74 Practice Problem Calculate the thermodynamic equilibrium constant at 25 °C for the following reaction: -66.19 Agl(s) Ag¹(aq) + (aq) disolution 77.12 -51.59 AG (kJ/mol) Substance at 25 °C Ag(s) 0 Favors reactants Ag (aq) 77.12 Agl(s) -66.19 Kc = 8.4 x 10-17 I(g) 70.21 1(aq) -51.59 1¯(g) -221.5 L (+) (-) (-) Nonspontan AG = -RT In(K) Put C hyd ove ds. As to fing
74 Practice Problem Calculate the thermodynamic equilibrium constant at 25 °C for the following reaction: -66.19 Agl(s) Ag¹(aq) + (aq) disolution 77.12 -51.59 AG (kJ/mol) Substance at 25 °C Ag(s) 0 Favors reactants Ag (aq) 77.12 Agl(s) -66.19 Kc = 8.4 x 10-17 I(g) 70.21 1(aq) -51.59 1¯(g) -221.5 L (+) (-) (-) Nonspontan AG = -RT In(K) Put C hyd ove ds. As to fing
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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![### Practice Problem: Thermodynamic Equilibrium Constant
#### Problem Statement:
Calculate the thermodynamic equilibrium constant at 25°C for the following reaction:
\[ \text{AgI}(s) \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \]
#### Data Table:
| Substance | \( \Delta G^\circ \) (kJ/mol) at 25°C |
|-----------|---------------------------------------|
| I\(_2\) (g) | -221.5 |
| I\(_2\) (aq) | -51.59 |
| AgI (s) | -66.19 |
| Ag\(^+\) (aq)| 77.12 |
| I\(^-\) (aq) | -51.59 |
#### Calculation:
Using the relationship:
\[ \Delta G^\circ = -RT \ln(K) \]
Where:
- \(R\) is the gas constant \( (8.314 \, \text{J/mol} \cdot \text{K} = 0.008314 \, \text{kJ/mol} \cdot \text{K}) \)
- \(T\) is the temperature in Kelvin (25°C + 273.15 = 298.15 K)
- \(K\) is the equilibrium constant
To find \(\Delta G^\circ\) for the reaction, we use the Gibbs free energy of formation (\( \Delta G^\circ_f \)) values:
\[ \Delta G^\circ = \sum \Delta G^\circ_f(\text{products}) - \sum \Delta G^\circ_f(\text{reactants}) \]
For the given reaction:
\[ \Delta G^\circ = [77.12 + (-51.59)] - (-66.19) = 77.12 - 51.59 + 66.19 = 91.72 \, \text{kJ/mol} \]
Given \(\Delta G^\circ = 91.72 \, \text{kJ/mol}\):
\[ 91.72 = -0.008314 \times 298.15 \ln(K) \]
\[ \ln(K) = -\frac{91.72}{0.008314 \times 298.15} = -37.08 \]
\[ K = e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb99220ad-d702-4943-b904-ec3abe399c73%2Ffcb22c7f-43fd-472c-9e0e-45bb807c2cd0%2F231jj17_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Practice Problem: Thermodynamic Equilibrium Constant
#### Problem Statement:
Calculate the thermodynamic equilibrium constant at 25°C for the following reaction:
\[ \text{AgI}(s) \rightleftharpoons \text{Ag}^+ (aq) + \text{I}^- (aq) \]
#### Data Table:
| Substance | \( \Delta G^\circ \) (kJ/mol) at 25°C |
|-----------|---------------------------------------|
| I\(_2\) (g) | -221.5 |
| I\(_2\) (aq) | -51.59 |
| AgI (s) | -66.19 |
| Ag\(^+\) (aq)| 77.12 |
| I\(^-\) (aq) | -51.59 |
#### Calculation:
Using the relationship:
\[ \Delta G^\circ = -RT \ln(K) \]
Where:
- \(R\) is the gas constant \( (8.314 \, \text{J/mol} \cdot \text{K} = 0.008314 \, \text{kJ/mol} \cdot \text{K}) \)
- \(T\) is the temperature in Kelvin (25°C + 273.15 = 298.15 K)
- \(K\) is the equilibrium constant
To find \(\Delta G^\circ\) for the reaction, we use the Gibbs free energy of formation (\( \Delta G^\circ_f \)) values:
\[ \Delta G^\circ = \sum \Delta G^\circ_f(\text{products}) - \sum \Delta G^\circ_f(\text{reactants}) \]
For the given reaction:
\[ \Delta G^\circ = [77.12 + (-51.59)] - (-66.19) = 77.12 - 51.59 + 66.19 = 91.72 \, \text{kJ/mol} \]
Given \(\Delta G^\circ = 91.72 \, \text{kJ/mol}\):
\[ 91.72 = -0.008314 \times 298.15 \ln(K) \]
\[ \ln(K) = -\frac{91.72}{0.008314 \times 298.15} = -37.08 \]
\[ K = e
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