When a 100 mL sample of 0.2345 M aqueous hydrochloric acid is added to 100.0 mL of 0.2345 M aqueous ammonia in a calorimeter whose heat capacity(minus water) is 0.480 kJ/°C, (continued in description)

Question

the temperature increase by (117 x 0.02345)°C. Assume that the volumes are additive, and that the

density of the solution is 1.000 g/mL.

The following reaction occurs when the two solutions are mixed. Calculate the molar heat of

reaction.

HCl(aq) + NH3(aq) NH4Cl(aq)

Answer 1

Not sure what the temperature increase is, but will assume it to be 117 x 0.02345º = 2.744º (4 sig. figs. for a temperature reading is a little suspect).

Reaction: HCl(aq) + NH₃(aq) ==> NH₄Cl(aq)

mols HCl present = 100 ml x 1 L / 1000 ml x 0.2345 mol / L = 0.02345 mols HCl

mols NH₃ present = 100 ml x 1 L / 1000 ml x 0.2345 mol / L = 0.02345 mols NH₃

As they are present in equal quantities there is no limiting reactant.

Total volume of reaction = 100 ml + 100 ml = 200 ml

Mass of reaction = 200 ml x 1 g/ ml = 200 g

Molar heat of reaction = heat absorbed by the solution + heat absorbed by the calorimeter / moles of reactant

heat absorbed by solution = q = mC∆T = (200 g)(4.184 J/gº)(2.744º) = 2296 J

heat absorbed by the calorimeter = Ccal x ∆T = 0.480 kJ/º x 2.744º = 1.317 kJ = 1317 J

Sum of heats = 2296 J + 1317 J = 3613 J

Divide this by 0.02345 mols of reactant to get J/mol and convert to kJ/mol to get the answer.

3613 J / 0.02345 mol x 1 kJ / 1000 J = 154.1 kJ/mol = molar heat of reaction