Whats the answer?

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

ΔH value indicates the amount of heat associated with the reaction involving the number of

moles of reactants and products

So, in the original equation 3CH3Br+NH3→(CH3)3N+3HBr ΔH=73KJ this indicates that when 3 moles of CH3BrL and 1 mole of NH3 change to 1 mole of (CH3)3N and 3 moles of HBr, 73 kJ of heat is released to the surroundings.

If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor.

Therefore,

6CH3Br+2NH→2(CH3)3 N+6HBr

If you multiply the coefficients of the original equation by 2 you get this equation. So multiply ΔH by 2, thus, ΔH for this equation is 73*2 =146 KJ

1/3(CH3)3 N+HBr→CH3Br+1/3NH3

First, this equation is the reverse of the original one so you must multiply ΔH by -1. Then, If you divide the coefficients of the original equation by 3 you get this equation. Therefore, you must devide ΔH by -3.

Thus, ΔH for this equation is 73/(-3) = -24.3 KJ

(CH3)3 N+3HBr→3CH3Br+NH3

This equation is the reverse of the original so multiply ΔH by (-1). Thus ΔH = -73 KJ