Question

The ΔHvap of a certain compound is 47.45 kJ·mol−1 and its ΔSvap is 59.15 J·mol−1·K−1. What is the boiling point of this compound?

Answer 1

Answer:

802.2 K

Explanation:

To solve this problem we can use the formula:

• Tb = ΔHvap / ΔSvap

Where Tb is the boiling point (in K).

We already know the values of both ΔHvap and ΔSvap, so we calculate Tb:

• ΔHvap = 47.45 kJ·mol⁻¹ = 47.45x10³J·mol⁻¹

Tb = 47.45x10³J·mol⁻¹/59.15 J·mol⁻¹·K⁻¹

• Tb = 802.2 K

So the boiling point of the compound is 802.2 K (or 529 °C)

Answer 2

The boiling point of the compound whose ΔHvap is 47.45 KJ is 802.2 K.

What is the boiling point?

The temperature at which a liquid's vapor pressure matches the pressure surrounding the liquid is known as the boiling point of that substance.

By the formula of colligative property

$Tb = \dfrac{\Delta Hvap}{\Delta Svap }$

Where Tb is the boiling point (in K).

ΔHvap is 47.45 kJ·mol⁻¹

ΔSvap is  J·mol−1·K⁻¹

Putting the values

$Tb = \dfrac{47.45}{59.15 } =802.2 K.$

Thus, the boiling point is 802.2 K or 529° C.

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