Question

The ΔHvap of nitrous oxide is 16.53 kJ · mol−1 and its ΔSvap is 89.51 J · mol−1 · K−1. What it the boiling point of nitrous oxide?

Answer 1

Answer:

$T_b=-88.48\°C$

Explanation:

Hello,

In this case, since the entropy of vaporization is defined in terms of the enthalpy of vaporization and the boiling point of the given substance, nitrous oxide, as shown below:

$\Delta _{vap}S=\frac{\Delta _{vap}}{T_b}$

Solving for the boiling point of nitrous oxide, we obtain:

$T_b=\frac{\Delta _{vap}H}{\Delta _{vap}S}=\frac{16.53\frac{kJ}{mol}*\frac{1000J}{1kJ} }{89.51\frac{J}{mol} } \\ \\T_b=184.67K$

Which in degree Celsius is also:

$Tb=184.67-273.15\\\\T_b=-88.48\°C$

Best regards.