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ΔG˚ = 19.1 kJ Kb for (CH3NH2) at 25˚C is 4.4x10-4 What is the value of...

ΔG˚ = 19.1 kJ Kb for (CH3NH2) at 25˚C is 4.4x10-4

What is the value of ΔG when [H+] = 1.6 ×10-8 M, [CH3NH3+] = 5.1x10-4, and [CH3NH2] = 0.130 M?

I got the answer as 49.1 kJ using the following equation:

ΔG= -RT ln Kb

ΔG= (-8.314 J/mol*K)(298.15 K)(ln(2.4519x10-9) = 49,146 J/mol = 49.1 kJ/mol

The answer wants it in kJ not kJ/mol. What am I doing wrong?

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Homework Answers

Answer #1

CH3NH2 (aq) + H2O (l) <-> CH3NH3+(aq) + OH-(aq)

we convert H+ concentration to OH-

as H+ * OH- = 10-14

OH-   = 10-14 /  H+ = 10-14 /1.6 ×10-8 = 0.625 * 10-6

Kb=  CH3NH3+ *  OH- / CH3NH2 = 5.1x10-4 mole * 0.625 * 10-6   mole /  0.130 mole = 24.5192 * 10-10 mole

G0= -RTlnK = - 8.313 * 298.15 K * 2.303 log  24.5192 * 10-10 mole

= 8.313J/mol*K * 298.15 K * 2.303 * 8.6104 mole = 49154.366 = 49.154 KJ

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