Chapter 18: Problem 122
\(K_{s p}\) for silver chloride at \(25.0^{\circ} \mathrm{C}\) is \(1.782 \times 10^{-10}\). At \(35.0^{\circ} \mathrm{C}, K_{s p}^{\circ}\) is \(4.159 \times 10^{-10} .\) What are \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction?
Short Answer
Expert verified
\(\Delta H^{\circ} \approx 21.2 \text{ kJ/mol}\), \(\Delta S^{\circ} \approx -122.9 \text{ J/mol K}\).
Step by step solution
01
Understand the Reaction
The solubility product, \(K_{sp}\), represents the equilibrium between the solid phase and its ions in solution. For silver chloride, the dissociation is \( \text{AgCl(s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \).
02
Use the Van't Hoff Equation
The Van't Hoff equation relates the change in the equilibrium constant to the change in temperature: \( \ln \left( \frac{K_2}{K_1} \right) = \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \). Identify \(K_1 = 1.782 \times 10^{-10}\) and \(K_2 = 4.159 \times 10^{-10}\), \(T_1 = 298.15 \text{ K}\), and \(T_2 = 308.15 \text{ K}\).
03
Solve for \(\Delta H^{\circ}\)
Calculate \( \ln \left( \frac{4.159 \times 10^{-10}}{1.782 \times 10^{-10}} \right) \) to find the left side of the equation. Then, use the known value of the gas constant \( R = 8.314 \text{ J/mol K} \) to solve for \( \Delta H^{\circ} \).
04
Calculate \(\Delta H^{\circ}\)
The value of \( \ln \left( \frac{4.159}{1.782} \right) \) is 0.852. Plug this value into the Van't Hoff equation which simplifies to \( 0.852 = \frac{-\Delta H^{\circ}}{8.314} \left( \frac{1}{308.15} - \frac{1}{298.15} \right) \). Solving for \( \Delta H^{\circ} \), calculate the temperature difference term first and multiply it by \( 8.314 \) to isolate \( \Delta H^{\circ} \).
05
Find the Temperature Difference Term
Compute \( \frac{1}{308.15} - \frac{1}{298.15} = -3.34 \times 10^{-4}\, \text{K}^{-1} \).
06
Solve for \(\Delta H^{\circ}\)
Substitute the temperature difference into the equation: \( 0.852 = \frac{-\Delta H^{\circ}}{8.314} (-3.34 \times 10^{-4}) \). Solving gives \( \Delta H^{\circ} = 21.2 \text{ kJ/mol} \).
07
Use the Gibbs Free Energy Equation
Now, use \( \Delta G^{\circ} = -RT \ln K_{sp} \) at \( T_1 \) and \( T_2 \) to find \( \Delta G^{\circ} \). Calculate \( \Delta G^{\circ}(298.15) = -8.314 \times 298.15 \times \ln(1.782 \times 10^{-10}) \) and similarly for \( T_2 \).
08
Compute \(\Delta G^{\circ}\) Values
For \( 298.15 \text{ K} \), \( \Delta G^{\circ} = 57.9 \text{ kJ/mol} \) and for \( 308.15 \text{ K} \), \( \Delta G^{\circ} = 55.2 \text{ kJ/mol} \).
09
Calculate \(\Delta S^{\circ}\) Using \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\)
Using the two \( \Delta G^{\circ} \) values, \(abla S^{\circ}\) can be found by rearranging the equation to \( abla S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} \) for each temperature. Calculate for both, then ensure \( \Delta S^{\circ} \) is consistent.
10
Final Calculation for \(\Delta S^{\circ}\)
Using temperatures 298.15 K and 308.15 K, calculate using \(\Delta H^{\circ} \approx 21.2 \text{ kJ/mol}\). Calculate \( \frac{21.2 - 57.9}{298.15} \) and \( \frac{21.2 - 55.2}{308.15} \); average these values to approximate consistent \(\Delta S^{\circ}\). The calculated \( \Delta S^{\circ} \) is approximately \(-122.9 \text{ J/mol K}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product
The solubility product, often represented as \(K_{sp}\), is a special type of equilibrium constant that applies to sparingly soluble salts. It describes the equilibrium between a solid substance and its ions dissolved in a solution. For silver chloride, \(\text{AgCl (s)} \rightleftharpoons \text{Ag}^+ \text{ (aq)} + \text{Cl}^- \text{ (aq)}\), the solubility product reflects how easily the solid dissolves into these ions in water.
- A low \(K_{sp}\) means the compound is less soluble in water.
- A high \(K_{sp}\) suggests more ions are released into the solution, meaning higher solubility.
Enthalpy Change
Enthalpy change, denoted as \(\Delta H\), represents the heat absorbed or released during a chemical reaction at constant pressure. It is a crucial concept when studying energy changes in reactions.
In this exercise, we determine \(\Delta H^{\circ}\) using the Van't Hoff equation, which connects it to temperature changes and equilibrium constants:
\[\ln \left( \frac{K_2}{K_1} \right) = \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\]
In this exercise, we determine \(\Delta H^{\circ}\) using the Van't Hoff equation, which connects it to temperature changes and equilibrium constants:
\[\ln \left( \frac{K_2}{K_1} \right) = \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\]
- Positive \(\Delta H\) indicates an endothermic reaction, which absorbs heat.
- Negative \(\Delta H\) indicates an exothermic reaction, releasing heat.
Equilibrium Constant
Equilibrium constants, including \(K_{sp}\), indicate the ratio of product concentrations to reactant concentrations at equilibrium.
They apply to reversible reactions, where forward and backward reactions occur simultaneously at equal rates:
The increase from \(1.782 \times 10^{-10}\) to \(4.159 \times 10^{-10}\) with temperature signifies increased solubility as the temperature rises.
They apply to reversible reactions, where forward and backward reactions occur simultaneously at equal rates:
- An equilibrium constant greater than 1 implies products are favored.
- An equilibrium constant less than 1 suggests the reactants are favored.
The increase from \(1.782 \times 10^{-10}\) to \(4.159 \times 10^{-10}\) with temperature signifies increased solubility as the temperature rises.
Gibbs Free Energy
Gibbs free energy, \(\Delta G\), assesses a reaction's spontaneity at constant temperature and pressure. A negative \(\Delta G\) implies a spontaneous process, while positive \(\Delta G\) means non-spontaneity.
In this exercise, we calculate \(\Delta G^{\circ}\) for silver chloride at two temperatures:
In this exercise, we calculate \(\Delta G^{\circ}\) for silver chloride at two temperatures:
- At \(298.15 \text{ K}\), \(\Delta G^{\circ}\) is \(57.9 \text{ kJ/mol}\)
- At \(308.15 \text{ K}\), \(\Delta G^{\circ}\) is \(55.2 \text{ kJ/mol}\)