Chapter 11, Problem 11.1P

Interpretation Introduction

Interpretation:

The heat of vaporization $\left(\Delta H_{\text{vap}}\right)$ of bromine is to be calculated.

Concept introduction:

Clausius-Clapeyron equation is used to estimate the vapor pressure at another temperature if vapor pressure at one temperature and enthalpy of vaporization is known.

The expression of Clausius-Clapeyron equation is as follows:

$\ln P_{\text{vap}}=\left(\frac{-\Delta H_{\text{vap}}}{\text{R}}\right)\left(\frac{1}{T}\right)+C$

Where,

• $P_{\text{vap}}$ is the vapor pressure.
• $\Delta H_{\text{vap}}$ is the heat of vaporization.
• $T$ is the temperature.
• $C$ is the constant.
• $\text{R}$ is universal gas constant.

For the two pressures and temperature, the Clausius-Clapeyron equation is modified as follows:

$\ln \frac{P{}_1}{P_2}=\left(\frac{\Delta H_{\text{vap}}}{\text{R}}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

The conversion factor to convert $°\text{C}$ to Kelvin is,

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$

The conversion factor to convert $\text{kJ}$ to $\text{J}$ is as follows:

$1\text{ kJ}=\text{1000 J}$

Answer to Problem 11.1P

Heat of vaporization $\left(\Delta H_{\text{vap}}\right)$ of bromine is $31.069\mathrm{kJ}/mol$.

Explanation of Solution

Given information:

The value of $P{}_1$ is $\text{400 mmHg}$, $P{}_2$ is $\text{760 mmHg}$, $T_1$ is $\text{41}°\text{C }$, $T_2$ is $\text{331}\text{.9 K}$ and $\text{R}$ is $\text{8}\text{.314 J/mol}\cdot \text{K}$.

The formula to convert degree Celsius to kelvin is as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$

The temperature in Celsius is $41°\text{C}$.

Substitute the value in the above formula to calculate $T_1\left(\text{K}\right)$.

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}\\ =41°\text{C}+273\\ =314\mathrm{K}\end{array}$

For the two pressures and temperature, the Clausius-Clapeyron equation is modified as follows:

$\ln \frac{P{}_1}{P_2}=\left(\frac{\Delta H_{\text{vap}}}{\text{R}}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Rearrange above equation to calculate $\Delta H_{\text{vap}}$, the equation is modified is as follows

$\Delta H_{\text{vap}}=\frac{\left(\ln \frac{P{}_1}{P_2}\right)\left(\text{R}\right)}{\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$

The value of $P{}_1$ is $\text{400 mmHg}$.

The value of $P{}_2$ is $\text{760 mmHg}$.

The value of $T_1$ is $\text{314 K}$.

The value of $T_2$ is. $\text{331}\text{.9 K}$

The value of $\text{R}$ is $\text{8}\text{.314 J/mol}\cdot \text{K}$.

Substitute these values in above equation.

$\begin{array}{c}\Delta H_{\text{vap}}=\frac{\left(\ln \left(\frac{400\text{mmHg}}{760\text{ mmHg}}\right)\right)\left(\text{8}\text{.314 J/mol}\cdot \text{K}\right)}{\left(\frac{1}{331.\text{9 K}}-\frac{1}{31\text{4 K}}\right)}\\ =31069\text{ J/mol}\left(\frac{1\mathrm{kJ}}{1000\text{J}}\right)\\ =31.069\mathrm{kJ}/mol\end{array}$

Conclusion

Heat of vaporization $\left(\Delta H_{\text{vap}}\right)$ of bromine is $31.069\mathrm{kJ}/mol$.