Heat of reaction

Question

KOH(s) → K+(aq) + OH– (aq) ΔHdissolution = –56 kJ
The temperature of 150 mL of water rises by 26.76 °C
after adding some amount of KOH. What is the molar
concentration of the solution?
(cwater= 4.186 J/g °C, dwater = 1 g/mL)
(neglect change in the volume by the addition of KOH)

Answer 1

q = mc∆T

q = heat of reaction = ?

m = mass of water = 150 ml x 1 g/ml = 150 g

c = specific heat of water = 4.186 J/g/deg

∆T = change in temperature = 26.76ºC

q = (150 g)(4.186 J/g/deg)(26.76.deg) = 16,803 J = 16.8 kJ

Assuming you meant the ∆H_dissolution = -56 kJ/mole, then you can find the moles as follows:

56 kJ/mole (x moles) = 16.8 kJ

x = 0.30 moles

molar concentration = 0.30 moles/150 ml x 1000 ml/L = 2 mole/L = 2 M