CHEMISTRY QUESTION

Question

A calorimeter that contains 3.6 L of water is used to measure the enthalpy of solution for 45 g of KOH. Calculate the molar enthalpy of solution that is associated with the reaction if the temperature of the water changes from 25.5 °C to 20.7 °C. [The molar mass of KOH is 56.10 g/mol]

Answer 1

First, we have to assume that the 3.6 L of water in the calorimeter has a density of 1.0 g/ml (see below). Given that assumption, we perform the following calculations.

q = mC∆T

q = heat = ?

m = mass of water = 3.6 L x 1000 ml / L = 3600 ml x 1.0 g /ml = 3600 g (assuming a density of 1 g/ml)

C = specific heat of water = 4.184 J/gº (a constant that you should know or can look up)

∆T = change in temperature = 25.5 - 20.7 = 4.8º

Solving for heat (q) we have...

q = (3600 g)4.184 J/gº)(4.8º)

q = 72,300 J (+ value because the reaction in endothermic)

Now, the question asks for the MOLAR enthalpy. What we calculated above is the enthalpy change for dissolving 45 g KOH, not 1 mole KOH. So, we do a simple conversion as follows:

45 g KOH x 1 mol KOH / 56.10 g = 0.802 mols KOH

∆H = 72,300 J / 0.802 mols = 90,150 J / mol = 90,200 J / mol

∆H = 90.2 kJ / mol