Indicate whether \(\Delta G\) increases, decreases, or stays the same for each of the following reactions as the partial pressure of \(\mathrm{O}_{2}\) is increased: $$ \begin{array}{l}{\text { (a) } 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)} \\ {\text { (b) } 2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)} \\ {\text { (c) } 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)}\end{array} $$

Short Answer

Expert verified
For all three reactions (a), (b), and (c), as the partial pressure of \(O_2\) increases, \(\Delta G\) will increase.

Step by step solution

01

(a) Reaction 1 Analysis

First, let's analyze the reaction, $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ Representing the partial pressure of each gas species by the respective symbols, the reaction quotient, Q, for this reaction is given by: $$ Q=\frac{p_{CO_2}^2}{p_{CO}^2 \cdot p_{O_2}} $$ Now, we will find the effect of increasing the partial pressure of O₂ on the reaction quotient Q.
02

(a) Reaction 1 Effect on ∆G

When the partial pressure of O₂ increases, the denominator in the expression for Q will increase. Therefore, Q will decrease. Looking at the equation for ∆G, as Q decreases, ∆G will increase. So, for reaction (a), as the partial pressure of O₂ increases, ∆G will increase.
03

(b) Reaction 2 Analysis

Now, let's analyze the reaction, $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ The reaction quotient, Q, for this reaction is given by: $$ Q=\frac{p_{O_2} }{1} $$ Note that since both reactant and product are liquids, their concentrations remain constant and are not part of the quotient (hence we have "1" in the denominator).
04

(b) Reaction 2 Effect on ∆G

As the partial pressure of O₂ increases, Q also increases. Looking at the equation for ∆G, as Q increases, ∆G will increase. So, for reaction (b), as the partial pressure of O₂ increases, ∆G will increase.
05

(c) Reaction 3 Analysis

Finally, let's analyze the reaction, $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ The reaction quotient, Q, for this reaction is given by: $$ Q= p_{O_2}^3 $$ Note that since both reactants are solids, their concentrations remain constant and do not appear in the reaction quotient.
06

(c) Reaction 3 Effect on ∆G

When the partial pressure of O₂ increases, Q will increase as well. Looking at the equation for ∆G, as Q increases, ∆G will increase once again. So, for reaction (c), as the partial pressure of O₂ increases, ∆G will increase. In summary, for all three reactions (a), (b), and (c), as the partial pressure of O₂ increases, ∆G will increase.

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy, often denoted as \(\Delta G\), is a critical concept in thermodynamics that indicates the spontaneity of a reaction at constant temperature and pressure. It combines the system's enthalpy (\(\Delta H\)), temperature (\(T\)), and entropy (\(\Delta S\)) into one value. The general equation for Gibbs free energy is given as \(\Delta G = \Delta H - T\Delta S\). When \(\Delta G\) is negative, the reaction is spontaneous; if positive, the reaction is non-spontaneous.

In the context of the exercise, increasing the partial pressure of \(\mathrm{O}_{2}\) shifts the direction of the reaction which can increase \(\Delta G\), indicating that the reaction has become less spontaneous under the new conditions. Particularly, the use of the reaction quotient provides a snapshot of the reaction's position relative to equilibrium at any given moment.
Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. Dalton's Law of Partial Pressures asserts that in a gas mixture, the total pressure is the sum of the partial pressures of the constituent gases. The significance of partial pressure lies in its direct effect on the reaction's progress and therefore its impact on \(\Delta G\).

Changes in partial pressure can change the concentrations of reactants and products as per the reaction's stoichiometry, thus affecting the reaction quotient \(Q\) and consequently the Gibbs free energy. This is what happens in the exercise examples, where changing the partial pressure of \(\mathrm{O}_{2}\) influences \(\Delta G\).

Reaction Quotient (Q)

Understanding Reaction Quotient

The reaction quotient, \(Q\), represents the ratio of product concentrations to reactant concentrations at any moment during a reaction, each raised to the power of their respective stoichiometric coefficients. Unlike the equilibrium constant \(K\), which applies when a reaction is at a dynamic equilibrium, \(Q\) applies at any point and can indicate the direction in which a reaction will proceed to achieve equilibrium.

If \(Q < K\), the reaction tends to produce more products; if \(Q > K\), the reaction will favor the reactants. A change in \(Q\), like with the increase of \(\mathrm{O}_{2}\) partial pressure in the exercise, can lead to a change in \(\Delta G\), providing insight into how a reaction's spontaneity may alter under new conditions.

Le Chatelier's Principle

Application of Le Chatelier's Principle

Le Chatelier's principle states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way as to minimize the stress. This could refer to changes in concentration, pressure, volume, or temperature. When the partial pressure of a reactant or product gas is altered, as in the provided exercise, the system shifts to counteract the pressure change, thus affecting the equilibrium position.

This principle is closely related to the changes in \(\Delta G\) discussed in the exercise as it describes the system's response to increased \(\mathrm{O}_{2}\) partial pressure. Increasing the partial pressure of \(\mathrm{O}_{2}\) shifts the equilibrium, which can be related to the reaction quotient and the subsequent increase in Gibbs free energy, demonstrating the reaction's shift toward or away from spontaneity based on these principles.

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Most popular questions from this chapter

(a) Is the standard free-energy change, \(\Delta G^{\circ},\) always larger than \(\Delta G ?(\mathbf{b})\) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0\) ? (c) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process, necessarily has a low activation barrier?

Using data from Appendix \(\mathrm{C}\) , calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at 298 \(\mathrm{K}\) under standard conditions. (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g)\) (c) \(6 \mathrm{Cl}_{2}(g)+2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 4 \mathrm{FeCl}_{3}(s)+3 \mathrm{O}_{2}(g)\) (d) \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Using \(S^{\circ}\) values from Appendix \(\mathrm{C},\) calculate \(\Delta S^{\circ}\) values for the following reactions. In each case, account for the sign of \(\Delta S^{\circ} .\) $$ \begin{array}{l}{\text { (a) } \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} \\ {\text { (b) } \mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)} \\ {\text { (c) } \mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } 2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)}\end{array} $$

(a) Can endothermic chemical reactions be spontaneous? (b) Can a process be spontaneous at one temperature and nonspontaneous at a different temperature? (c) Water can be decomposed to form hydrogen and oxygen, and the hydrogen and oxygen can be recombined to form water. Does this mean that the processes are thermodynamically reversible? (d) Does the amount of work that a system can do on its surroundings depend on the path of the process?

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

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