Question

hen 170. mL of 0.209 M NaCl(aq) and 170. mL of 0.209 M AgNO₃(aq), both at 21.6°C, are mixed in a coffee cup calorimeter, the temperature of the mixture increases to 24.2°C as solid AgCl forms.

NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

This precipitation reaction produces 3.68 ✕ 10³ J of heat, assuming no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and density of the solutions are the same as those for water (4.18 J/g·°C, and 0.997 g/mL, respectively). Using this data, calculate ΔH in kJ/mol of AgNO₃(aq) for the given reaction.

Answer 1

NaCl(aq) + AgNO3(aq) --> AgCl(s) + NaNO3(aq)

no of mol of AgNO3 = Molarity *Volume

                           = 0.209*0.170 (1L=1000 mL)

                           = 0.03553 mol

heat released(q) = m*s*DT

m = mass of mixture = 340*0.997 = 338.98 g

s = specific heat = 4.18 j/g.c

DT = 24.2-21.6 = 2.6 c

   q = 338.98 *4.18*2.6

     = 3684.034 joule

     = 3.684 kJ

deltaH rxn= -q/n

       = - 3.684 /0.03553

       = -103.68 kj/mol