Question

In a constant-pressure calorimeter, 65.0 mL of 0.830 M H2SO4 was added to 65.0 mL of 0.270 M NaOH. The reaction caused the temperature of the solution to rise from 21.71 °C to 23.55 °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

H2SO4 + 2NaOH ----> Na2SO4 + 2H2O

Moles of H2SO4 = 0.83 x 0.065 = 0.05395 moles

Moles of NaOH = 0.27 x 0.065 = 0.01755 moles

According to the stoichiometry of the reaction 1 mole of H2SO4 reacts with 2 moles of NaOH

Moles of NaOH = 0.01755

Therefore, moles of H2SO4 that will react = 0.01755 / 2 = 0.008775 moles

2 moles of NaOH produces 2 moles of H2O

=> Moles of H2O produced = 0.01755

Total heat released for 0.01755 moles of H2O produced = m Cp (T2 - T1)

where,

m = mass of solution

Volume = 65 + 65 = 130 mL

=> m = 130 x 1 = 130 g

Cp = 4.184

T2 =23.55

T1 = 21.71

=> Heat = 130 x 4.184 x (23.55 - 21.71) = 1000.8 J

=> For 0.01755 moles of H2O produced, heat released = 1000.8 J

=> For 1 mole of H2O produced, heat released = 1000.8 / 0.01755 = 57026.4 J

=> delta H = - 57.03 kJ / mol (-ve since heat is released)