Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of 0.100 M AgNO3 solution is mixed with a 100.0 mL sample of 0.200 M NaCl solution, the temperature in the calorimeter rises to 25.30 °C. Determine the ∆H°rxn in kJ/mol AgCl for the reaction as written below. The density of the final solution is 1.00 g/mL and heat capacity of the final solution is 4.18 J/goC.

To solve this question we need to calculate the heat absorbed by the cup calorimeter and by the water in the solutions. We will also need to calculate the amount in moles produced by the reaction since we want to know the ∆H°rxn in kJ/mol AgCl .

mol AgNO₃ = 100 mL x 1L/1000 mL x 0.100 mol/L = 0.01 mol

mol NaCl = 100 mL x 1L/1000 mL x 0.200 mol/L = 0.02 mol

Therefore our limiting reagent is the 0.01 mol AgNO₃ and 0.01 mol AgCl will be produced according to the stoichiometry of the reaction:

AgNO₃ + NaCl ⇒ AgCl + NaNO₃

Heat absorbed by the water:

qw = m(H₂O) x c x ΔT where m (H₂O) = 200 g ( the density of final solution is 1 g/ml)

c = specific heat of water = 4.18 J/gºC

ΔT = change in temperature = (25.30 - 24.60 ) ºC = 0.7ºC

qw = 200 g x 4.18 J/gºC x 0.7 ºC = 585.20 J

Heat Absorbed by the calorimeter :

q cal = C cal x ΔT = 15.5 J/ºC x 0.7ºC = 10.85 J

Total Heat released by the combustion = qw + qcal = 585.20 J +10.85 J

= 596.05 J

We have to change the sign to this quantity since it is an exotermic reaction ( ΔT is positive ) and have the ∆Hrxn

∆H rxn = -596.05 J

but this is not what we are being asked since this heat was released by the formation of 0.0100 mol of AgCl so finally

∆H°rxn = -596.05 J /0.01 mol = -59,605 J/mol x 1 kJ/1000J = -59.61 kJ/mol