Two solutions, initially at 24.60°C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°

c. When a100.0 mL volume of 0.100 M AgNO3 solution is mixed with a 100.0 mL sample of 0.200 M NaClsolution, the temperature in the calorimeter rises to 25.30°

c. Determine the? H°rxn for thereaction as written below. Assume that the density and heat capacity of the solutions is the sameas that of water. NaCl (aq) + AgNO3 (aq) ? AgCl (s) + NaNO3 (aq) ? H°rxn = ?

The solution is as follows:

Since the mass of the calorimeter is given, let's take the heat effects of the calorimeter as negligible.

Compute for sensible heat from 24.60°C (297.6 K) to 298 K. The heat capacity of water is 4.18 J/g·K.

ΔH₁ = ∫[ (0.1 mol/L AgNO₃) (0.1 L) (169.87 g/mol AgNO₃) (4.18 J/g·K) dT + (0.2 mol/L NaCl) (0.1 L) (58.44 g/mol NaCl) (4.18 J/g·K) dT]

Take the integral from limits 297.6 K to 298.8 K (25.30°C).

ΔH₁ = 4,289.12 J

Compute for the heat of reaction at room temperature:

ΔH₂ = ∑ (Heat of formation of products*stoichiometric coefficient) - ∑ (Heat of formation of reactants*stoichiometric coefficient)

ΔH₂ = [ (-127 kJ/mol AgCl) (1 mol) + (-467 kJ/mol NaNO₃) (1 mol) ] - [ (-123.02 kJ/mol AgNO₃) (1 mol) + (-407.27 kJ/mol NaCl) (1 mol) ]

ΔH₂ = - 63.71 kJ or - 63,710 J

Thus,

ΔHrxn = ΔH₁ + ΔH₂ = 4,289.12 J + - 63,710 J

ΔHrxn = 59,420.88 J or 59.42 kJ