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Chapter 14: Problem 74

Consider the reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{PCl}_{5}(g)\) Use forward and reverse rate meters to represent the forward and reverse reaction rates for: (a) The initial equilibrium (b) Moments after you disturb the equilibrium by adding \(\mathrm{PCl}_{5}\) (c) The restored equilibrium (d) Which way did the reaction shift to get from the initial equilibrium, part (a), to the restored equilibrium, part (c)?

Short Answer

Expert verified
(a) At the initial equilibrium, both forward and reverse rate meters are equal. (b) After adding more \(\mathrm{PCl}_5\), the reverse rate meter shows a higher rate than the forward rate meter as the reaction tries to reduce the excess product. (c) At the restored equilibrium, both rate meters are once again equal, but at different levels from the initial equilibrium. (d) The reaction shifts in the reverse direction from part (a) to part (c) to reduce the excess \(\mathrm{PCl}_{5}\) and restore equilibrium.

Step by step solution

01

General concept

In an equilibrium, the rate at which a reaction occurs in the forward direction is equal to the rate at which the reverse reaction occurs. This results in no net change in the concentration of the reactants and products over time. <ActionResult#:
02

(a) Initial Equilibrium

At the initial equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This means that the forward rate meter and the reverse rate meter will be at the same level, showing equal rates.
03

(b) Moments after adding PCl5

When we add more \(\mathrm{PCl}_5\) to the reaction, we disturb the equilibrium. In this case, the reverse reaction will have a higher rate than the forward reaction, as the system tries to reduce the excess concentration of the product. The reverse rate meter will show a higher rate than the forward rate meter.
04

(c) Restored Equilibrium

As the reaction proceeds, the consumption of the excess \(\mathrm{PCl}_5\) will eventually lead to a new equilibrium. At this point, the forward and reverse reaction rates will once again be equal, so the forward and reverse rate meters will show the same level, but these levels might be different from the initial equilibrium.
05

(d) Shift from initial to restored equilibrium

The reaction shifted in the reverse direction to reach the restored equilibrium, from part (a) to part (c). This is because the addition of more \(\mathrm{PCl}_5\) caused a temporary imbalance, and the reverse reaction worked to consume the extra product and restore the equilibrium. In this process, the concentrations of the reactants (\(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\)) increased and the concentration of the product (\(\mathrm{PCl}_{5}\)) decreased to achieve a new balanced state.

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Most popular questions from this chapter

In the Haber process, nitrogen gas reacts with hydrogen gas to form gaseous ammonia (see Problem 14.76). (a) Write the equilibrium expression for the reaction. (b) The equilibrium constant at a certain temperature is \(1.5 \times 10^{-2}\). If the equilibrium concentrations of hydrogen and nitrogen are both \(0.20 \mathrm{M}\) at this temperature, what is the equilibrium concentration of ammonia?

If you added \(\mathrm{SO}_{3}\) to a vessel in which the reaction \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftarrows 2 \mathrm{SO}_{3}\) is at equilibrium, which way would the reaction shift?

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