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Chapter 6: Problem 69

The temperature at which the reaction $$ \mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) $$ is at equilibrium is \(\ldots \ldots \ldots\) Given \(\Delta \mathrm{H}=30 \mathrm{~kJ}\) mol and \(\Delta S=0.066 \mathrm{~kJ} \mathrm{k}^{\prime} \mathrm{mol}^{\prime}\). (1) \(462.12 \mathrm{~K}\) (2) \(362.12 \mathrm{~K}\) (3) \(262.12 \mathrm{~K}\) (4) \(562.12 \mathrm{~K}\)

Short Answer

Expert verified
462.12 K

Step by step solution

01

- Understand the given data

Identify the given values: \( \Delta \mathrm{H}=30 \mathrm{~kJ} / \mathrm{mol} \) and \( \Delta S=0.066 \mathrm{~kJ} / (\mathrm{mol} \cdot \mathrm{K}) \)
02

- Recall the condition for equilibrium

For a reaction to be at equilibrium, the Gibbs free energy change, \( \Delta G \), must be zero. The relationship is given by: \[ \Delta G = \Delta H - T \Delta S \] At equilibrium, \( \Delta G = 0 \) which implies \( \Delta H = T \Delta S \)
03

- Solve for temperature (T)

Using the equation \[ \Delta H = T \Delta S \], solve for \( T \): \[ T = \frac{ \Delta H}{ \Delta S } \] Substitute the given values: \[ T = \frac{30 \mathrm{~kJ} / \mathrm{mol}}{0.066 \mathrm{~kJ} / (\mathrm{mol} \cdot \mathrm{K})} \]
04

- Calculate the value of T

Perform the division: \[ T = \frac{30}{0.066} = 454.55 \]Round to one of the given options closest to 454.55, which is 462.12 \( K \)

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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 is a thermodynamic potential used to predict whether a process will occur spontaneously at constant pressure and temperature. It is defined as: \[ \text{Gibbs free energy} (\text{G}) = \text{enthalpy} (\text{H}) - \text{temperature} (\text{T}) \times \text{entropy} (\text{S}) \]
In this equation:
  • \textbf{Enthalpy (H)} represents the total energy of the system.
  • \textbf{Temperature (T)} is in Kelvins and must always be positive.
  • \textbf{Entropy (S)} measures the disorder or randomness of the system.
When \
Enthalpy Change (螖H)
Enthalpy (\text{螖H}) refers to the heat content of a system and is a measure of the total energy in the system. It accounts for energy stored in chemical bonds and during phase changes.
  • Enthalpy change is given by \( \text{螖H} = \text{H}_{\text{products}} - \text{H}_{\text{reactants}} \).
    If \( \text{螖H} > 0 \), the process is endothermic (heat is absorbed).
    If \( \text{螖H} < 0 \), the process is exothermic (heat is released).

In the given problem, the enthalpy change for the decomposition of \
Entropy Change (螖S)
Entropy (\text{螖S}) is a measure of randomness or disorder in a system. In a reaction, the entropy change can help predict the feasibility of the process.
  • A positive entropy change (\text{螖S} > 0) indicates an increase in disorder.
  • A negative entropy change (\text{螖S} < 0) indicates a decrease in disorder.

In the given exercise, the entropy change, \( \text{螖S} = 0.066 \text{kJ/(mol鈰匥)} \), measures how the disorder or randomness of the system changes during the decomposition reaction of \

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Most popular questions from this chapter

Equal volumes of \(1 \mathrm{M} \mathrm{HCl}\) and \(1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are neutralized by dilute \(\mathrm{NaOH}\) solution and \(\mathrm{X}\) and \(\mathrm{Y}\) kcal of heat are liberated, respectively. Which of the following is true? (1) \(\mathrm{X}=\mathrm{Y}\) (2) \(\mathrm{X}=0.5 \mathrm{Y}\) (3) \(\mathrm{X}=0.4 \mathrm{Y}\) (4) None

If \(\Delta H\) is the change in cathalpy and \(\Delta U\) is the change in internal cnergy, then (1) \(\Delta H\) is always less than \(\Delta U\) (2) \(\Delta H\) is always greatur than \(\Delta U\) (3) \(\Delta H<\Delta U\) only if the number of moles of gascous products is greater than the number of moles of gascous reactants (4) \(\Delta H<\Delta U\) only if the number of moles of gascous products is less than the number of moles of gascous reactants

Ileat of solution is defined as (1) Heat required in dissolving 1 mole in excess of water. (2) Heat evolved when 1 mole is dissolved in excess of water. (3) Change in heat content of the system when I mole of the solute is dissolved in excess of water so that further dilution of solution does not bring any heat change. (4) None of the above.

A mong the following, the wrong statement is (1) Entropy decreases during the crystallization of a solute from solution. (2) At a certain temperuture \(T\), the endothermic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) proceeds almost to completion if \(\Delta S>0\).(3) In a spontaneous irreversible process the total entropy of the system and surroundings increases. (4) When the value of entropy is greater, then the ability to work is minimum,

A gaseous system changes from state \(\mathrm{A}\left(P_{1}, V_{1}, T_{1}\right)\) to \(\mathrm{B}\left(P_{2}, V_{2}, T_{2}\right)\) to \(\mathrm{C}\left(P_{3}, V_{3}, T_{3}\right)\) and finally from C to \(\mathrm{A}\). The whole process may be called (1) Reversible process (2) Cyclic process (3) Isobaric process (4) Spontancous process
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