When 20.1 mL of 0.500 M H2SO4 is added to 20.1 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume...

Question

that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.) Do not answer in scientific notation.

I got -112 kJ /mol H₂O which was wrong.

Answer 1

The reaction is H2SO4 + 2KOH -----> 2 H2O + K2SO4.

First calculate the number of moles of each reactant there are.

H2SO4: 0.5 moles / 1000 mL x 20.1 mL = 0.01005 moles KOH: 1.0 moles / 1000mL x 20.1 mL = 0.0201 moles.

As the coefficients in front of KOH and H2O are the same, 0.0201 moles of H2O are also produced.

The heat evolved for the reaction is = mass solution x heat capacity water x Δtemp.

ΔH = 40.2 g x 4.18 j/g x 6.67°C = 1120.8 j to produce 0.0201 moles G=H2O.

To find the ΔH per mole of H2O, you must divide 1120.8 j by 0.0201 mole H2O = 1120.8 / 0.0201 = 55761 j or 56 kj / mole H2O. This is 1/2 your answer.