Heat of reaction

Question

When 9.85 grams of potassium hydroxide(KOH) are dissolved in 150.0 grams of water at 25.0 °C in an insulated container, the temperature of the water increases to 40.1 °C. Assuming that the specific heat of the solution is 4.184 J/(g °C) and that no heat is gained or lost by the container, what is the (🔼H) of solution of KOH in kJ/mol?

Answer 1

q = mC∆T

q = heat = ?

m = mass = 150 ml x 1 g / ml = 150 g (assuming a density of 1 g /ml)

C = specific heat = 4.184 J/gº

∆T = change in temperature = 40.1º

q = (150 g)(4.184 J/gº)(40.1º)

q = 25,167 J

This is the heat generated from 9.85 g of KOH. To find ∆H in kJ/mole, we need to first find moles KOH...

moles KOH = 9.85 g KOH x 1 mol KOH / 56.1 g = 0.1756 mols

∆H = 25,167 J / 0.1756 moles = -142,320 J/mol = -142 kJ/mol (negative sign since heat is generated)