Question

Question 10 of 10 > A coffee cup calorimeter with a heat capacity of 6.20 JrC was used to measure the change in enthalpy of a precipitation reaction. A 50.0 mL solution of 0.370 M AgNO, was mixed with 50.0 mL of 0.300 M KBr. After mixing, the temperature was observed to increase by 2.9 °C. Calculate the enthalpy of reaction, AH per mole of precipitate formed (AgBr). Assume the specific heat of the product solution is 4.18 J/ (g C) and that the density of both the reactant solutions is 1.00 g/mL Calculate the theoretical moles of precipitate formed from AgNO, and KBr.

Answer 1

The balanced equation is AgNO3 + KBr ------> AgBr (s) + KNO3 (aq) Number of moles of AgNO3 = M*V = 0.370 M * 50.0 mL = 18.5 mmol = 0.0185 mol Number of moles of KBr = M*V = 0.300 M * 50.0 mL = 15 mmol = 0.015 mol Theoretical moles of precipitate formed from AgNO3 = 0.0185 mol Theoretical moles of precipitate formed from KBr = 0.015 mol KBr is the Limiting reagent. Total volume of the solution = 50.0 + 50.0 = 100 mL density of the solution = 1.00 g/mol Mass of the solution = 100 ml * 1.00 g/mL = 100 g Heat capacity of solution = 4.18 J.g 1.°C-1. Change in temperature = AT = 2.9 °C Heat energy associated with this rise in temperature Acontents = m*C* AT = 100 g x 4.18 J.g 1.°C-1.x 2.9 °C = 1212.2 J Heat associated with the calorimeter ges= C* AT = 6.2 J/°C * 2.9 °C = 17.98 J solution = - (q.contents + acal) = -(1212.2 + 17.98 ) = - 1230.18 J Number of moles of AgBr = 0.015 mol AH for reaction = heat energy/moles of AgBr = - 1230.18 J / 0.015 mol =-82012 J/mol =- 82.012 kJ/mol =-82.01 kJ/mol AH Solution = - 82.01 kJ/mol