{(1-x)^{2}}\) (c) \(\frac{x^{2}}{(2+x)^{2}}\) (d) \(\frac{x^{2}}{1-x^{2}}\)

Consider the reactions \(\alpha\) (i) \(2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2}(g)\); Eqm. Constant \(=K_{1}\) (ii) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) ;\) Eqm. Constant \(=K_{2}\) (iii) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) ;\) Eqm. Constant \(=K_{3}\) Which of the following relation is correct? (a) \(K_{3}=\frac{K_{1}}{K_{2}}\) (b) \(K_{3}=\frac{K_{1}^{2}}{K_{2}^{2}}\) (c) \(K_{3}=K_{1} K_{2}\) (d) \(K_{3}=\sqrt{K_{1}} \cdot K_{2}\)

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