Question

Calculate the change in entropy that occurs in the system when 2.70 mole of diethyl ether (C4H6O) condenses from a gas to a liquid at its normal boiling point (34.6∘C). ΔHvap = 26.5 kJ/mol

Answer 1

Answer:233 Joules/K

Explanation:

∆H= 26.5KJmol-1

Kelvin temperature = 34.6 + 273 = 307.6 K

No of moles= 2.7 moles

2.70 mole x 26.5 kJ/mole = 71.55kilojoules

∆S=71.55 kilojoules / 307.6 K = 0.233 kilojoules/ K

Convert to JK-1

0.233 kilojoules/ K x 1000 Joules/kilojoule = 233 Joules/K

Answer 2

The change in entropy that occurred in the system is 232.6 J/K.

The given parameters:

• number of moles of the diethyl ether  compound, n = 2.7 moles
• heat of vaporization, ΔHvap = 26.5 kJ/mol
• temperature, T = 34.6 ⁰C = 273 + 34.6 = 307.6 K

The energy released in the process is calculated as follows;

$\Delta H = n \Delta H_{vap}\\\\ \Delta H = 2.7 \times (26.5 \times 10^3)\\\\ \Delta H = 71,550 \ J$

The change in entropy that occurred in the system is calculated as follows;

$\Delta S = \frac{\Delta H}{T} \\\\\Delta S = \frac{71,550}{307.6} \\\\\Delta S = 232.6 \ J/K$

Thus, the change in entropy that occurred in the system is 232.6 J/K.

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