Coffee cup calorimeter with a heat capacity of 3.60j

Question

A coffee cup calorimeter with a heat capacity of 3.60 J/∘C was used to measure the change in enthalpy of a precipitation reaction. A 50.0 mL solution of 0.330 M AgNO3 was mixed with 50.0 mL of 0.250 M KCl. After mixing, the temperature was observed to increase by 1.95 ∘C. Calculate the enthalpy of reaction, Δ𝐻rxn, per mole of precipitate formed (AgCl). Assume the specific heat of the product solution is 4.18 J / (g⋅∘C) and that the density of both the reactant solutions is 1.00 g/mL.

Answer 1

The heat of the reaction (∆H_rxn) is the heat generated and which is then absorbed by both the water present (mC∆T) and the calorimeter itself (C_cal x ∆T). That is, we must take into account the heat absorbed by the calorimeter. Before we find the ∆H_rxn per mole of AgCl, we will first find the heat generated under the conditions described.

heat = q = mC∆T + Cal∆T

m = mass of the contents = 50.0 ml + 50.0 ml = 100 ml x 1.00 g/ml = 100.0 g

C = 4.18 J/gº

∆T = change in temperature = 1.95º

C_cal = 3.60 J/º

q = (100.0 g)(4.18 J/gº)(1.95) + (3.60 J/gº)(1.95º) = 815.1 + 7.02

q = 822.1 J = ∆H

This is also ∆H because q = ∆H when pressure is constant.

Now to find ∆H per mol AgCl formed. We will write the correctly balanced equation for the reaction taking place:

AgNO₃(aq) + KCl(aq) ==> KNO₃(aq) + AgCl(s) Note that 1 mol AgCl is formed from 1 mole AgNO₃ and 1 mole KCl. So let us now find moles of each reactant present:

moles AgNO₃ = 50.0 ml x 1 L/1000 ml x 0.330 mol/L = 0.0165 mol AgNO₃

moles KCl = 50.0 ml x 1 L/1000 ml x 0.250 mol/L = 0.0125 mol KCl

We see that there are fewer moles KCl than AgNO₃ making KCl LIMITING. And since we get 1 mol AgCl per mol of each reactant, the most AgCl we can make is 0.0125 moles AgCl.

∆H_rxn/mol AgCl = 822.1 J/0.0125 moles = 65,768 J/mole = 65.8 kJ/mole