Delta H of soln

Question

When a 3.00g sample of KBr is dissolved in water a calorimeter that has a total heat capacity of 1.32 kJ • K^-1, the temperature decreases by 0.380 K. Calculate the molar heat of solution of KBr.

Answer 1

q = C_cal∆T

q = heat = ?

C_cal = calorimeter constant = 1.32 kJ/º

∆T = change in temperature = 0.380º

Solving for q, we have ...

q = (1.32 kJ/º)(0.380º)

q = 0.502 kJ

This is the heat absorbed by dissolving 3.00 g of KBr. Since heat is absorbed (temperature decreases), this is an endothermic process, and q has a + value. Since the question asks for the MOLAR heat of solution, we must convert 3.00 g KBr to moles, and then calculate molar heat of solution, as follows:

molar mass KBr = 119.0 g / mole

moles of KBr = 3.00 g x 1 mol / 119 g = 0.252 mols

molar heat of solution = 0.502 kJ / 0.0252 mols = +19.9 kJ / mole