Question

For 2-methylbutane, the ∆h° of vaporization is 25.22 kj/mol and the ∆s° of vaporization is 84.48 j/mol･k. at 1.00 atm, what is the boiling point of 2-methylbutane, in k?

Answer 1

To find the boiling point of 2-methylbutane, use the Clausius-Clapeyron equation and divide the enthalpy of vaporization by the entropy of vaporization, which yields a boiling point of approximately 298.65 K at 1.00 atm.

Step-by-step explanation:

The boiling point of a substance is the temperature at which its vapor pressure is equal to the surrounding atmospheric pressure. To find the boiling point of 2-methylbutane at 1.00 atm, we can use the Clausius-Clapeyron equation, which relates the enthalpy of vaporization (∆H°) and the entropy of vaporization (∆S°) to the temperature:

∆H° = T∆S°

We are given that the ∆H° of vaporization for 2-methylbutane is 25.22 kJ/mol and the ∆S° of vaporization is 84.48 J/mol·K. To obtain the boiling point in Kelvin, we need to divide the enthalpy by the entropy:
T = ∆H° / ∆S°

To ensure SI units are consistent, convert ∆H° from kJ/mol to J/mol by multiplying by 1000:
T = (25.22 kJ/mol × 1000 J/kJ) / 84.48 J/mol·K

T = 25220 J/mol / 84.48 J/mol·K

T ≈ 298.65 K, which is the boiling point of 2-methylbutane at 1.00 atm.