Question

A 102.2 mL sample of 1.00 M NaOH is mixed with 51.1 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 23.05 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 32.10 °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.

Calculate the enthalpy change per mole of H2SO4 in the reaction.

Answer 1

The balance reaction of NaOH and H2SO4 is as follows:

2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) + 2H2O(l)

Given that;

A 102.2 mL sample of 1.00 M NaOH is mixed with 51.1 mL of 1.00 M H2SO4

Number of mole = molarity * volume

Moles NaOH: 0.1022 L x 1.00M = 0.1022 moles base
Moles H2SO4: 0.0511 L x 1.00M = 0.0511 moles acid
Mole ratio base to acid 2:1,
0.051 1 moles acid needs x2 base, there is enough
for both to completely react into products, so none left

If no heat lost, q = mass x specific heat x delta T

Total volume of solution = 102.2 +51.1 =153.3 ml

= 0.1533 L

Mass of solution = density * volume

= 1.00 g/ ml *153.3 ml

= 153.3 g

and the amount of heat release from this recation is calculated as follows:

q = mcdT

m of solution, c = specific heat and dt temperature change

q = 153.3g x 4.18 J/g C x (32.10-23.05)
q = 5799.1857 J

= 5.799KJ

here; Moles H2SO4: 0.0511 moles
Enthalpy change per mole:

total heat / number of mole of H2SO4 = [5.799KJ ]/ 0.0511 mol

= 113.487 kJ/mol