Question

The salt ammonium bromide is soluble in water. When 6.17 g of NH4Br is dissolved in 104.00 g of water, the temperature of the solution decreases from 25.00 to 22.68 °C. Based on this observation, calculate the enthalpy of dissolution of NH4Br (in kJ/mol).

Assume that the specific heat of the solution is 4.184 J/g °C and that the heat absorbed by the calorimeter is negligible.

Answer 1

Q=c(m(NH4Br)+m(H2O))(t2-t1)= 4.184 J/g°C(6.17g+104.00g)(22.68°C-25.00°C)=-1069.407J

M(NH4Br)=80+14+4=98g/mol

For 1 mole of NH4Br:

H=98·1069.407/6.17=16986J/mol=17kJ/mol