Question

Chemistry: Calorimetry and Heats of Reaction

Mass of water in styrofoam cup: 100g

Mass of KCl: 1.650g

Initial Temperature: 20.6degC

Final Temperature: 19.6degC

q=(75g)(1.0 cal/g x degC)(-1degC)= -75cal

qreaction= -qcontents

-75 cal = 75 cal --> 0.075kCal

1.650g KCl = 0.2214 mol KCl

0.075kCal/0.02214mol = 3.39kCal mol KCl

Is this the standard heat of solution?

If this is the standard heat of solution, how would I determine the percent error of my result?

I was given this equation:

Percent error = |lit value - exptl value/lit value| x 100

with this information:

KCl(s) --> K+(g) + Cl-(g) dH=167.1kCal/mol

K+(g) + Cl-(g --> dH=163.0 kCal/mol

KCl(s) --> K+(aq) + Cl-(aq) dH=??

I know this question was super long. I'm sorry I'm confused and I'm not really sure where to begin.

Answer 1

This is simple concept based question

First we will calcualte the enthalpy of solution from the given data [experimental value]

the specific heat of water = 1cal / g 0C

mass of water taken = 100 grams

temperature change = 1 ⁰C

Heat absrobed by water = Mass of water X specific heat of water x change in temperature = 100 X 1 X 1 = 100 cal

heat absrobed by water in Kcal = 0.1 KCal

The amount of KCl dissolved = 1.650 grams

Molecular weight of KCl = 74.55 g / mole

So when 1.650 grams of KCl is dissolved the heat liberated = 0.1 Kcal

so when 1 gram of KCl will be dissolved the heat liberated = 0.1 / 1.650 Kcal

so when 74.55 gram (1 mole) of KCl will be dissolved the heat liberated = 0.1 X 74.55 / 1.650 Kcal = 4.518 Kcal

Other information for literature values are

KCl(s) + H2O --> K+(g) + Cl-(g) dH=167.1kCal/mol ......(1) [Lattice energy]

K+(g) + Cl-(g --> K+(aq) + Cl-(aq) dH= -163.0 kCal/mol ....(2) [heat of hydration]

KCl(s) + H2O --> K+(aq) + Cl-(aq) dH=?? ...(3) [enthalpy of dissolution]

Delta H(3) = Delta H (1) + Delta H(2) = 167.1 - 163 = 4.1 Kcal / mole

The percentage of error will be

|lit value - exptl value/lit value| x 100 = |4.1- 4.518/4.1| x 100 = 10.195 %