Question

Consider the dissolution of CaCl₂: CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq) ∆H = -81.5 kJ/mol A 15.0-g sample of CaCl₂ is dissolved in 145 g water with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/°C･g.

Answer 1

Answer : The final temperature of the solution is, $41.4^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = 81.5 kJ/mol

q = heat released = ?

m = mass of $CaCl_2$ = 15.0 g

Molar mass of $CaCl_2$ = 110.98 g/mol

$\text{Moles of }CaCl_2=\frac{\text{Mass of }CaCl_2}{\text{Molar mass of }CaCl_2}=\frac{15.0g}{110.98g/mole}=0.135mole$

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

$81.5kJ/mol=\frac{q}{0.135mole}$

$q=11.0kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 11.0 kJ = 11000 J

m = mass of solution = 15.0 + 145 = 160.0 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $25.0^oC$

$T_2$ = final temperature = ?

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

$11000J=160.0g\times 4.18J/g^oC\times (T_2-25.0)$

$T_2=41.4^oC$

Thus, the final temperature of the solution is, $41.4^oC$

Answer 2

When a 15.0-g sample of CaCl₂ is dissolved in 145 g water, both substances at 25.0°C, the final temperature of the solution is 41.4 °C.

Let's consider the thermochemical equation for the dissolution of CaCl₂:

CaCl₂(s) → Ca²⁺(aq) + 2 Cl⁻(aq)    ∆H = -81.5 kJ/mol

We can calculate the heat released by the dissolution (Qd) of 15.0 g of CaCl₂ considering that:

• 81.5 kJ are released by the dissolution of 1 mole of CaCl₂.

• The molar mass of CaCl₂ is 110.98 g/mol.

$Qd = 15.0 g \times \frac{1mol}{110.98g} \times \frac{(-81.5kJ)}{mol} = -11.0 kJ$

According to the law of conservation of energy, the sum of the heat released by the dissolution (Qd) and the heat absorbed by the solution (Qs) is zero.

$Qd + Qs = 0\\\\Qs = -Qd = 11.0 kJ$

We can calculate the final temperature of the solution using the following expression.

$Qs = c \times m \times (T2-T1)\\\\T2 = \frac{Qs}{c \times m} + T1 = \frac{11.0 \times 10^{3}J }{\frac{4.18J}{\° C.g } \times (15.0g+145g)} + 25.0\° C = 41.4 \° C$

where,

• c: specific heat capacity of the solution
• m: mass of the solution

When a 15.0-g sample of CaCl₂ is dissolved in 145 g water, both substances at 25.0°C, the final temperature of the solution is 41.4 °C.

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