Question

When 4.51 g of CaCl2 dissolved in 50.00 mL of water in a coffee cup calorimeter, the temperature of the solution rose from 22.6°C to 25.8°C.
Specific heat of the solution is equal to the specific heat of water = 4.18 J/gºC.
Density of the solution is equal to the density of water = 1.00 g/mL.

What is qsolution?

What is qreaction ?

What is ÎHrxn in kJ/mol of CaCl2 ?

Answer 1

Explanation:

The heat gained by the solution = q

$q=mc\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat of solution= $4.18 J/^oC$

Mass of the solution(m) = mass of water + mass of calcium chloride

Mass of water = ?

Volume of water = 50.00 mL

Density of water = 1.00 g/mL

Mass = Density × Volume

m = 1.00 g/mL × 50.00 mL = 50.00 g

Mass of the solution (m) = 50.00 g + 4.51 g =54.51 g

$T_{final}$ = final temperature = $25.8 ^oC$

$T_{initial}$ = initial temperature = $22.6 ^oC$

Now put all the given values in the above formula, we get:

$q=54.51 g\times 4.18 J/g^oC\times (25.8-22.6)^oC$

$q=729.126 J$

The heat gained by the solution is 729.126 J.

Heat energy released during the reaction  = q'

q' = -q  ( law of conservation of energy)

q' = -729.126 J

The heat energy released during the reaction is -729.126 J.

Moles of calcium chloride, n = $\frac{4.51 g}{111 g/mol}=0.04063 mol$

$\Delta H_{rxn}=\frac{q'}{n}=\frac{-729.126 J}{0.04063 mol}=-17,945.23 J/mol= -17.945 kJ/mol$

The ΔH of the reaction is -17.945 kJ/mol.