Using the Drago-Wayland equation, calculate H for the reactions of trimethylamin

Question

Using the Drago-Wayland equation, calculate H for the reactions of trimethylamine and BF3 and of trimethylamine and B(CH3)3. Comment on the calculated values in terms of the structures of BF3 and B(CH3)3. Discuss the values in terms of HSAB theory.

(CH3)3N: CB=11.54 kcal/mol EB=.808 kcal/mol

BF3: CA=1.62 kcal/mol EA=9.88 kcal/mol

B(CH3)3: CA=1.70 kcal/mol EA=6.14 kcal/mol

I know I'm using the correct formula and the correct values but I'm getting huge numbers for H that I know aren't right. I converted to Joules by multiplying by 4.184 so maybe that's where I went wrong? Any help would be appreciated.

Explanation / Answer

-delta H = Ea*Eb + Ca*Cb

H for the reaction of trimethylamine and BF3

-delta H = 9.88*0.808 + 1.62*11.54 = 26.67 Kcal/mol

delta H = -26.67 Kcal/mol = -111.58 KJ

delta H for the reaction of trimethylamine and B(CH3)3

-delta H = 4.96 + 19.61 = 24.579KCal = 102.839 KJ/mol

delta H = -102.839 KJ/mol

I think u have taken 0.808 Kcal/mol as 808 Kcal ....

Any doubts please comment

Bond energies of B-F is greater than bond energy of B-C

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