How much heat was released by the solution? What is the enthalpy of the reaction?

Question

Potassium nitrate, KNO3, has a molar mass of 101.1 g/mol. In a constant-pressure calorimeter, 34.1 g of KNO3 is dissolved in 279 g of water at 23.00 °C.

KNO3(s)−→−−H2OK+(aq)+NO−3(aq)

The temperature of the resulting solution decreases to 20.90 °C. Assume that the resulting solution has the same specific heat as water, 4.184 J/(g·°C), and that there is negligible heat loss to the surroundings.

How much heat was released by the solution?

1. 𝑞soln= ? kJ 

What is the enthalpy of the reaction?

1. Δ𝐻rxn= ? kJ/mol

Answer 1

q = mC∆T

q = heat = ?

m = mass of water = 279 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 23.00º - 20.90º = 2.10º (note: temp decreased meaning an endothermic process).

Solving for q = heat released by solution, we have...

q = (279 g)(4.184 J/gº)(2.10º) = 2451 J

Converting to kJ we have...

q = 2.45 kJ = heat released by solution

Enthalpy of reaction:

The reaction KNO₃(s) ==>K⁺(aq) + NO₃^-(aq) shows ONE mole of KNO₃ dissolving in water. The enthalpy of reaction therefore should be heat per 1 mole of KNO₃. In the calculation above, we solved for the heat released by dissolution of 34.1 g of KNO3.

molar mass KNO3 = 101 g / mole

moles KNO3 dissolved = 34.1 g x 1 mol / 101 g = 0.3376 moles

heat released = 2.45 kJ (see above calculations)

Enthalpy of reaction = 2.45 kJ / 0.3376 mols = 7.26 kJ / mole (3 sig. figs.)