Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: slu1986 on September 04, 2008, 07:24:03 PM
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While mercury is very useful in barometers, mercury vapor is toxic. Given that mercury has a heat of vaporization of 56.9 kJ/mol and its normal boiling point is 356.7 °C, calculate the vapor pressure in mm Hg at room temperature, 25 °C.
A. 4.29 X 10^-3 mm Hg <----this is the right answer but I don't understand how to solve the problem and get that answer.
B. 2.36 mm Hg
C. 62.8 mm Hg
D. 751 mm Hg
If someone could guide me in the right direction on how to solve this problem I would appreciate it. :)
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http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation
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I have the Clausius-Clapeyron equation..but I don't understand how to set up the equation when your not given the vapor pressure. The problem is asking to calculate the vapor pressure in mm Hg at room temp 25 °C.
It gives the heat of vaporization of mercury = 56.9 kJ/mol and its normal boiling pt = 356.7 °C.
the clausius-clapeyron equation is
ln P vap = (- Delta H Vap/R) 1/T + C
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The boiling point is defined as the temperature at which the vapor pressure equals 1 atm.
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I'm sorry but I still have no idea how to set up the problem.
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Ok this is how I set up and worked the problem..I came up with an answer close to the answer my teacher got.
ln P2 = ln (760) + (56900 J/mol/8.314 J/mol *K) (1/629.7 K - 1/298 K)
ln (760) = 6.63 56900 J/mol / 8.314 J/mol * K = 6844 1/630 K - 1/298 K = -0.00177
6.63 + -12.1 = -5.47
e^-5.47 = 0.00421 mm Hg = 4.21 * 10^-3 mm Hg
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Looks to me like you are rounding intermediate results. You shouldn't. Otherwise your approach is correct now. If it helps, I can't get 4.29x10-3 as well, more like 4.24 or 4.28 depending on whether I use 273 or 273.15 to convert deg C to K.