AP Chemistry Thermodynamics

Question

The solid in question 2 was aluminum sulfate, AL2(SO4)3. Calculate the molar heat of solution, delta H solution (in kJ/mol) for aluminum sulfate. Hint: The units for molar heat of solution are kilojoules per mole (kJ/mol). First determine the heat released per gram of solid.

Answer 1

Give us the information provided in question 2. To find heat released, we need such info as mass of material, change in temperature, etc.

Edited Response with additional information provided:

q = mC∆T

q = heat = ?

m = mass = 25.0 g + 60.0 g = 85.0 g (assuming the density of water is 1.0 g/ml)

C = specific heat = 4.184 J/gº (assuming it is the same as the specific heat of water)

∆T = change in temperature = 25.3 - 21.4 = 3.9º

q = (85.0 g)(4.184 J/gº)(3.9º) = 1387 J

This is the heat generated from 25.0 g of Al₂(SO₄)₃. To find the ∆H in kJ/mol, we need to find the moles.

Molar mass Al₂(SO₄)₃ = 342 g / mol

25.0 g x 1 mol / 342 g = 0.0731 mols

∆H = 1387 J / 0.0731 mols = 18,974 J / mol= 18.974 kJ / mol = 19.0 kJ / mol (3 sig. figs.)