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Chapter 15: Problem 48

Consider this equilibrium process: $$ \begin{aligned} \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \\\ \Delta H^{\circ} &=92.5 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Predict the direction of the shift in equilibrium when (a) the temperature is raised, (b) more chlorine gas is added to the reaction mixture, \((\mathrm{c})\) some \(\mathrm{PCl}_{3}\) is removed from the mixture, (d) the pressure on the gases is increased, (e) a catalyst is added to the reaction mixture.

Short Answer

Expert verified
The shift in equilibrium will be: (a) to the right, (b) to the left, (c) to the right, (d) to the left, and (e) no shift

Step by step solution

01

Consider the effect of temperature

From the question, \(\Delta H^\circ = 92.5 \: kJ/mol\). This means the reaction is endothermic because \(\Delta H\) is positive. When the temperature increases, according to Le Chatelier鈥檚 principle, the system will shift in the direction that uses up the extra heat, which is towards the endothermic direction. So, the system will shift to the right.
02

Consider the effect of more chlorine gas

Adding a reactant to a reaction at equilibrium will lead to a shift towards the direction that uses up the added reactant. In this case, the system will shift to the left to consume the added chlorine gas.
03

Consider the effect of removing \(\mathrm{PCl}_3\)

Removing a product from a reaction at equilibrium will lead to a shift towards the product to replace what was removed. In this case, the system will shift to the right to produce more \(PCl_3\).
04

Consider the effect of pressure

Increasing the pressure on the gases causes the system to shift towards the side with fewer gas molecules in order to reduce the pressure. In this case, there are two moles of gas on the right side and one mole on the left side, therefore, the system will shift to the left.
05

Consider the effect of a catalyst

Adding a catalyst to a reaction at equilibrium speeds up both the forward and reverse reactions equally, without favouring one direction over the other. Therefore, the addition of a catalyst does not cause the system to shift in one direction or the other.

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Key Concepts

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

Equilibrium
In a chemical reaction, equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction. This doesn't mean the concentrations of reactants and products are equal; instead, they remain constant over time. For the reaction \( \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \), equilibrium is dynamic and responds to changes according to Le Chatelier鈥檚 Principle. If a change occurs in temperature, concentration, or pressure, the system shifts to counteract the change, while keeping the equilibrium constant \( K \) the same unless the temperature changes.
Endothermic Reaction
An endothermic reaction absorbs heat, indicated by a positive \( \Delta H \). In the case of \( \mathrm{PCl}_{5}(g) \) dissociating into \( \mathrm{PCl}_{3}(g) \) and \( \mathrm{Cl}_{2}(g) \), \( \Delta H^{\circ} = 92.5 \mathrm{~kJ/mol} \). Raising the temperature adds energy to the system. According to Le Chatelier鈥檚 Principle, the system will shift towards the endothermic direction, which consumes the added heat. Here, the reaction will shift to the right, increasing the concentration of \( \mathrm{PCl}_{3} \) and \( \mathrm{Cl}_{2} \).
Catalyst Effect
Catalysts speed up the rate of both the forward and reverse reactions equally, without affecting the equilibrium position. This is because catalysts lower the activation energy needed for the reactions to proceed. In the reaction \( \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \), adding a catalyst will not change the equilibrium concentrations of the substances. However, the reaction will reach equilibrium faster, which can be beneficial in industrial processes.
Pressure Effects
Pressure affects gaseous reactions at equilibrium according to Le Chatelier鈥檚 Principle. When pressure increases, the system shifts towards the side with fewer gas molecules to reduce pressure. For \( \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \), there is one mole of gas on the left and two on the right. Increasing pressure shifts the equilibrium to the left, favoring the formation of \( \mathrm{PCl}_{5}(g) \), which has fewer gas molecules. Changes in pressure do not affect reactions involving only solids or liquids.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are: \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.491 \mathrm{M}\). The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(4.63 \times\) \(10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color.]

Heating solid sodium bicarbonate in a closed vessel established this equilibrium: \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system, (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system, (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.

Define equilibrium. Give two examples of a dynamic equilibrium.

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.2 at \(1650^{\circ} \mathrm{C}\). Initially \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.80 \mathrm{~mol}\) \(\mathrm{CO}_{2}\) are injected into a \(5.0-\mathrm{L}\) flask. Calculate the concentration of each species at equilibrium.

The dissociation of molecular iodine into iodine atoms is represented as $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ At \(1000 \mathrm{~K}\), the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(I_{2}\) in a 2.30 - \(L\) flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?
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