Question

At 0.00 °C, hexane, C6H14, has a vapor pressure of 45.37 torr. Its ΔHvap is 30.1 kJ mol-1. What is the vapor pressure (in torr) of hexane at 49 °C?

R=8.314 J K-1 mol​

Answer 1

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Answer 2

Given: 0.00°C the vapour pressure in 45.37torr. ΔH is $30.1kJ/mol$.

To find: vapour pressure at 49°C

Step-by-step explanation:

Step 1 of 1

The vapour pressure can be calculated by using temperature by Clausius-Clayperon equation:

$ln(\frac{P_{1} }{P_{2} } )=\frac{delH_{vap} }{R} [\frac{1}{T_{2} }-\frac{1}{T_{1} } ]$

Substituting the values, we will get:

$ln(\frac{45.37*760 }{P_{2} } )=\frac{30100}{8.314} [\frac{1}{322 }-\frac{1}{273} ]\\ln(\frac{34,481.2}{P_{2} } )=-2.018\\2.303log\frac{34,481.2}{P_{2} } =-2.018\\log\frac{34,481.2}{P_{2} } =-0.8762\\\frac{34,481.2}{P_{2} } =0.1329\\\frac{34,481.2}{0.1329 } =P_{2}\\\\P_{2}=2,59,452.22atm\\P_{2}=341.38torr$

Vapour pressure at 49°C is 341.38torr.