The equilibrium constant for the equation CH3COOH(aq) + H2O(l) H3O + (aq) + CH3COO-(aq) is Ka = 1.8 Times 10-5 M at 25.0 degree C. Calculate the value of DeltaGdegreerXn at 25.0 degree C for acetic acid when [CH3COO-] = [H3O+] = [CH3COOH] = 1.00 M Will the acid spontaneously dissociate under these conditions? Calculate the value of DeltaGrxn at 25 degree C for acetic acid when [CH3COO-]=[H3OO + ] = 4.56 Times 10-5 M and [CH3COOH] = 1.140 M Will the acid spontaneously dissociate under these conditions?
The Correct Answer and Explanation is :
Step 1: Definitions and Formulae
The equilibrium constant ((K_a)) and Gibbs free energy change ((\Delta G)) are related by the equation:
[
\Delta G^\circ_{\text{rxn}} = -RT \ln K_a
]
Where:
- (R = 8.314 \, \text{J/mol·K}), the gas constant.
- (T = 298.15 \, \text{K}), the temperature in Kelvin.
- (K_a = 1.8 \times 10^{-5}), the equilibrium constant.
Under non-standard conditions, the reaction Gibbs free energy ((\Delta G_{\text{rxn}})) is given by:
[
\Delta G_{\text{rxn}} = \Delta G^\circ_{\text{rxn}} + RT \ln Q
]
Where:
- (Q = \frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}), the reaction quotient.
Part 1: When ([\text{CH}_3\text{COOH}] = [\text{H}_3\text{O}^+] = [\text{CH}_3\text{COO}^-] = 1.00 \, \text{M})
- Calculate (\Delta G^\circ_{\text{rxn}}):
[
\Delta G^\circ_{\text{rxn}} = -RT \ln K_a = -(8.314)(298.15) \ln(1.8 \times 10^{-5})
]
[
\Delta G^\circ_{\text{rxn}} \approx 26920 \, \text{J/mol} = 26.92 \, \text{kJ/mol}
] - Calculate (Q):
[
Q = \frac{(1.00)(1.00)}{1.00} = 1.00
] - Calculate (\Delta G_{\text{rxn}}):
[
\Delta G_{\text{rxn}} = \Delta G^\circ_{\text{rxn}} + RT \ln Q = 26.92 + 0 = 26.92 \, \text{kJ/mol}
]
Since (\Delta G_{\text{rxn}} > 0), the dissociation is non-spontaneous under these conditions.
Part 2: When ([\text{CH}_3\text{COO}^-] = [\text{H}_3\text{O}^+] = 4.56 \times 10^{-5} \, \text{M}) and ([\text{CH}_3\text{COOH}] = 1.140 \, \text{M})
- Calculate (Q):
[
Q = \frac{(4.56 \times 10^{-5})(4.56 \times 10^{-5})}{1.140}
]
[
Q \approx 1.82 \times 10^{-9}
] - Calculate (\Delta G_{\text{rxn}}):
[
\Delta G_{\text{rxn}} = \Delta G^\circ_{\text{rxn}} + RT \ln Q
]
[
\Delta G_{\text{rxn}} = 26.92 + (8.314)(298.15) \ln(1.82 \times 10^{-9})
]
[
\Delta G_{\text{rxn}} \approx 26.92 – 50.87 \approx -23.95 \, \text{kJ/mol}
]
Since (\Delta G_{\text{rxn}} < 0), the dissociation is spontaneous under these conditions.
Conclusion
- When ([\text{CH}_3\text{COOH}] = [\text{H}_3\text{O}^+] = [\text{CH}_3\text{COO}^-] = 1.00 \, \text{M}), dissociation is non-spontaneous.
- When ([\text{CH}_3\text{COO}^-] = [\text{H}_3\text{O}^+] = 4.56 \times 10^{-5} \, \text{M}) and ([\text{CH}_3\text{COOH}] = 1.140 \, \text{M}), dissociation is spontaneous.
This demonstrates the importance of the reaction quotient ((Q)) in determining spontaneity, as it compares actual concentrations to equilibrium concentrations.