Calculate the enthalpy change (ΔH) for: HCl(aq) + NH3(aq) → NH4Cl(aq)
The reaction of 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NH3 causes a temperature rise of 6.4 ˚C in the resulting 100.0 mL of solution. The NH4Cl solution has a density of 1.005 g/mL and a specific heat of 3.97 Jg-1˚C-1.
(My book says ΔH should be reported with no digits beyond the first uncertain digit, which I don't understand?)
My Work:
1) q=(100.5g)*(3.97Jg^-1C^-1)*(6.4C)=-2553.504J
2) ΔH=-2553.504J/.050moles *1 mole NH3=-51070.08J/1molrxn*1 kj/1000J=-510kj/molrxn
Did I do it right? What should I use for digits for ΔH based on what my book says? Thanks again so much!
1 answer
Yes, you did it right. The answer should be reported as -510 kJ/mol.