calorimetry / thermodynamics problem

Question

Potassium nitrate, KNO₃, has a molar mass of 101.1 g/mol. In a constant-pressure calorimeter,  26.5g of KNO₃ is dissolved in 243 g of water at 23.00ºC.

KNO₃ (s) → K⁺ (aq) + NO₃^- (aq)

The temperature of the resulting solution decreases to 20.80ºC. Assume that the resulting solution has the same specific heat as water, 4.184 J/(g•ºC), and that there is negligible heat loss to the surroundings.

a. How much heat was released by the solution?

b. What is the enthalpy of the reaction?

Answer 1

q = mC∆T

q = heat = ?

m = mass = 243 g (the mass of KNO3 is ignored in introductory courses.)

C = specific heat = 4.184 J/gº

∆T = change in temperature = 2.2º

(a) q = (243 g)(4.184 J/gº)(2.2º) = 2237 J of heat "released" by the surroundings. This would be endothermic.

(b) The enthalpy of the reaction would be 2237 J. But if you want the standard enthalpy of reaction then you would have to divide by moles of KNO3 present. This is not a well-written question, in my opinion.

mole KNO3 = 26.5 g x 1 mol/101.1 g = 0.262 moles KNO3

∆H/mol = 2237 J / 0.262 mol = 8538 J/mol = 8.54 kJ/mol