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Calculate enthalpy of reaction?
When 29.5 mL of 0.675 M H2SO4 is added to 29.5 mL of 1.35 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume the total volume is the sum of the volumes and the density and specific heat capacity of the solution are the same as for water.)
a) ΔH per mole of H2SO4 reacted in kJ/mol
b)ΔH per mole of KOH reacted in kJ/mol
1 Answer
- Roger the MoleLv 75 years ago
H2SO4 + 2 KOH → K2SO4 + 2 H2O
(29.5 mL) x (0.675 M H2SO4) = 19.9125 mmol H2SO4
(29.5 mL) x (1.35 M KOH) = 39.825 mmol KOH
The two reactants are present in their stoichiometric quantities, so there is no excess, and either can be considered the limiting reactant.
(4.184 J/g·°C) x (29.5 g + 29.5 g) x (30.17 - 23.50) °C = 1646.5 J
(1646.5 J) / (19.9125 x 10^-3 mol H2SO4) = 82687 J/mol = -82.7 kJ/mol H2SO4
(1646.5 J) / (39.825 x 10^-3 mol KOH) = 41343 kJ/mol = -41.3 kJ/mol KOH
The minus signs do not come from the calculations, they are the convention for exothermic processes.
