Question

A calorimeter contains 23.0 mL of water at 11.5 ∘C . When 1.90 g of X (a substance with a molar mass of 76.0 g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)→X(aq)

and the temperature of the solution increases to 25.5 ∘C .

Calculate the enthalpy change, ΔH, for this reaction per mole of X.

Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.

Answer 1

Initial temperature of water =

Final temperature of solution,

Hence, the change in temperature due to dissolution is

Volume of water = 23.0 mL

Given that density of water = 1.00 g/mL ,we can calculate the mass of water as follows:

We have also added 1.90 g of X.

Hence, total mass of the solution, m = mass of water + mass of X = 23.0 g + 1.90 g= 24.9 g

Given that the specific heat of the solution is  

Now, we can calculate the heat absorbed by the solution as follows:

Note that the temperature is increasing by the dissolution process. Hence, it is an exothermic reaction.

Hence, the sign of enthalpy change will be negative and its magnitude will be equal to the amount of heat absorbed by the solution assuming no heat is lost to the surrounding or calorimeter.

Hence, we can write

for dissolution of 1.90 g of X.

It is given that the molar mass of X = 76.0 g/mol

Hence, the number of moles of X taken is

Hence, the enthalpy change per mole of X can be calculated by dividing the enthalpy change above by the number of moles calculated.

Hence,

Hence, the enthalpy change of the dissolution per mole of X dissolved is -58.3 kJ/mol (rounded to three significant figures).

Note: some calculations ignore the contribution of mass of X to the mass of solution. If that is the case, take the mass of the solution as 23 g and recalculate to find that is -53.8 kJ/mol.