A 92.2 mL sample of 1.00 M NaOH is mixed with 46.1 mL of 1.00 M H2SO4 in a large

Question

A 92.2 mL sample of 1.00 M NaOH is mixed with 46.1 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 22.25 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 32.30 °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.

Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. Remember to include phases in the balanced chemical equation.

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

Explanation / Answer

Balanced chemical equation for the reaction

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

Moles of NaOH = Volume(in L) * moalrity = 0.0922 L * 1.0 M = 0.0922 mole

Moles of H2SO4 = Volume(in L) * moalrity = 0.0461 * 1.0 M = 0.0461 mole

From the balanced equation, base to acid ration is 2:1. Hence no reactant is left over

Mass of the solution = total volume * density of the mixed solution

                                    = (92.2 + 46.1) mL * 1.00 g/mL

Mass of the solution = 138.3 g

q= mass * specific heat * deltaT

q= 138.3g * 4.18 J/(g·°C) * (32.30-22.25)

q= 5809.8 J

moles of H2SO4 = 0.0461

Enthalpy change per mole of H2SO4 = [5809.8 J]/0.045mol

                                                                = 126.03 kJ/mol

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