Question

Question #5:

PART A:

In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 8.90 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol.

QUESTION #8:

PART A:

A calorimeter contains 18.0 mL of water at 12.0 ∘C . When 2.50 g of X (a substance with a molar mass of 69.0 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.

Express the change in enthalpy in kilojoules per mole to three significant figures.

Assume that the specific heat of the solution formed in the calorimeter is the same as that for pure water: Cs=4.184 J/g⋅∘C.

Express your answer with the appropriate units.

Answer 1

Question #5:

QUESTION #8:

Total volume of the solution = 18.0 + 2.50 = 20.50 mL

The density of the solution is assumed to be 1 g/mL.

Hence, the total mass of the solution = 20.50 mL x 1 g/mL = 20.50 g

The temperature change = 28.0 -12.0 = 16.0 ^oC

The enthalpy change is the product of mass, the specific heat of the resulting solution and the temperature change.

q = m c delta T

q= 20.5 g x 4.18 J/g^oC x 16.0 ^oC

q = 1371 J

This is enthalpy change for 2.50 g of X. The enthalpy change for 1 mole of X will be

1371 J x 69.0 g/mole / 2.50 g = 37840.7 J = 37.8 kJ/mole

Since heat is released in this reaction, the sign of the enthalpy change must be negative.

So it would be -37.8kJ/mol.