Chemistry, 14.09.2019 09:30 victoriadorvilu
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
Answers: 1
= 100 mm Hg or 
= 0.13157 atm
= 
= (1080 + 273) K = 1357 K
= 
= (1220 + 273) K = 1493 K
= 600 mm Hg or 
= 0.7895 atm
![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}]](/tpl/images/0231/2242/21278.png)
![log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]](/tpl/images/0231/2242/46d38.png)
= 
J/mol
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The vapor pressure of substance x is 100. mm hg at 1080.°c. the vapor pressure of substance x increa...
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