calorimetry ...

Question

4.41 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the dissolution of MgSO₄. (The specific heat of water is 4.18 J/g･°C and the density of the water is 1.00 g/mL). You can assume that the specific heat of the solution is the same as that of water.

Answer 1

First, notice that the temperature of the water increased, meaning this is an EXOTHERMIC reaction. Thus, the final answer will have a negative sign.

∆H = mC∆T

∆H = enthalpy (heat) of reaction = ?

m = mass =100.0 ml x 1.0 g/ml = 100.0 g (this is mass of water, not mass of MgSO4)

C = specific heat = 4.18 J/g/deg

∆T = 6.70 degrees

∆H = (100.0 g)(4.184 J/g/deg)(6.70 deg)

∆H = -2801 Joules

This is the enthalpy for 4.41 g MgSO₄ but the question asks for the kJ/mole, so we need to find moles.

4.41 g MgSO₄ x 1 mole MgSO₄/120.4 g = 0.0366 moles

∆H = 2801 kJ/0.0366 moles = 76,530 kJ/mole = -76.5 kJ/mole