Enthalpy and Calorimetry

Question

The molar enthalpy of a solution of ammonium chloride is +14.8 KJ/mol. What would the final temperature of a soliton in which 40g of ammonium chloride is added to 200.00mL of water, initially the temperature is 25 degrees C

Answer 1

Since the ∆Hsoln is positive, this means the solution process is endothermic and the temperature of the solution will increase.

We will find the moles of ammonium chloride going into solution and then use the molar enthalpy of solution to find the amount of heat that is generated. Once we find the amount of heat, we can find the final temperature of the solution.

moles of NH4Cl = 40 g x 1 mol / 53.5 g = 0.748 moles

∆Hsoln = 0.748 mols x 14.8 kJ / mol = 11.1 kJ of heat

Next we use q = mC∆T to find the change in temperature.

q = heat = 11.1 kJ

m = mass = 200.0 g assuming a density of water of 1.0 g/ml = 0.200 kg

C = specific heat of water = 4.184 J/gº = 4.184 kJ/kgº

∆T = ?

Solving for ∆T ...

∆T = q / (m)(C) = 11.1 kJ / (0.200 Kg)(4.184 kJ/kgº)

∆T = 13.3º

Final temperature of solution = 25º + 13.3 = 38.3ºC