Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K. %3D 1A +3B →5C + 4D +3E AHF°(kJ/mol) A -19.68 s°(J/mol/K) 13.41 B -9.29 10.92 C 0.83 19.81 D 15.35 9.48 E 3.32 0.15 Remember, pK = -log(K) Report your answer to two decimal points.
Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K. %3D 1A +3B →5C + 4D +3E AHF°(kJ/mol) A -19.68 s°(J/mol/K) 13.41 B -9.29 10.92 C 0.83 19.81 D 15.35 9.48 E 3.32 0.15 Remember, pK = -log(K) Report your answer to two decimal points.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K.**
Reaction:
\[ \text{1A + 3B} \rightarrow \text{5C + 4D + 3E} \]
| Substance | \(\Delta H_f^0\) (kJ/mol) | \(S^0\) (J/mol/K) |
|-----------|-----------------|----------------|
| A | -19.68 | 13.41 |
| B | -9.29 | 10.92 |
| C | 0.83 | 19.81 |
| D | 15.35 | 9.48 |
| E | 3.32 | 0.15 |
**Remember,**
\[ \text{pK} = -\log(K) \]
**Report your answer to two decimal points.**
---
**Explanation:**
1. **\(\Delta H_f^0\) column**: Lists the standard enthalpy of formation for each substance in kJ/mol.
2. **\(S^0\) column**: Lists the standard entropy for each substance in J/mol/K.
To calculate the equilibrium constant (K), use the Gibbs free energy change (\(\Delta G^0\)) for the reaction:
\[ \Delta G^0 = \Delta H^0 - T\Delta S^0 \]
Where:
- \(\Delta H^0\) and \(\Delta S^0\) are determined from the reaction data.
- \(T\) is the temperature in Kelvin (298 K for 25°C).
Calculate \(\Delta G^0\) and then use the relation:
\[ \Delta G^0 = -RT \ln(K) \]
to solve for \(K\) and subsequently, \(\text{pK} = -\log(K)\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84d84543-76fc-42a7-8001-3bf35a9f5c0a%2F406e995f-3954-4e2d-a116-bc36290006ea%2F667d0c4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K.**
Reaction:
\[ \text{1A + 3B} \rightarrow \text{5C + 4D + 3E} \]
| Substance | \(\Delta H_f^0\) (kJ/mol) | \(S^0\) (J/mol/K) |
|-----------|-----------------|----------------|
| A | -19.68 | 13.41 |
| B | -9.29 | 10.92 |
| C | 0.83 | 19.81 |
| D | 15.35 | 9.48 |
| E | 3.32 | 0.15 |
**Remember,**
\[ \text{pK} = -\log(K) \]
**Report your answer to two decimal points.**
---
**Explanation:**
1. **\(\Delta H_f^0\) column**: Lists the standard enthalpy of formation for each substance in kJ/mol.
2. **\(S^0\) column**: Lists the standard entropy for each substance in J/mol/K.
To calculate the equilibrium constant (K), use the Gibbs free energy change (\(\Delta G^0\)) for the reaction:
\[ \Delta G^0 = \Delta H^0 - T\Delta S^0 \]
Where:
- \(\Delta H^0\) and \(\Delta S^0\) are determined from the reaction data.
- \(T\) is the temperature in Kelvin (298 K for 25°C).
Calculate \(\Delta G^0\) and then use the relation:
\[ \Delta G^0 = -RT \ln(K) \]
to solve for \(K\) and subsequently, \(\text{pK} = -\log(K)\).
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