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.
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!