Question

A generic solid, X, has a molar mass of 67.9 g/mol. In a constant-pressure calorimeter, 16.8 g of X is dissolved in 269 g of water at 23.00 °C.

X(s) ------ X(aq)

The temperature of the resulting solution rises to 26.40 °C. Assume the solution has the same specific heat as water, 4.184 J/(g·°C), and that there\'s negligible heat loss to the surroundings.

1. How much heat was absorbed by the solution?

2. What is the enthalpy of the reaction?

Answer 1

1) Heat absorbed by the solution

Heat absorbed = m Cp (T2 - T1)

where,

m = Mass = 269 + 16.8 = 285.8 g

Cp = 4.184

T2 = 26.4

T1 = 23

=> Heat absorbed = 285.8 x 4.184 x (26.4 - 23) = 4065.7 J

2) Enthalpy

Moles of X = 16.8 / 67.9 = 0.2474 moles

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

For 0.2474 moles of X, heat evolved = 4065.7 J

=> For 1 mole of X, Heat evolved = 4065.7 / 0.2474 = 16432.2 J

=> Enthalpy of reaction = - 16432.2 J / mol or - 16.43 kJ / mol (-ve since heat is being evolved)