Question

In a constant-pressure calorimeter, 75.0 mL of 0.830 M H2SO4 was added to 75.0 mL of 0.400 M NaOH. The reaction caused the temperature of the solution to rise from 21.85 °C to 24.57 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·K, respectively), what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Volume of H₂SO₄ = 75 mL.

Concentration of H₂SO₄ = 0.830 M

Volume of NaOH = 75 mL.

Concentration of NaOH = 0.400 M

Initial temperature of solution = 21.85˚C

Final temperature of solution = 24.57˚C

Change in temperature = (24.57 – 21.85)˚C = 2.72˚C = 2.72 K

Density of the solution = 1 g/mL.

Total volume of the solution = (75 + 75) mL = 150 mL.

Mass of the solution = volume*density = (150 mL)*(1 g/1 mL) = 150 g

Specific heat capacity of solution = 4.184 J/g.K

Heat of the reaction = (mass of solution)*(specific heat of solution)*(change in temperature) = (150 g)*(4.184 J/g.K)*(2.72 K) = 1707.072 J = 1.707 kJ.

This is half the solution; we need to calculate the heat of neutralization per mole of water produced. Now write down the molar equation of reaction:

H₂SO₄ + 2 KOH -------> K₂SO₄ + 2 H₂O

The neutralization reaction is

H⁺ (aq) + OH^- (aq) -------> H₂O

From the molar equation, we see that there is a 1:1 molar ratio between KOH used and H₂O produced (actually 2:2 which reduces to 1:1). Therefore, moles KOH = (75 mL)*(1 L/1000 mL)*(0.400 mole/L) = 0.03 mole.

Therefore, moles H₂O produced = 0.03 mole.

The heat of neutralization will be negative since the reaction is exothermic (the reaction liberates heat and thus the temperature increases); therefore, the heat of neutralization per mole of water produced = (-1.707 kJ)/(0.03 mole) = -56.9 kJ/mol (ans).