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Chapter 2: Problem 48

A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) \(2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}\) Which option best describes what will immediately occur to the reaction rates if the pressure on the system is increased after it has reached equilibrium? (A) The rate of both the forward and reverse reactions will increase. (B) The rate of the forward reaction will increase while the rate of the reverse reaction decreases. (C) The rate of the forward reaction will decrease while the rate of the reverse reaction increases. (D) Neither the rate of the forward nor reverse reactions will change.

Short Answer

Expert verified
(A) The rate of both the forward and reverse reactions will increase.

Step by step solution

01

Understand the provided information

The exercise provides a chemical equation that represents an equilibrium reaction where Hydrogen sulfide gas decomposes into Hydrogen gas and Sulfur gas. We also have the value of the equilibrium constant \(K_{c} = 1.0 \times 10^{-6}\). The question asks what happens to the rates of both the forward and reverse reactions if the pressure on the system is increased after it has achieved equilibrium.
02

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. In the case of pressure changes, the reaction will shift towards the side with fewer gas molecules to reduce the pressure. In this equation, there are 2 moles of gas on either side of the reaction: \(2 \mathrm{H}_{2} \mathrm{S}(g) \leftrightarrow 2\mathrm{H}_{2}(g) + \mathrm{S}_{2}(g)\). Therefore, an increase in pressure does not favor the forward nor the reverse reaction over the other.
03

Analyze the effect on reaction rates

Since the pressure change does not shift the reaction, the rates of the forward and reverse reactions will increase to the same extent. This is because increasing the pressure means more molecules are colliding per unit area per unit time in the system, hence the rates of both the forward and reverse reactions will increase. Therefore, the correct answer is (A) The rate of both the forward and reverse reactions will increase.

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Most popular questions from this chapter

20.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is placed in a beaker and titrated with a solution of \(1.0 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) resulting in the creation of a precipitate. How much \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to reach the equivalence point? (A) 10.0 mL (B) 20.0 mL (C) 30.0 mL (D) 40.0 mL

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20.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is placed in a beaker and titrated with a solution of \(1.0 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) resulting in the creation of a precipitate. What will happen to the conductivity of the solution after additional \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is added past the equivalence point? (A) The conductivity will increase as additional ions are being added to the solution. (B) The conductivity will stay constant as the precipitation reaction has gone to completion. (C) The conductivity will decrease as the solution will be diluted with the addition of additional \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\). (D) The conductivity will stay constant as equilibrium has been established.
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