Question

A calorimeter contains 35.0 mLmL of water at 15.0 ∘C∘C . When 2.20 gg of XX (a substance with a molar mass of 56.0 g/molg/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq)X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 26.0 ∘C∘C . Calculate the enthalpy change, ΔHΔHDelta H, for this reaction per mole of XX. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)J/(g⋅∘C)], that density of water is 1.00 g/mLg/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings. Express the change in enthalpy in kilojoules per mole to three significant figures.

Answer 1

Answer:

ΔH rx = -43.5 kJ / mol

Explanation:

In water, Xdissolves thus:

X(s) + H₂O(l) → X(aq) + H₂O(aq)

It is possible to find the heat in dissolution process using coffee cup calorimeter equation:

Q = -m×C×ΔT

Where Q is heat, m is mass of solution (35.0g -density 1g/mL- + 2.20g = 37.2g), C is specific heat of solution (4.18J/g°C), and ΔT is change in temperature (26.0°C-15.0°C = 11.0°C)

Replacing:

Q = -37.2g×4.18J/g°C×11.0°C

Q = -1710J = -1.71kJ

As enthalpy is the change in heat per mole of reaction, moles of X that reacted were:

2.20g X × (1mol / 56.0g) = 0.0393 moles

As heat produced per 0.0393moles was -1.71kJ, heat per mole of X is:

-1.71kJ / 0.0393mol = -43.5 kJ / mol = ΔH rx