Question

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C. Determine the molar heat of vaporization of substance X using the derived form of the Clausius-Clapeyron equation given below. (Include the sign of the value in your answer.) ____ kJ/mol

Answer 1

Explanation:

The given data is as follows.

         $P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

         $T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

         $T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

         $P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                   $log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

            $log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

          $log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

                0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

              $\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

                                   = $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.