the heat of vaporization $delta h_{v}$ of carbon disulfide ($cs_{2}$) is 26.74 kj/mol. calculate the change…

the heat of vaporization $delta h_{v}$ of carbon disulfide ($cs_{2}$) is 26.74 kj/mol. calculate the change in entropy $delta s$ when 3.3 g of carbon disulfide condenses at -78.55 $^{circ}$c. be sure your answer contains a unit symbol. round your answer to 2 significant digits.
Answer
Explanation:
Step1: Calculate moles of CS₂
The molar mass of CS₂ is $M = 12.01+2\times32.07=76.15\ g/mol$. The number of moles $n$ of CS₂ with a mass $m = 3.3\ g$ is $n=\frac{m}{M}=\frac{3.3\ g}{76.15\ g/mol}\approx0.0433\ mol$.
Step2: Convert temperature to Kelvin
The temperature $T=-78.55^{\circ}C$. Converting to Kelvin, $T = - 78.55+273.15 = 194.6\ K$.
Step3: Calculate heat change during condensation
The heat of vaporization $\Delta H_v = 26.74\ kJ/mol$. For condensation, $\Delta H=-n\Delta H_v$. Substituting $n = 0.0433\ mol$ and $\Delta H_v=26.74\ kJ/mol$, we get $\Delta H=-0.0433\ mol\times26.74\ kJ/mol=-1.16\ kJ=-1160\ J$.
Step4: Calculate change in entropy
The formula for entropy change is $\Delta S=\frac{\Delta H}{T}$. Substituting $\Delta H=- 1160\ J$ and $T = 194.6\ K$, we have $\Delta S=\frac{-1160\ J}{194.6\ K}\approx - 6.0\ J/K$.
Answer:
$-6.0\ J/K$