a. Write a balanced chemical equation for the acid dissociation of acetic acid, CH3COOH, in water. (1 point)
CH3COOH + H2O --> CH3COO- + H3O+
b. Using the thermodynamic data in appendix B, calculate ΔrxnGo and Ka for this reaction at 298 K. Do not include water as a reactant in your equation.
G0f (kJ mol-1)
CH3COOH (aq) -396.46
CH3COO- (aq) -369.31
H3O+ (aq) 0
H2O(l) -285.8
Grxn = Gproducts - GReactants = (-369.31 + 0)-(-396.46) = 27.15
First of all, shouldn't Grxn be negative if this is spontaneous?
Subbing into lnK = -G/RT
lnK = -27.15/(8.314)(298) =
lnK = -0.01095
K = 0.989
Ka for acetic acid is 1.8x10-5
What am I doing wrong?