Question

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, what is the change in temperature (°C) of the solution? (The specific heat capacity of the solution is 4.184 J/g･°C and the density of the solution is 1.02 g/mL).

Answer 1

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908