Question

the heat of vaporization $\delta h_v$ of diethyl ether $\left(\left(\text{ch}_3\
ight)_2\left(\text{ch}_2\
ight)_2\text{o}\
ight)$ is 26.7 kj/mol. calculate the change in entropy $\delta s$ when 138. g of diethyl ether condenses at 34.6 $^\circ$c.
be sure your answer contains a unit symbol. round your answer to 3 significant digits.

Explanation:

Step1: Find molar mass of diethyl ether

Diethyl ether formula: \((\text{CH}_3)_2(\text{CH}_2)_2\text{O}\). Molar mass: \(2\times(12.01 + 3\times1.008)+2\times(12.01 + 2\times1.008)+16.00\)
\(= 2\times15.034 + 2\times14.026 + 16.00 = 30.068 + 28.052 + 16.00 = 74.12\space\text{g/mol}\).

Step2: Calculate moles of diethyl ether

Moles \(n=\frac{\text{mass}}{\text{molar mass}}=\frac{138.\space\text{g}}{74.12\space\text{g/mol}}\approx1.862\space\text{mol}\).

Step3: Determine heat for condensation

Condensation is reverse of vaporization, so \(\Delta H = -n\times\Delta H_v\) (negative because heat is released).
\(\Delta H = -1.862\space\text{mol}\times26.7\space\text{kJ/mol}\approx -49.715\space\text{kJ} = -49715\space\text{J}\) (convert to joules for entropy units).

Step4: Convert temperature to Kelvin

\(T = 34.6^\circ\text{C} + 273.15 = 307.75\space\text{K}\).

Step5: Calculate entropy change

Entropy change formula: \(\Delta S=\frac{\Delta H}{T}\) (at constant temperature, for phase change).
\(\Delta S=\frac{-49715\space\text{J}}{307.75\space\text{K}}\approx -161.5\space\text{J/K}\) (rounded to 3 significant digits).

Answer:

\(-162\space\text{J/K}\) (or \(-161\space\text{J/K}\) depending on rounding precision; more accurately, \(-162\space\text{J/K}\) when using precise calculations)