Question

# When 23.8 mL of 0.500 M H2SO4 is added to 23.8 mL of 1.00 M KOH...

When 23.8 mL of 0.500 M H2SO4 is added to 23.8 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 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.)

?: kJ/mol H2O

Total volume = 23.8 mL + 23.8 mL = 47.6 mL

Density of the solution = 1 g/mL (Assumed same as that of water)

Therefore, mass of the solution = Volume x Density = 47.6 g

Heat gained by the calori meter, qreaction = mcT = 47.6 g x 4.184 J/goC x (30.17- 23.50) = 1328.4 J

Heat released during the reaction, qneutralization = -1328.4 J

Moles of H2SO4 used = 0.5 M x (0.0238 L) = 0.0119 moles

Moles of KOH used = 1 M x (0.0238 L) = 0.0238 moles

H2SO4 + 2KOH K2SO4 + 2H2O

Thus, one mole of H2SO4 produces two moles of H2O

Therefore, 0.0119 moles of H2SO4 will produce 0.0119 x 2 = 0.0238 moles of H2O

Similarly, 0.0238 moles of KOH will produce 0.0238 moles of H2O

Therefore,

H = -qneutralization/ Moles of H2O = -1328.4 J/ 0.0238 = -55.8 kJ/mol

Enthalpy of the reaction = -55.8 kJ/mol

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