A calormeter has a mass of 200g and a specific heat capacity of 0.420 J g-¹. Into it are put 50.0 cm³ of 1.25 mol dm-³ hydrochloric acid and 50.0cm³ of 1.25 mol dm-³ potassium hydroxide solution at...

Question

A calormeter has a mass of 200g and a specific heat capacity of 0.420 J g-¹. Into it are put 50.0 cm³ of 1.25 mol dm-³ hydrochloric acid and 50.0cm³ of 1.25 mol dm-³ potassium hydroxide solution at the same temperature. The temperature of the calorimeter and contents rises by 10.0⁰C. Calculate the standard enthalpy of normalization.

Aassume the specific heat capacity of all solutions is 4.18 J g-¹ K-¹ and the density of all solutions is 1 g cm-³

Answer 1

Ccal = specific heat capacity of the calorimeter = 0.420 J/g x 200 g = 84 J/º

Csoln = 4.18 J/gº

m = mass of solution = 50 cm3 + 50 cm3 = 100 cm3 x 1 g/cm3 = 100 g (ignoring mass of solutes)

∆T = change in temperatuer = 10.0º

q = mC∆T + Ccal∆T

q = (100 g)(4.18 J/gº)(10.0º) + (84 J/º)(10.0º) = 4180 J + 840 J

q = 5020 J

This is the heat of neutralization for the conditions indicated. To find the standard ∆H of neutralization, we need to normalize this to the production of ONE MOLE OF H2O

HCl + KOH = H2O + KCl

mole HCl = 0.05 L x 1.25 mol/L = 0.0625 mole HCl

mole KOH = 0.05 L x 1.25 mol/L = 0.0625 mole KOH

Moles H2O formed = 0.0625

Standard enthalpy of neutralization = ∆Hº = 5020 J/0.0625 moles x 1 kJ/1000 J = 80.3 kJ/mole