Molar Heat of KBr

Question

when a 4.00 gram sample of KBr is dissolved in water in a calorimeter that has a total heat capacity of 3.185kJ*K^-1 , the temperature decreases by 0.210 K. Calculate the molar heat of solution of KBr.

Answer 1

Harley,

 

This is a question about calorimetry.  We are given a temperature change and the total heat capacity of the calorimeter.  We are also given the amount ( in grams) of KBr being dissolved.  Our goal is to calculate the enthalpy change per  mole of KBr.

 

First, we will calculate the enthalpy change and then convert that heat into a molar quantity.

 

To solve, we will use the equation

 

 ΔH = -C_cal x ΔT  where C_cal is the heat capacity of the calorimeter and ΔT is the temperature change. Notice the negative sign in front of C_cal!  

 

So, using the values given

 

ΔH = -(3.185 kJ K^-1) x (-0.210 K)   Notice that ΔT is negative because we are told the temperature decreased.

 

ΔH = 0.669 kJ, but we need to convert to a molar quantity, so

 

(0.669 kJ/4.00 g KBr) x (119.0 g KBr/1 mol KBr)

 

Grams cancel and we get 0.669 kJ/ 0.0336 mol KBr = 19.9 kJ/mol

 

I hope this helps!