Consider the equilibrium \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\). (a) What happens to the partial pressure of \(\mathrm{SO}_{3}\) when the partial pressure of \(\mathrm{SO}_{2}\) is decreased? (b) If the partial pressure of \(\mathrm{SO}_{2}\) is increased, what happens to the partial pressure of \(\mathrm{O}_{2}\) ?

Short Answer

Expert verified
When the partial pressure of \(SO_2\) is decreased, the partial pressure of \(SO_3\) decreases. If the partial pressure of \(SO_2\) is increased, the partial pressure of \(O_2\) decreases.

Step by step solution

01

Understanding Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle will guide us in predicting the behavior of the system when stress (change in partial pressure) is applied.
02

Analyzing the Effect of Decreased Partial Pressure of \(SO_2\)

When the partial pressure of \(SO_2\) is decreased, the equilibrium will shift to the left, in order to produce more \(SO_2\) and counteract the decrease. As a result, the production of \(SO_3\) slows down and its partial pressure decreases.
03

Analyzing the Effect of Increased Partial Pressure of \(SO_2\)

When the partial pressure of \(SO_2\) is increased, the equilibrium will shift to the right to produce more \(SO_3\) and use up the excess \(SO_2\). This increases the consumption of \(O_2\) and consequently, the partial pressure of \(O_2\) decreases.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction and its reverse process proceed at the same rate, leading to no overall change in the amounts of the reactants and products. This state is dynamic, meaning that the molecules continue to react, but because the reaction rates in both directions are equal, the concentrations remain constant over time.

In the context of the exercise concerning the reaction \(2\text{SO}_{2}(\text{g}) + \text{O}_{2}(\text{g}) \rightleftharpoons 2\text{SO}_{3}(\text{g})\), the system reaches equilibrium when the rate at which the reactants turn into the products (sulfur dioxide and oxygen forming sulfur trioxide) is equal to the rate at which the products decompose back into reactants.

Understanding this concept is crucial for interpreting the changes in partial pressures as described in the exercise solution. It provides a baseline to predict how the system readjusts to maintain that finely balanced state when subjected to external changes.
Partial Pressure
Partial pressure is the pressure that a single gas component in a mixture of gases would exert if it alone occupied the entire volume. It's a way to describe the contribution of individual gas species to the total pressure of the mixture.

In the equilibrium scenario presented, the partial pressures of \(\text{SO}_{2}\), \(\text{O}_{2}\), and \(\text{SO}_{3}\) come into play significantly. For instance, if the problem asks what happens to the partial pressure of \(\text{SO}_{3}\) when \(\text{SO}_{2}\)'s partial pressure decreases, it implies that we're looking at how the individual gas's behavior impacts the overall system. The understanding of partial pressures is key to predicting changes in equilibrium conditions, such as those prompted in the exercise, and it's an important concept in chemical thermodynamics.
Equilibrium Shift
An equilibrium shift is the movement of a reaction mixture's composition towards the products or reactants side in response to an external stress, as described by Le Chatelier's Principle. When a chemical system at equilibrium experiences a change in concentration, temperature, volume, or pressure, the system will adjust accordingly to minimize that change and restore a new equilibrium state.

In the exercise, decreasing the partial pressure of \(\text{SO}_{2}\) causes the equilibrium to shift to the left, increasing its production and decreasing \(\text{SO}_{3}\)'s partial pressure as the system seeks to counteract the disturbance. Conversely, increasing \(\text{SO}_{2}\)'s partial pressure would cause the equilibrium to shift to the right, using more \(\text{O}_{2}\) and resulting in its partial pressure decreasing. This demonstrates how equilibrium can dynamically adjust to changes, a central concept when predicting the behavior of chemical systems.

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Most popular questions from this chapter

Calculate the standard Gibbs free energy of each of the following reactions: (a) \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}), K=6.8\) at 1200 . \(\mathrm{K}\) (b) \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) (s) \(\rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}), K=1.1 \times 10^{-12}\) at \(298 \mathrm{~K}\)

(a) Calculate the reaction Gibbs free energy of \(\mathrm{I}_{2}(\mathrm{~g}) \rightarrow\) \(2 \mathrm{I}(\mathrm{g})\) at \(1200 . \mathrm{K}(K=6.8)\) when the partial pressures of \(\mathrm{I}_{2}\) and \(\mathrm{I}\) are \(0.13\) bar and \(0.98\) bar, respectively. (b) What is the spontaneous direction of the reaction? Explain briefly.

State whether the following statements are true or false. If false, explain why. (a) A reaction stops when equilibrium is reached. (b) An equilibrium reaction is not affected by increasing the concentrations of products. (c) If one starts with a higher pressure of reactant, the equilibrium constant will be larger. (d) If one starts with higher concentrations of reactants, the equilibrium concentrations of the products will be larger.

A 25.0-g sample of ammonium carbamate, \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)\), was placed in an evacuated \(0.250\) - L flask and kept at \(25^{\circ} \mathrm{C}\). At equilibrium, \(17.4 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) was present. What is the value of \(K_{c}\) for the decomposition of ammonium carbamate into ammonia and carbon dioxide? The reaction is \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) .\)

Use the following data, which were collected at \(460 .^{\circ} \mathrm{C}\) and are equilibrium partial pressures, to determine \(K\) for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}):\) \(\begin{array}{lll}P_{\mathrm{H}_{2}}(\text {bar}) & P_{\mathrm{I}_{2}}(\text {bar}) & P_{\mathrm{HI}}(\text { bar }) \\ 6.47 \times 10^{-3} & 0.594 \times 10^{-3} & 0.0137 \\ 3.84 \times 10^{-3} & 1.52 \times 10^{-3} & 0.0169 \\\ 1.43 \times 10^{-3} & 1.43 \times 10^{-3} & 0.0100\end{array}\)

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