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

Question

A 103.2 mL sample of 1.00 M NaOH is mixed with 51.6 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 22.25 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 32.50 °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.

A. Create a balanced equation

B. Is there any NaOH or H₂SO₄ left in the cup?

C. What is the enthalpy changer per mole of H₂SO₄ in the reaction

Answer 1

There is a lot here, so we'll just go step by step.

A. Create a balanced equation:

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

B. Is there any NaOH or H₂SO₄ left?:

Since we are given the amounts of both reactants, we must figure out which, if either, is limiting. One way

to do this is to divide moles of each by the corresponding coefficient in the balanced equation, and

whichever value is less represents the limiting reactant:

For NaOH: 103.2 ml x 1 L / 1000 ml x 1.0 mol / L = 0.1032 moles NaOH (÷2->0.0516)

For H₂SO₄: 51.6 ml x 1 L / 1000 ml x 1.0 mol / L = 0.0516 mols H₂SO₄ (÷1->0.0516)

Since they are the same, neither is limiting and they are present in stoichiometric amounts.

No NaOH nor H₂SO₄ is left

C. What is the enthalpy change per mol of H₂SO₄?:

q = mC∆T

q = heat = ?

m = mass of solution = 103.2 ml + 51.6 ml = 154.8 ml x 1 g / ml = 154.8 g*

C = specific heat of solution = 4.18 J/gº

∆T = change in temperature = 32.50º - 22.25º = 10.25º

*154.8 g does not include the mass of the NaOH and H₂SO₄. This would be a more advanced calc.

q = (154.8 g)(4.18 J/gº)(10.25º)

q = 6632 J (this is for the production of 0.0516 mols H₂SO₄)

∆H/mol H₂SO₄ = 6632 J / 0.0516 mols = 128535 J / mole = 129 kJ/mol (3 sig.figs.)