Question

In a constant-pressure calorimeter, 55.0 mL of 0.840 M H2SO4 was added to 55.0 mL of 0.260 M NaOH. The reaction caused the temperature of the solution to rise from 21.91 °C to 23.68 °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 solution = 55 + 55 = 110 mL

mass of solution = 110 x 1 = 110 g

Q = m Cp dT

    = 110 x 4.184 x (23.68 - 21.91)

Q = 814.6 J

moles of H2SO4 = 55 x 0.840 / 1000 = 0.0462

moles of NaOH = 55 x 0.260 / 1000 = 0.0143

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

    1                2

0.0462        0.0143

here limiting reagent is NaOH.

ΔH = - Q / n

       = - 814.6 x 10^-3 / 0.0143

ΔH = - 56.97 kJ/mol