Question

When 3.02 g NH₄Cl solid was dissolved in 20.05 mL of water, the resulting 23.07 g solution temperature decreased from 19.80 ^oC to 11.25 ^oC. Calculate the enthalpy change ?H in kJ, when 1 mol of NH₄Cl dissolves in water. The specific heat of the solution is 4.18 J/g ^oC.

Answer 1

Given that;

The specific heat of the solution is 4.18 J/g ^oC

Density of water =1.0 g/ ml

Therefore 20.05 ml water= 20.05 g

Temperature change = 11.25-19.80

= - 8.55

Step1 Heat energy = 20.05 x4.18x[T2-T1]

= 20.05x4.18x- 8.55 J
= -716.57 J
Step2 Molar mass of NH4Cl = 53.5 g/ mol

Number of moles = amount in g/ molar mass

= 3.02 g/ 53.5 g/ mole

= 0.0564 moles

Step3 Enthalpy change for 1 mole = total heat /* number o fmoles

= -716.57 J/ 0.0564 moles

= -12705.14 J/ mole

= - 12.7 KJ/ mole