Question

# When 27.7 mL of 0.500 M H2SO4 is added to 27.7 mL of 1.00 M KOH...

When 27.7 mL of 0.500 M H2SO4 is added to 27.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ?H 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.)

Total volume of solution = 27.7 + 27.7 = 55.4 mL

Density of solution = 1g/mL

Mass of solution = volume x density

= 55.4 mL x 1g/mL

= 55.4 g

Specific heat capacity Cp = 4.184 J/g·°C

The given reaction is exothermic because the temperature rises from 23.5 c to 30.17 c

?H = - m x Cp x (T2-T1)

= - 55.4 g x 4.184 J/g·°C x (30.17 - 23.50)C

= - 1546.06 J

Negative sign shows that the heat is released

The balanced reaction

H2SO4 + 2KOH = K2SO4 + 2H2O

Moles of H2SO4 = molarity x volume

= 0.5 mol/L x 0.0277 L = 0.01385 mol

Moles of KOH = 1 mol/L x 0.0277 L = = 0.0277 mol

From the stoichiometry of the reaction

1 mol H2SO4 required = 2 mol KOH

0.01385 Mol H2SO4 required = 2*0.01385 = 0.0277 mol KOH

All moles will be consumed

Moles of H2O produced = 0.0277 mol

H = - (1546.06 J x 1kJ/1000 J) / 0.0277 mol H2O

= - 55.814 kJ/mol H2O

#### Earn Coins

Coins can be redeemed for fabulous gifts.