A calorimeter contains 31.0 mL of water at 13.5 ∘C . When 2.30 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 29.5 ∘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. Express the change in enthalpy in kilojoules per mole to three significant figures.

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jufsanabriasa jufsanabriasa · Answer 1 · 2020-10-14T01:38:35+02:00

Answer:

-73.6kJ/mol

Explanation:

The solution is increasing its temperature from 13.5°C to 29.5°C, that means the enthalpy change is < 0

The released heat is determined using the calorimeter equation:

Q = -C×m×ΔT

Where Q is heat,

C is specific heat of the solution (4.18J/g°C)

m is mass of the solution (31.0mL = 31.0g + 2.30g = 33.3g)

And ΔT is change in temperature (29.5°C - 13.5°C = 16°C)

Replacing:

Q = -4.18J/g°C×33.3g×16°C

Q = -2227J are released

This heat is released when 2.30g are dissolved, that is:

-2227J / 2.30g × (76.0g / mol) = -73591J/mol =