Question

A 68.0 mL sample of 1.0 M NaOH is mixed with 49.0 mL of 1.0 M H₂SO₄ in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 21.2°C. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until reaction is complete. Assume that the density of the mixed solutions is 1.0 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g •°C), and that no heat is lost to the surroundings. The ΔΗ_rxn for the neutralization of NaOH with H₂SO₄ is –114 kJ/mol H₂SO₄.

1.What is the maximum measured temperature in the Styrofoam cup?

________°C

Answer 1

1.0 mol H2SO 2 mol H* 1 mol H2SO 1L 49.0 mLx Initi al moles of H X X 1000mL 1L =0.098 mol H 1mol OH 1L 68.0 mLx 1.0 mol NaOH Initi al moles of OH 1000mL 1L 1mol NaOH =0.068 mol OH 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 will be 1mol H200.098 mol H2O 1m ol H mol H2O 1mol OH 0.098 mol H*x 0.068 mol H20 0.068 mol OH x Since NaOH acts as the limiting reactant, 0.068 moles of HO produced No of moles of H,O formed 0.068 mol -114 kJ Enthalpy of neutralization = 0.068 mol HOx- mol H2O =7.752 kJ heat absorbed,^alrm = -(-7.752 kJ) =7.752 kJ 7752 J (49.0 68.0) mL Volume of solution 117 mL The heat absorbed by calorimeter contents (water )is 1.00 g 4.18J *(7-21.2C 117 mLx mL alrim A.18(7- 21.2°c) 7752 J (117 g)x. g °C 7752 J 15.8 C T 21.2° C 4.18J (117 g)x g oC 37.0° C T 21.2° C+15.8° C