Question

When 1.045 g of K2O is added to 50.0 mL of water at 25.0 ∘C in a calorimeter, the temperature of the water increases to 41.5 ∘C.

Assuming that the specific heat of the solution is 4.18 J/(g⋅∘C)J/(g⋅∘C) and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔHΔH in kilojoules/mol for the reaction

K2O(s)+H2O(l)→2KOH(aq)K2O(s)+H2O(l)→2KOH(aq)

Express the change in enthalpy in kilojoules per mole.

Answer 1

Density of water at 25 oC = 0.997 g/mL

Mass of K2O = 94.2 g/mol

We have Qrxn = m x Cp x dT

Mass of 50 mL of water = 0.997 x 50 g = 49.85 g

Here, m = 49.85 + 1.045 = 50.895 g, Cp = 4.18 J/g.oC, dT = 41.5 – 25 = 16.5 oC

Qrxn = 50.895 x 4.18 x 16.5 = 3.51 kJ

dH = -Qrxn = -3.51 kJ

Number of moles of K2O in 1.045 g = 1.045/94.2 = 0.0111 mol

dH = -3.51 kJ/0.0111 mol = -316.22 kJ/mol