A generic solid, X, has a molar mass of 60.8 g/mol. In a constant‑pressure calorimeter, 17.7 g of X is dissolved in 333 g of water at 23.00 °C. X(s)⟶X(aq)

Question

The temperature of the resulting solution rises to 27.30 °C. Assume the 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 absorbed by the solution?

What is the enthalpy of the reaction?

Answer 1

q = mC∆T

q = heat = ?

m = mass of water = 333 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 27.30º - 23.00º = 4.3º

q = (333 g)(4.184 J/gº)(4.3º) = 5991 J of heat was absorbed by the solution

Enthalpy of reaction = ∆Hrxn = kJ/mol

kJ = 5991 J x 1 kJ/1000 J = 5.99 kJ

moles = 17.7 g x 1 mol / 60.8 g = 0.291 mols

∆Hrxn = 5.99 kJ / 0.291 mols = 20.6 kJ/mol