Thermochemistry Problem

Question

When 125 mL of 0.220 M NaCl(aq) and 125 mL of 0.220 M AgNO₃(aq), both at 21.6°C, are mixed in a coffee cup calorimeter, the temperature of the mixture increases to 24.3°C as solid AgCl forms.

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

This precipitation reaction produces 2.81 ✕ 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.

Due to the online nature of my chemistry assignment, there is no textbook and the internet has not been very forthcoming. Any help is appreciated.

Answer 1

The first step here is to determine what q(rxn) or the heat absorbed/released by the reaction is in order to determine ΔH (KJ/mol) is for AgNO3.

The question here already gives you the heat produced by the reaction otherwise you would have to calculate it by first determining the heat absorbed by the solution itself with equation:

q(soln) = m × C (s,soln) × ΔT

m = (0.997 g/ml)(250 mL) = 249.25 g [Multiply density of solution by total volume]

C (s,soln) = 4.18 J/g * °C the specific heat capacity of the solution

ΔT = 24.3 - 21.6 = 2.7 °C

q (soln) = 2810 J

Then since the solution absorbed heat that means that the reaction itself released heat so...

q (rxn) = - q(soln)

The heat released by the reaction is simply the same heat absorbed by the solution. The negative sign indicates the release of heat.

q (rxn) = -2810 J (convert to KJ)

q (rxn) = -2810 J × (1 kJ/1000 J) = -2.81 KJ

Use the enthalpy equation to acquire the final answer...

ΔH (rxn) = q (rxn) / mol AgNO3

Now the mols of AgNO3 in the intital reaction must be determined, this can easily be done using the formula for molarity:

M = mol (solute)/L (solution) rearrange to mol (solute) = (M)(L (solution))

mol AgNO3 = (0.220 M)(0.125 L) = 0.0275 mol AgNO3

So...

ΔH (rxn) = (-2.81 KJ) / (0.0275 mol AgNO3)

ΔH (rxn) = -102 KJ / mol AgNO3 [Answer]