   Chapter 5, Problem 96IL

Chapter
Section
Textbook Problem

Insoluble PbBr2(s) precipitates when solutions of Pb(NO3)2(aq) and NaBr(aq) are mixed.Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)ΔrH° = ?To measure the enthalpy change, 200. mL. of 0.75 M Pb(NO3)2(aq) and 200. mL of 1.5 M NaBr(aq) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by 2.44 °C. Calculate the enthalpy change for the precipitation of PbBr2(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 PbBr2 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 Pb(NO3)2=1g/mL× 200mL =200g

Mass of NaBr=1g/mL × 200mL =200g

The amount of AgCl formed.

200mL Of 0.75M Pb(NO3)2 will contain (0.200L×0.75mol/L )0.15mol of Pb(NO3)2

200mL Of 1.5M NaBr will contain 0.3mol of NaBr,

So we assume to get 0

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