Question

A. What was the ?H value obtained for NH3 + HCl

Answer 1

Just applying Hess's Law you get:

[eqn 1] ? NaOH + HCl ? NaCl + H2O ... ?H = -60 kJ/mol

[eqn 2] ? NH3 + HCl ? NH4Cl ... ?H = -38 kJ/mol

required reaction is NaOH + NH4Cl ? NaCl + NH3 + H2O ... ?H =?

[eqn 1] is OK the way it is

BUT [eqn 2] has to be flipped to get stuff on the right sides:

so [eqn 1] ? NaOH + HCl ? NaCl + H2O ... ?H = -60 kJ/mol

[flipped eqn 2] ? NH4Cl ? NH3 + HCl ... ?H = 38 kJ/mol ... [don/t forget to change the sign for ?H]

now [eqn 1] + [flipped eqn 2] ? NaOH + NH4Cl ? NaCl + NH3 + H2O ... ?H = -60 + 38 = -22 kJ/mol ... [the HCls cancel each other]

so from the given data the expected enthalpy change for the required reaction is -22 kJ/mol

not sure what's with the given -11kJ/mol ... Is that the answer they have given?

From what I could find the correct ?H for the reaction should be about -3.9 kJ/mol ... so both answers seem to be way off ... but that's likely b/c a lot of heat could be lost to the surroundings during the lab messing up the results