Whats the answer for this?

Question

A chemist measures the enthalpy change ΔH during the following reaction:

3CH3Br+NH3→(CH3)3N+3HBr ΔH=73KJ

Use this information to complete the table below. Round each of your answers to the nearest kJ/mol.

6CH3Br+2NH→2(CH3)3 N+6HBr

1/3(CH3)3 N+HBr→CH3Br+1/3NH3

(CH3)3 N+3HBr→3CH3Br+NH3

Answer 1

3CH₃Br + NH₃ ==> (CH₃)₃N + 3HBr ... ΔH = 73KJ

6CH₃Br + 2NH₃ ==> 2(CH₃)₃N + 6HBr This is 2x the original. Multiply 73 kJ by 2

1/3(CH₃)₃N + HBr ==> CH₃Br + 1/3 NH₃ This is the reverse of the original and is 1/3 so divide 73 kJ by 3 and change the sign

(CH₃)₃N + 3HBr ==> 3CH₃Br + NH₃ This is the reverse of the original so simply change the sign