Question

A generic solid, X, has a molar mass of 75.3 g/mol. In a constant-pressure calorimeter, 16.9g of X is dissolved in 267g of water at 23.00 degrees Celsius.

X(s) --> X(aq)

The temperature of the resulting solution rises to 24.50 degrees Celsius. Assume the solution has the same specific heat as water, 4.184J/(g.C) and that's negligible heat loss to the surroundings.

How much heat was absorbed by the solution?

What is the enthalpy of the reaction?

Thanks!

Answer 1

1) Heat absorbed by the solution

Heat absorbed = m Cp (T2 - T1)

where,

m = Mass = 267 + 16.9 = 283.9 g

Cp = 4.184

T2 = 24.5

T1 = 23

=> Heat absorbed = 283.9 x 4.184 x (24.5 - 23) = 1781.76 J

2) Enthalpy

Moles of X = 16.9 / 75.3 = 0.2244 moles

Enthalpy of a reaction is defined as Heat evolved per mole of the reaction

For 0.2244 moles of X, heat evolved = 1781.76 J

=> For 1 mole of X, Heat evolved = 1781.76 / 0.2244 = 7940.1 J

=> Enthalpy of reaction = - 7940.1 J / mol or - 7.94 kJ / mol (-ve since heat is being evolved)