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

The change in entropy (∆S) when 207 grams of toluene boils at 110.6 °C is 0.223 kJ/K. This was calculated by first converting the mass of toluene to moles, using the heat of vaporization to find total heat absorbed, and then applying the formula ∆S = q/T with the temperature in Kelvin.

Step-by-step explanation:

To calculate the change in entropy (∆S) when 207 grams of toluene boils at 110.6 °C, we first need to convert grams to moles using the molar mass of toluene, which is about 92.14 g/mol. Next, we use the heat of vaporization (∆1) provided, 38.1 kJ/mol, to find the total heat absorbed during vaporization. Since entropy is a measure of disorder and vaporization increases disorder, the entropy change will be positive. Lastly, we apply the formula ∆S = q/T, where q is the total heat absorbed and T is the temperature in Kelvin.

Firstly, calculate the moles of toluene: 207 g / 92.14 g/mol = 2.247 mol. Then, calculate the total heat absorbed: 2.247 mol × 38.1 kJ/mol = 85.61 kJ. Finally, convert the temperature to Kelvin: 110.6 °C + 273.15 = 383.75 K. Now, calculate the entropy change: ∆S = 85.61 kJ / 383.75 K = 0.223 kJ/K, or 223 J/K when converted to joules.

Answer 2

Answer: The change in entropy change when 207 g of toluene boils at
`110.6^(o)C` is 223
`J/K`.

Step-by-step 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}}`

=
`(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 * 38.1 KJ/mol`

= 85.725 KJ

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

`\Delta S = (\Delta H)/(\Delta T)`

=
`(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`.