Question

Question #a165f

Answer 1

The enthalpy is `-"3.6 kJ/mol"`.

Explanation:

The answer can be found out using the Clausius Clapeyron equation:

`ln(P_2/P_1) = -(DeltaH_(vap)) / R (1/T_2 - 1/T_1)`

Here, `P_2 = "290 torr"` and `P_1 = "242 torr"`.

Don't forget to change temperature to absolute Kelvin (Celsius + 273 approximately).

So, `T_2 = "323 K"` and `T_1 = "373 K"`

`R = "8.314 J/mol"cdot"K"`

`ln("290 torr"/"242 torr") = -(DeltaH_(vap)) / ("8.314 J/mol"cdot"K") (1/"323 K" - 1/"373 K")`

`= -"3624.9 J/mol" = -"3.6 kJ/mol"`

Since the enthapy is negative , the liquid could be condensing!