Question

Enter your answer in the provided box. When 20.6 mL of 0.500 M H2SO4 is added to 20.6 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate AH 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

Answer 1

0.500 mol HS04 X 1L Initi al moles of H 20.6 mLx 1000mL 1L = 0.0103 mol H2SO 1L 1.00 mol KOH Initi al moles of OH 20.6 mLx 1000mL 1L = 0.0206 mol KOH Equation: HS0 2KOH - 4 The two reactants are in equal amounts that are in stoichiometric proportions as per reaction. So the moles of water produced from either of initial moles wil1 be 2 mol H2O 1m ol H2SO = 0.0206 mol H20 0.0103 mol H2SO4*. 2mol H2O = 0.0206 molH20 0.0206 mol KOHx 2mol KOH Thus, 0.0206 moles of HO produced Volume of water = (20.6 +20.6) mL = 41.2 mL The heat absorbed by calorimeter contents (water is 1.00 g 4.184J x(30.17-23.50) C 41.2 mLx mL X calarim g °C 1.00 g 4.184J (6.67°c) 41.2 mLx mL -X g °C = 1150 J 1150 kJ heat of reaction ,q -9carin = -1.150 kJ -1.150 kJ Heat of neutralization per mole of water , AH 0.0206 mol H20 = -55.8 kJ/molH2O = 56 kJ/mol H20