The value of Kb for methylamine CH3NH2 is 44 x 104 at 25 oC

The value of Kb for methylamine (CH3NH2) is 4.4 x 10-4 at 25 oC. what is the value ofG when [H+] 6.7 x 10 M, [CH3NHs\'1-2.4x 103 M, and (CH3NH2) 0.098 M 0 +23.28 kJ O -23.28 kJ O There is not enough information to solve this problem

Solution

The balanced equation

CH3NH2(aq) + H2O(l) = CH3NH3+(aq) + OH-(aq)

Step 1

Calculate the standard gibbs free energy change of reaction

G° = -RT ln(K)
G° = - (8.314 J/K*mole) (25 + 273.15 K) ln (4.4 x 10^-4)
G° = 19.1 kJ/mole.

Step 2

Calculate the Gibbs energy change of reaction at different concentrations

G = G° + RT ln(Q)

From the reaction

Q = [OH-] [CH3NH3+] / [CH3NH2]

[H+] [OH-] = 1.0 x 10^-14

[OH-] = (1.0 x 10^-14) / (6.7 x 10^-9 M)

= 1.493 x 10^-6

substitute the given values into the equation to find Q:

Q = (1.493 x 10^-6) (2.4 x 10^-3) / (0.098)

= 3.656 x 10^-8

For 1 mol -
G = 1 mol x 19.1 kJ/mole + 1 mol x (8.314 J/K*mole) x (1kJ/1000J) x (25 + 273.15 K) x ln(3.656 x 10^-8)

G = - 23.28 kJ

Option B is the correct answer