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

`T_b=-88.48\°C`

Step-by-step 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=(\Delta _(vap))/(T_b)`

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

`T_b=(\Delta _(vap)H)/(\Delta _(vap)S)=(16.53(kJ)/(mol)*(1000J)/(1kJ) )/(89.51(J)/(mol) ) \\ \\T_b=184.67K`

Which in degree Celsius is also:

`Tb=184.67-273.15\\\\T_b=-88.48\°C`

Best regards.