Calculate the enthalpy change per mole of H2SO4 in the reaction

Question

A 105.2 mL sample of 1.00 M NaOH is mixed with 52.6 mL of 1.00 M H₂SO₄ in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 22.45 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 31.30 °C. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g·°C), and that no heat is lost to the surroundings.

Answer 1

heat = mass x specific heat x change in temperature

q = mC∆T

q = heat = ?

m = mass = 105.2 ml + 52.6 ml = 157.8 ml x 1.00 g/ml = 157.8 g

C = specific heat = 4.18 J/gº

∆T = change in temperature = 31.30º - 22.45º = 8.85º

q = (157.8 g)(4.18 J/g)(8.85º) = 5837 J ... this is the heat generated by reaction as given. To find heat per 1 mole of H2SO4, we first need to find out how many moles of H2SO4 gave us the 5837 J.

mole H2SO4 = 52.6 ml x 1 L/1000 ml x 1.00 mol/L = 0.0526 moles H2SO4 present

moles NaOH = 105.2 ml x 1 L/1000 ml x 1.00 mol/L = 0.1052 moles NaOH

Since the mol ratio NaOH : H2SO4 = 2:1, all the H2SO4 and NaOH reacted.

∆H per mol H2SO4 = 5837 J/0.0526 mole H2SO4 = 110,970 J/mol = 111 kJ/mol H2SO4