A calorimeter contains 35.0mL of water at 14.0?C . When 1.30g of X (a substance

Question

A calorimeter contains 35.0mL of water at 14.0?C . When 1.30g of X (a substance with a molar mass of 64.0g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)?X(aq)

and the temperature of the solution increases to 28.5?C .

Calculate the enthalpy change, ?H, 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.

Explanation / Answer

We use the equation, Q = mcdT

m = mass of water = 35.0 g (since density of water = 1.0 g/mL)

c = 4.18 J/g oC

dT = Tf-Ti = 28.5 oC - 14.0 oC = 14.5 oC

Plugging in the values, we get

Q = 35.0 g x 4.18 J/g oC x 14.5 oC

Q = 2121.35 J

--------------------

Given, 1.30 g of X

Moles of X = Mass / Molar mass = 1.30 g / 64.0 g/mol = 0.0203 mol

Thus, 0.0203 mol of X gives an energy of 2121.35 J

1 mol of X gives 2121.35 / 0.0203 = 104500 J

Thus, enthalpy Delta H of X = 104500 J/mol

Delta H of X = 104500 / 1000 = 1045 kJ/mol

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