Question

For pentane, the ∆H° of vaporization is 26.22 kJ/mol and the ∆S° of vaporization is 87.88 J/mol･K. At 1.00 atm and 201 K, what is the ∆G° of vaporization for pentane, in kJ/mol?

Answer 1

At 1.00 atm and 201 K, the ∆G° of vaporization for pentane is calculated using the formula ∆G° = ∆H° - T∆S° and the provided values, resulting in 8.56 kJ/mol.

Step-by-step explanation:

The student asked about the ∆G° of vaporization for pentane at a certain temperature and pressure. To find the Gibbs free energy (∆G°) of vaporization, we use the equation:

∆G° = ∆H° - T∆S°

Here, ∆H° is the enthalpy of vaporization, T is the temperature in Kelvin, and ∆S° is the entropy of vaporization. We are given:

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First, we need to convert the temperature to Kelvin if it is not already, and ∆S° to kJ/mol·K by dividing by 1000. The temperature is already in Kelvin, and the conversion for ∆S° is:

∆S° = 87.88 J/mol·K ÷ 1000 = 0.08788 kJ/mol·K

Substitute the values into the equation to get:

∆G° = 26.22 kJ/mol - (201 K)(0.08788 kJ/mol·K) = 26.22 kJ/mol - 17.66388 kJ/mol = 8.56 kJ/mol

Therefore, at 1.00 atm and 201 K, the ∆G° of vaporization for pentane is 8.56 kJ/mol.

Answer 2

Answer:8.56 KJ/mol

Step-by-step explanation:

∆G = ∆H - T∆S

∆H = 26.22 KJ/mol

T = 201 K

∆S = 87.88 --> .08788 KJ/mol

26.22 - 201 * .08788

26.22 - 17.66388 = 8.55612