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Estimate the boiling point of Bromine. The enthalpy of vaporization of bromine is 30.9kJ/mol The standard entropy values for liquid and gaseous bromine are 152J/mol^ast K and 245J/mol^ast K respectively. Br_(2(liq.));Br_(2(g)) 4)

Вопрос

Estimate the boiling point of Bromine. The enthalpy of vaporization of bromine is
30.9kJ/mol The standard
entropy values for liquid and gaseous bromine are
152J/mol^ast K and 245J/mol^ast K respectively.
Br_(2(liq.));Br_(2(g))
4)

Estimate the boiling point of Bromine. The enthalpy of vaporization of bromine is 30.9kJ/mol The standard entropy values for liquid and gaseous bromine are 152J/mol^ast K and 245J/mol^ast K respectively. Br_(2(liq.));Br_(2(g)) 4)

Решения

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Гаврила
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мастер · Репетитор 5 лет

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To estimate the boiling point of bromine, we can use the Clausius-Clapeyron equation, which relates the change in pressure to the change in temperature during a phase transition:<br /><br />$\frac{dP}{dT} = \frac{L}{T \Delta S}$<br /><br />where $P$ is the pressure, $T$ is the temperature, $L$ is the enthalpy of vaporization, and $\Delta S$ is the change in entropy.<br /><br />Given:<br />- Enthalpy of vaporization, $L = 30.9 kJ/mol$<br />- Standard entropy values:<br /> - Liquid bromine, $S_{liq} = 152 J/mol \cdot K$<br /> - Gaseous bromine, $S_{gas} = 245 J/mol \cdot K$<br /><br />We need to convert the enthalpy of vaporization to the same units as the entropy values:<br />$L = 30.9 kJ/mol = 30900 J/mol$<br /><br />Now, we can calculate the change in entropy:<br />$\Delta S = S_{gas} - S_{liq} = 245 J/mol \cdot K - 152 J/mol \cdot K = 93 J/mol \cdot K$<br /><br />Next, we can integrate the Clausius-Clapeyron equation to find the boiling point:<br />$\int_{T_1}^{T_2} \frac{dT}{T} = \frac{L}{\Delta S} \int_{P_1}^{P_2} \frac{dP}{P}$<br /><br />Assuming standard pressure (1 atm) and using the ideal gas law, we can simplify the integral:<br />$\int_{T_1}^{T_2} \frac{dT}{T} = \frac{L}{\Delta S} \int_{P_1}^{P_2} \frac{dP}{P} = \frac{L}{\Delta S} \ln\left(\frac{P_2}{P_1}\right)$<br /><br />Since we are estimating the boiling point, we can assume $P_1$ to be the standard pressure (1 atm) and $P_2$ to be the vapor pressure at the boiling point. The vapor pressure of bromine at its boiling point is approximately 2.5 atm.<br /><br />Now, we can solve for the boiling point:<br />$\ln\left(\frac{2.5}{1}\right) = \frac{30900}{93} \cdot \frac{1}{T_2}$<br /><br />$T_2 = \frac{30900}{93 \cdot \ln(2.5)}$<br /><br />$T_2 \approx 158.8 K$<br /><br />Therefore, the estimated boiling point of bromine is approximately 158.8 K.
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