Question

A calorimeter contains 22.0 mL of water at 14.0 ∘C . When 2.50 g of X (a substance with a molar mass of 82.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 28.0 ∘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

Answer:

The enthalpy change in the the reaction is -47.014 kJ/mol.

Explanation:

$X(s)+H_2O(l)\rightarrow X(aq)$

Volume of water in calorimeter = 22.0 mL

Density of water = 1.00 g/mL

Mass of the water in calorimeter = m

$m=1.00 g/mL\times 22.0 mL=22 g$

Mass of substance X = 2.50 g

Mass of the solution = M = 2.50 g + 22 g = 24.50 g

Heat released during the reaction be Q

Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C

Specific heat of the solution is equal to that of water :

c = 4.18J/(g°C)

$Q=Mc\times \Delta T$

$Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ$

Heat released during the reaction is equal to the heat absorbed by the water or solution.

Heat released during the reaction =-1.433 kJ

Moles of substance X= $\frac{2.50 g}{82.0 g/mol}=0.03048 mol$

The enthalpy change, ΔH, for this reaction per mole of X:

$\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol$