Chapter 12, Problem 12.126P

Interpretation Introduction

Interpretation:

The temperature at the flash point of $\mathrm{n}\text{-hexane}$ is to be calculated.

Concept introduction:

Clausius-Clapeyron equation is used to find the vapor pressure at one temperature when vapor pressure at another temperature and heat of enthalpy is given. It determines the change in vapor pressure with a change in temperature. The expression of the Clausius-Clapeyron equation is as follows:

$\ln \left(\frac{P_2}{P_1}\right)=\frac{-\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$        (1)

Here,

$P_1\text{ and }{P}_2$ are the vapor pressure.

$T_1\text{ and }{T}_2$ are the temperature.

$R$ is the gas constant.

$\Delta H_{\text{vap}}$ is the heat of vaporization.

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.2$        (2)

Answer to Problem 12.126P

The temperature at flash point of $\mathrm{n}\text{-hexane}$ is $-30.2°\text{C}$.

Explanation of Solution

Substitute $20.0°\text{C}$ in equation (2) to calculate the temperature $T_1$ in Kelvin as follows:

$\begin{array}{c}{T}_1\left(\text{K}\right)=20.0°\text{C}+273.2\\ =293.2\text{ K}\end{array}$

Substitute $68.7°\text{C}$ in equation (2) to calculate the temperature $T_2$ in Kelvin as follows:

$\begin{array}{c}{T}_2\left(\text{K}\right)=68.7°\text{C}+273.2\\ =341.9\text{ K}\end{array}$

Rearrange the equation (1) to calculate $\Delta H_{\text{vap}}$.

$\Delta H_{\text{vap}}=-\frac{\left(R\right)\ln \left(\frac{P_2}{P_1}\right)}{\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$        (3)

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $293.2\text{ K}$ for $T_1$, $341.9\text{ K}$ for $T_2$ and $121\text{mmHg}$ for $P_1$ and $1\text{atm}$ for $P_2$ in the equation (3).

$\begin{array}{c}\Delta H_{\text{vap}}=-\frac{\left(8.314\text{ J/K}\cdot \text{mol}\right)\ln \left(\frac{1\text{atm}}{121\text{mmHg}}\left(\frac{760\text{mmHg}}{\text{1}\text{atm}}\right)\right)}{\left(\frac{1}{341.9\text{ K}}-\frac{1}{293.2\text{ K}}\right)}\\ =3.1447\times {10}^4\text{ J/mol}\\ \approx 3.14\times {10}^4\text{J/mol}\end{array}$

The LFL of $\mathrm{n}\text{-hexane}$ is $1.1\%$ and therefore the mole fraction of $\mathrm{n}\text{-hexane}$ is $0.011$. The formula to calculate the pressure of hexane at flash point is as follows:

$P_{\text{flash point}}=\left(X_{\text{hexane}}\right)\left(P_{\text{Total}}\right)$        (4)

Substitute $0.011$ for $X_{\text{hexane}}$ and $1\text{atm}$ for $P_{\text{Total}}$ in the equation (4)

$\begin{array}{c}{P}_{\text{flash point}}=\left(0.011\right)\left(1\text{atm}\right)\left(\frac{760\text{mmHg}}{1\text{atm}}\right)\\ =8.36\text{mmHg}\\ \approx 8.4\text{mmHg}\end{array}$

The expression to calculate the temperature at flash point is as follows:

$\frac{1}{T_{{}_{\text{flash point}}}}=\left(\frac{\ln \left(\frac{P_{\text{flash point}}}{P_2}\right)}{\frac{-\Delta H_{\text{vap}}}{R}}+\frac{1}{T_2}\right)$        (5)

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $341.9\text{ K}$ for $T_2$, $8.4\text{mmHg}$ for $P_{\text{flash point}}$, $1\text{atm}$ for $P_2$ and $3.1447\times {10}^4\text{ J/mol}$ for $\Delta H_{\text{vap}}$ in the equation (5).

$\begin{array}{c}\frac{1}{T_{{}_{\text{flash point}}}}=\left(\frac{\ln \left(\frac{8.4\text{mmHg}}{1\text{atm}}\left(\frac{\text{1}\text{atm}}{760\text{mmHg}}\right)\right)}{\frac{-3.1447\times {10}^4\text{ J/mol}}{8.314\text{ J/K}\cdot \text{mol}}}+\frac{1}{341.9\text{ K}}\right)\\ =\frac{1}{242.961\text{K}}\\ T_{{}_{\text{flash point}}}=242.961\text{K}\\ \approx 243.0\text{K}\end{array}$

Rearrange the equation to convert the temperature at flash point from Kelvin to $°\text{C}$ is as follows:

$T_{{}_{\text{flash point}}}\left(°\text{C}\right)=T_{{}_{\text{flash point}}}\left(\text{K}\right)-273.2$        (6)

Substitute $243.0\text{K}$ for $T_{{}_{\text{flash point}}}\left(\text{K}\right)$ in the equation (6).

$\begin{array}{c}{T}_{{}_{\text{flash point}}}\left(°\text{C}\right)=243.0\text{K}-273.2\\ =-30.2°\text{C}\end{array}$

Conclusion

The temperature at the flash point of $\mathrm{n}\text{-hexane}$ is $-30.2°\text{C}$.