   Chapter 5, Problem 95IL

Chapter
Section
Textbook Problem

Insoluble AgCl(s) precipitates when solutions of AgNO3(aq) and NaCl(aq) are mixed.AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) ΔrH° = ?To measure the energy evolved in this reaction, 250. mL of 0.16 M AgNO3(aq) and 125 mL of 0.32 M NaCl(aq) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises from 21.15 °C to 22.90 °C. Calculate the enthalpy change for the precipitation of AgCl(s), in kJ/mol. (Assume the density of the solution is 1.0 g/mL and its specific heat capacity is 4.2 J/g · K.)

Interpretation Introduction

Interpretation:

The enthalpy change for the precipitation of AgCl has to be calculated.

Concept Introduction:

Standard enthalpy of the reaction,ΔrHo, is the change in enthalpy that happens when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states.

Enthalpy of the reaction,ΔrH, is the change in enthalpy that happens when matter is transformed by a given chemical reaction

Heat energy required to raise the temperature of 1g of substance by 1K.Energy gained or lost can be calculated using the below equation.

q=C×m×ΔT

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity, ΔT= change in temperature.

Explanation

Given,

Density of the solution=1g/mL

Mass of AgNO3=1g/mL× 250mL =250g

Mass of NaCl=1g/mL × 125mL =125g

The amount of AgCl formed.

250mL of 0.16M AgNO3 will contain(0.250L×0.16mol/L )=0.04mol of AgNO3

125mL of 0.32M NaCl will contain 0.04mol of NaCl,

So we assume to get 0.04mol of AgCl

Assume qr+qsol=0

qsol Can be calculated from q=C×m×ΔT

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