When a 13.00-g sample of solid potassium hydroxide dissolves in 200.0 g of water in a coffee-cup calorimeter (the following Figure), the temperature rises from 21.6 o C to 40.6 o C.

Question

Calculate ΔH(in kJ/mol KOH) for the solution process

KOH(s)→K⁺(aq) + ^-OH (aq)

Answer 1

No figure included but that’s ok. Assuming constant pressure, ΔH is q and q = mCΔT

q = heat

m = mass = 200 g (in intro courses they will use 200 g for water only but in advanced courses they use 213g)

C = specific heat of water = 4.184 J/g/deg

ΔT = change in temp = 19 deg

q = (200g)(4.184)(19) = 15899 J = 15.9 kJ

13.00 g KOH x 1 mol/56.11g = 0.2317 moles KOH

ΔH = 15.9 kJ/0.2317 moles = -68.6 kJ/mole (3 sig figs)

(Note the negative sign because the temperature increased making this exothermic)