/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 From the values given for \(\Del... [FREE SOLUTION] | 91Ó°ÊÓ

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From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ},\) calculate \(\Delta G^{\circ}\) for each of the following reactions at \(298 \mathrm{~K}\). If the reaction is not spontaneous under standard conditions at \(298 \mathrm{~K}\), at what temperature (if any) would the reaction become spontaneous? $$ \begin{array}{l} \text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\ \text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} \end{array} $$

Short Answer

Expert verified
For reaction (a), the calculated value of \(\Delta G^{\circ} = -764.82 \, \mathrm{kJ}\) indicates that the reaction is spontaneous at \(298 \mathrm{~K}\). For reaction (b), the calculated value of \(\Delta G^{\circ} = 161.662 \, \mathrm{kJ}\) indicates that the reaction is not spontaneous at \(298 \mathrm{~K}\), but it becomes spontaneous at a temperature of approximately \(3196 \mathrm{~K}\).

Step by step solution

01

(a) Calculate \(\Delta G^{\circ}\) for the reaction

Using the given values for the reaction: $$ \Delta H^{\circ} = -844 \mathrm{~kJ} $$ $$ \Delta S^{\circ} = -165 \mathrm{~J} / \mathrm{K} $$ We will plug these values into the formula for \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\): $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = -844\times10^3 \, \mathrm{J} - (298 \, \mathrm{K})(-165 \, \mathrm{J/K}) $$ Now we'll calculate the value of delta G: $$ \Delta G^{\circ} = -764820 \, \mathrm{J} = -764.82 \, \mathrm{kJ} $$
02

(a) Analyze the spontaneity of the reaction

Since the value of \(\Delta G^{\circ} < 0\), the reaction is spontaneous at \(298 \mathrm{~K}\). Therefore, there is no need to calculate the temperature at which the reaction becomes spontaneous.
03

(b) Calculate \(\Delta G^{\circ}\) for the reaction

Using the given values for the reaction: $$ \Delta H^{\circ} = 572 \mathrm{~kJ} $$ $$ \Delta S^{\circ} = 179 \mathrm{~J} / \mathrm{K} $$ We will plug these values into the formula for \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\): $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = 572\times10^3 \, \mathrm{J} - (298 \, \mathrm{K})(179 \, \mathrm{J/K}) $$ Calculating the value of delta G: $$ \Delta G^{\circ} = 161662 \, \mathrm{J} = 161.662 \, \mathrm{kJ} $$
04

(b) Analyze the spontaneity of the reaction

Since the value of \(\Delta G^{\circ} > 0\), the reaction is not spontaneous at \(298 \mathrm{~K}\). We need to find the temperature at which the reaction becomes spontaneous.
05

(b) Find the temperature for spontaneous reaction

To find the temperature at which \(\Delta G^{\circ} = 0\), we will plug the value of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) into the formula, and solve for T: $$ 0 = 572\times10^3 \, \mathrm{J} - T(179 \, \mathrm{J/K}) $$ Now solve for T: $$ T = \frac{572\times10^3 \, \mathrm{J}}{179 \, \mathrm{J/K}} \approx 3196 \, \mathrm{K} $$ So, the reaction becomes spontaneous at a temperature of approximately \(3196 \mathrm{~K}\).

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Key Concepts

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

Spontaneity of Reactions
In chemistry, the spontaneity of reactions refers to whether a chemical reaction can occur without any external influence, like added energy. The key indicator for this is the Gibbs Free Energy change, denoted as \( \Delta G \). A reaction is considered spontaneous if \( \Delta G < 0 \). This means that the process can happen on its own under given conditions, releasing free energy.
For non-spontaneous reactions, where \( \Delta G > 0 \), external energy would be required to initiate the reaction. For instance, in the exercise we reviewed, reaction (a), where \( \Delta G = -764.82 \text{ kJ} \), is spontaneous, suggesting it will occur naturally without extra input at \( 298 \text{ K} \).
Meanwhile, reaction (b) had \( \Delta G = 161.662 \text{ kJ} \), indicating it is non-spontaneous at standard temperature, requiring either a change in temperature or conditions to become spontaneous.
Thermodynamic Calculations
Thermodynamic calculations form the core of determining reaction spontaneity and involve using the Gibbs Free Energy equation:

\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]

This formula incorporates three variables: enthalpy change (\( \Delta H \)), temperature (\( T \)), and entropy change (\( \Delta S \)). Since enthalpy and entropy often have units that differ (kJ for \( \Delta H \) and J/K for \( \Delta S \)), you need to convert them to matching units, typically Joules, for accurate calculations.
In our example problem, for each reaction at \( 298 \text{ K} \), we calculated \( \Delta G^{\circ} \) by ensuring unit consistency and applying given values to the equation. The calculations give insight into whether energy is absorbed or released in the process, helping us understand readiness to occur spontaneously.
Solving thermodynamic problems accurately requires handling these conversions with care and precision to understand the behavior of chemical reactions under specific conditions.
Enthalpy and Entropy
Enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) are thermodynamic properties fundamental to understanding chemical reactions.
Enthalpy involves the heat content in a system. A negative \( \Delta H \) (exothermic reaction) releases heat, lowering system energy, more likely leading to spontaneity. For example, reaction (a) with \( \Delta H = -844 \text{ kJ} \) is spontaneous at standard temperature, suggesting that it naturally releases free energy.
On the other hand, entropy reflects the degree of disorder or randomness in a system. Reactions with increased entropy (positive \( \Delta S \)) tend to be spontaneous since systems naturally progress towards more disorder. In our exercise, reaction (b)'s \( \Delta S = 179 \text{ J/K} \) indicates increased entropy, but the high enthalpy prevents spontaneity at \( 298 \text{ K} \).
Hence, analyzing both \( \Delta H \) and \( \Delta S \) helps predict reaction behavior. When balancing endothermic reactions with significant entropy increases and exothermic reactions despite entropy changes, these concepts crucially inform thermodynamic spontaneity and feasibility.

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Most popular questions from this chapter

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(\mathrm{CO}(g)+\mathrm{H}_{2}(g) \longrightarrow C(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{O}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{HCl}(g)+\mathrm{NH}_{3}(g)\) (d) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g)\)

A system goes from state 1 to state 2 and back to state \(1 .\) (a) Is \(\Delta E\) the same in magnitude for both the forward and reverse processes? (b) Without further information, can you conclude that the amount of heat transferred to the system as it goes from state 1 to state 2 is the same or different as compared to that upon going from state 2 back to state \(1 ?(\mathbf{c})\) Suppose the changes in state are reversible processes. Is the work done by the system upon going from state 1 to state 2 the same or different as compared to that upon going from state 2 back to state \(1 ?\)

The potassium-ion concentration in blood plasma is about \(5.0 \times 10^{-3} \mathrm{M}\), whereas the concentration in muscle-cell fluid is much greater \((0.15 \mathrm{M})\). The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to \(\mathrm{K}^{+}\). (a) What is \(\Delta G\) for the transfer of \(1 \mathrm{~mol}\) of \(\mathrm{K}^{+}\) from blood plasma to the cellular fluid at body temperature \(37^{\circ} \mathrm{C} ?\) (b) What is the minimum amount of work that must be used to transfer this \(\mathrm{K}^{+} ?\)

The reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{\circ}\). Did your classmate make a mistake in the calculation? Explain.

(a) What is the difference between a state and a microstate of a system? (b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state? (c) In a particular spontaneous process, the number of microstates available to the system decreases. What can you conclude about the sign of \(\Delta S\) surr?

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