the heat of vaporization $delta h_{v}$ of benzene ($c_{6}h_{6}$) is 44.3 kj/mol. calculate the change in…

the heat of vaporization $delta h_{v}$ of benzene ($c_{6}h_{6}$) is 44.3 kj/mol. calculate the change in entropy $delta s$ when 831. g of benzene condenses at 80.1 $^{circ}$c. be sure your answer contains a unit symbol. round your answer to 3 significant digits.

the heat of vaporization $delta h_{v}$ of benzene ($c_{6}h_{6}$) is 44.3 kj/mol. calculate the change in entropy $delta s$ when 831. g of benzene condenses at 80.1 $^{circ}$c. be sure your answer contains a unit symbol. round your answer to 3 significant digits.

Answer

Explanation:

Step1: Calculate the number of moles of benzene

The molar mass of benzene ($C_6H_6$) is $M=(6\times12.01 + 6\times1.01)\text{ g/mol}=78.12\text{ g/mol}$. The number of moles $n=\frac{m}{M}$, where $m = 831\text{ g}$. So $n=\frac{831\text{ g}}{78.12\text{ g/mol}}=10.64\text{ mol}$.

Step2: Calculate the heat released during condensation

The heat of vaporization $\Delta H_v = 44.3\text{ kJ/mol}$. When benzene condenses, the heat released $q=-n\Delta H_v$. Substituting the values, $q=- 10.64\text{ mol}\times44.3\text{ kJ/mol}=-471.35\text{ kJ}=-471350\text{ J}$.

Step3: Convert the temperature to Kelvin

The temperature $T = 80.1^{\circ}C=(80.1 + 273.15)\text{ K}=353.25\text{ K}$.

Step4: Calculate the change in entropy

The formula for the change in entropy $\Delta S=\frac{q}{T}$. Substituting $q=-471350\text{ J}$ and $T = 353.25\text{ K}$, we get $\Delta S=\frac{-471350\text{ J}}{353.25\text{ K}}=-1334.2\text{ J/K}\approx - 1330\text{ J/K}$.

Answer:

$-1330\text{ J/K}$