Chemistry: Heat of vaporization

Question

The boiling point of liquid D is 45.5 °C at 1 atm (760.0 torr). Its vapor pressure at 25.0 °C is 145 torr. What is its ∆Hvap in kJ/mol?

Answer 1

We are given that P₁= 145 torr and for temperate we will covert from celsius to kelvin so we get

T₁= 25 C= 25+273.15= 298.15 K

Now for P₂= 760 torr and T₂= 273.15+45.5=318.65 K

We also know R is our gas constant at 8.314 J/k*mol

Now we would use the following formula to solve this problem:

We have the Clausius-Clapeyron equation-

ln (P2/ P1)=-∆Hvap/ R [1/T2 - 1/T1]

Putting the values in the equation,

ln( 760/ 145) = - ∆Hvap/ 8.314 J/k*mol [ 1/318.65- 1/298.15]

ln( 5.2413) = - ∆Hvap/ 8.314 J/k*mol [- 2.158 * 10^-4 K^-1]

1.65658= - ∆Hvap* 2.585 * 10^-5 K^-1]

Now after rearranging the problem we get:

∆Hvap=1.65658/ 2.585 * 10^-5 J/ mol

∆Hvap=6.4026 *10³ J/ mol

∆Hvap=6.4026 kJ/ mol

Hence, the enthalpy of vaporization of the liquid is 6.4026 kJ/mol.