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Consider the following equilibrium process: \(\mathrm{PCl}_{5}(g) \rightleftarrows \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \quad \Delta H^{\circ}=92.5 \mathrm{~kJ} / \mathrm{mol}\) Predict the direction of the shift in equilibrium when (a) the temperature is raised, (b) more chlorine gas is added to the reaction mixture, (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
(a) Right; (b) Left; (c) Right; (d) Left; (e) No shift.

Step by step solution

01

Analyze Temperature Change

The reaction \[\mathrm{PCl}_{5}(g) \rightleftarrows \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\] is endothermic, as indicated by the positive \(\Delta H^{\circ}\) of 92.5 kJ/mol. For an endothermic reaction, raising the temperature shifts the equilibrium to the right, favoring the formation of products (\(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\)).
02

Analyze Addition of Chlorine Gas

According to Le Châtelier's principle, adding more \(\mathrm{Cl}_2\) to the reaction mixture will cause the equilibrium to shift to the left to counteract the increase in concentration of \(\mathrm{Cl}_2\). This shift favors the formation of \(\mathrm{PCl}_{5}\).
03

Analyze Removal of PCl3

Removing some \(\mathrm{PCl}_{3}\) decreases its concentration in the reaction mixture. In response, the equilibrium will shift to the right to produce more \(\mathrm{PCl}_{3}\) and counteract the change.
04

Analyze Pressure Increase

Increasing the pressure in a gaseous equilibrium shifts the equilibrium towards the side with fewer moles of gas. Here, the reaction moves to the left, favoring \(\mathrm{PCl}_5\), which has fewer moles of gas (1 mole compared to 2 moles on the right side).
05

Analyze Addition of Catalyst

Adding a catalyst does not shift the equilibrium position. It only speeds up the rate at which equilibrium is achieved both forward and backward, without favoring either side.

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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 is a cornerstone concept in understanding how chemical equilibria respond to changes in the system. In simple terms, it states that when a change is imposed on a system at equilibrium, the equilibrium will shift in a direction that counteracts the change. This principle can help predict how different factors like concentration, temperature, and pressure will affect the equilibrium position of a chemical reaction. For example:
  • Adding or removing reactants or products will shift the equilibrium to restore balance.
  • Changing the temperature will shift the equilibrium depending on the reaction's heat dynamics (endothermic or exothermic).
  • Pressure changes will affect gaseous equilibria based on the number of moles involved.
Understanding these shifts can help in controlling reactions in various industrial processes and laboratory settings. By manipulating conditions, chemists can optimize the yield of desired products.
Endothermic Reactions
Endothermic reactions absorb heat from their surroundings, which can be identified by a positive enthalpy change (\(\Delta H^{\circ}\)). In the context of Le Chatelier's Principle, increasing the temperature in an endothermic reaction will drive the equilibrium toward the products. This is because the system absorbs additional warmth and compensates by forming more products.As seen in the equilibrium process of \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\), raising the temperature favors the production of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\). Thus, when working with endothermic reactions, understanding the temperature's effect is crucial for manipulating the reaction's direction to increase efficiency or yields.
Effect of Pressure on Equilibrium
Pressure changes can significantly impact the equilibrium position of gaseous reactions. According to Le Chatelier's Principle, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas. This helps reduce the change in pressure by decreasing the gas volume.In the reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\), increasing pressure shifts equilibrium towards the left, favoring \(\mathrm{PCl}_{5}\) because it has fewer moles (1 mole versus a total of 2 moles on the product side). Similarly, decreasing the pressure would favor the product side with more moles. Understanding this concept allows chemists to predict and control the direction of a reaction simply by modifying the pressure.
Effect of Catalysts on Equilibrium
Catalysts are unique in that they increase the rate of reaction without being consumed themselves and without altering the position of equilibrium. They work by providing an alternative reaction pathway with a lower activation energy, accelerating both forward and backward reactions equally. For the equilibrium reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\), adding a catalyst would quicken the approach to equilibrium but would not shift the equilibrium towards products or reactants. This makes catalysts invaluable in increasing the efficiency of chemical processes, particularly where rapid establishment of equilibrium is desired without altering the reaction conditions or final concentrations of products and reactants.

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Most popular questions from this chapter

Heating solid sodium bicarbonate in a closed vessel establishes the following equilibrium: $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftarrows \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g) $$ 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.

Write the equation for the reaction that corresponds to each of the following reaction quotients: (a) \(Q_{\mathrm{c}}=\frac{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{~S}_{2}\right]}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}\) (b) \(Q_{\mathrm{c}}=\frac{\left[\mathrm{NO}_{2}\right]^{2}\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{NClO}_{2}\right]^{2}}\) (c) \(Q_{\mathrm{c}}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{4}}\) (d) \(Q_{\mathrm{c}}=\frac{[\mathrm{NO}]^{2}\left[\mathrm{Br}_{2}\right]}{[\mathrm{NOBr}]^{2}}\)

Water is a very weak electrolyte that undergoes the following ionization (called autoionization): $$ \mathrm{H}_{2} \mathrm{O}(l) \stackrel{k_{1}}{\stackrel{\mathrm{m}_{-1}}} \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (a) If \(k_{1}=2.4 \times 10^{-5} \mathrm{~s}^{-1}\) and \(k_{-1}=1.3 \times 10^{11} / M \cdot \mathrm{s}\) calculate the equilibrium constant \(K\) where \(K=\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right] /\left[\mathrm{H}_{2} \mathrm{O}\right] .\) (b) Calculate the product \(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right],\) \(\left[\mathrm{H}^{+}\right],\) and \(\left[\mathrm{OH}^{-}\right] .\) (Hint : Calculate the concentration of liquid water using its density, \(1.0 \mathrm{~g} / \mathrm{mL}\).)

Consider the reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g) $$ If the equilibrium partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) are \(0.15,0.33,\) and 0.050 atm, respectively, at \(2200^{\circ} \mathrm{C}\) what is \(K_{P}\) ?

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction: $$2 \mathrm{NH}_{3}(g) \rightleftarrows \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$$ is 0.83 at \(375^{\circ} \mathrm{C}\). A \(14.6-\mathrm{g}\) sample of ammonia is placed in a 4.00-L flask and heated to \(375^{\circ} \mathrm{C}\). Calculate the concentrations of all the gases when equilibrium is reached.

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