100.0 mL of 1.25 M AgNO3

Question

100.0 mL of 1.25 M AgNO₃ is added to 100.0 mL of 1.75 M NaCl in a constant pressure calorimeter. The temperature goes from 22.4 ^oC to 31.9 ^oC.

 

Calculate Δ𝐻ΔH of the reaction per mole of AgCl produced. Assume the density of the solution is 1.00 g/mL and that the specific heat is 4.184 J/g^oC

Answer 1

AgNO3(aq) + NaCl(aq) ==> AgCl(s) + NaNO3(aq) .. balanced equation

moles AgNO3 = 100.0 ml x 1 L / 1000 ml x 1.25 mol / L = 0.125 mols

moles NaCl = 100.0 ml x 1 L / 1000 ml x 1.75 mol / L = 0.175 mols

Based on a 1:1 mol ratio of AgNO3 : NaCl, the AgNO3 is the limiting reactant (0.125 mols)

q = mC∆T

q = heat = ?

m = mass = 100 ml + 100 ml = 200 ml x 1 g / ml = 200 g (ignoring masses of solutes)

C = specific heat = 4.184 J/gº

∆T = change in temperature = 31.9º - 22.4º = 9.5º

Solving for q:

q = (200 g)(4.184 J/gº)(9.5º)

q = 7950 J

This is the heat generated from the production of 0.125 mols of AgCl.

∆H / mole AgCl = 7950 J / 0.125 mols AgCl = 63,600 J / mol = 63.6 kJ / mol AgCl (3 sig. figs.)