Question

In a constant-pressure calorimeter, 60.0 mL of 0.780 M H2SO4 was added to 60.0 mL of 0.490 M NaOH. The reaction caused the temperature of the solution to rise from 23.53 °C to 26.87 °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 + 2 NaOH → Na2SO4 + 2 H2O

(60.0 mL) x (0.780 M H2SO4) = 46.8 mmol H2SO4

(60.0 mL) x (0.490 M NaOH) = 29.4 mmol NaOH

29.4 mmoles of NaOH would react completely with 29.4 x (1/2) = 14.7 mmoles of H2SO4, but there is more H2SO4 present than that, so H2SO4 is in excess and NaOH is the limiting reactant.

(29.4 mmol NaOH) x (2 mol H2O / 2 mol NaOH) = 29.4 mmol H2O = 0.0294 mol H2O

(4.186 J/g·°C) x (60.0 g + 60.0 g) x (26.87 - 23.53)°C = 1677.75 J gained by the solution

(1677.75 J) / (0.0294 mol H2O) = 57066 J = 57.066 kJ/mol H2O