Question

The heat of vaporization Δ1, of toluene (C6H5CH3) is 38.1 kJ/mol. Calculate the change in entropy AS when 207. g of toluene boils at 1 10.6 °C. Be sure your answer contains a unit symbol. Round your answer to 3 significant digits.

Answer 1

Answer: The change in entropy change when 207 g of toluene boils at $110.6^{o}C$ is 223 $J/K$.

Explanation:

It is given that mass of toulene is 207 g and its molar mass is 92.14 g/mol. So, its moles will be calculated as follows.

     No. of moles = $\frac{mass}{\text{molar mass}}$

                           = $\frac{207 g}{92.14 g/mol}$

                           = 2.25 moles

As we are given that heat of vaporization for 1 mol is 38.1 kJ/mol. So, heat of vaporization for 2.25 moles will be calculated as follows.

            $2.25 moles \times 38.1 KJ/mol$

            = 85.725 KJ

Now, we know that the relation between enthalpy change and entropy change is as follows.

           $\Delta S = \frac{\Delta H}{\Delta T}$

                        = $\frac{85.725 kJ}{(110.6 + 273) K}$

                        = 0.223 $kJ/K$

or,                     = 223 $J/K$         (as 1 kJ = 1000 J)

Thus, we can conclude that the change in entropy change when 207 g of toluene boils at $110.6^{o}C$ is 223 $J/K$.