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March 29, 2017

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Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1 degrees celsius. At what temperature does benzene boil when the external pressure is 480 torr? Can someone show me an equation and how to solve this?

  • Chemistry - ,

    Use the Clausius-Clapeyron equation. Remember the definition of boiling point; i.e., the temperature at which the vapor pressure of the liquid equals atmospheric pressure. So you know p1 and p2 and T1. T2 is the only unknown.

  • Chemistry - ,

    I know im using that equation but would you set it up like this
    1/t2= (760mm hg/480torr)/(30720j/8.314)+(1/353.25degrees celsius) when i do it out i get a number that does not make sense.

  • Chemistry - ,

    I didn't work it out but I suspect the problem is that you are trying to solve for 1/t2 from the get go. It is far simpler if you plug the numbers into the equation and solve that way.
    ln(P2/p1) = DHvap/R (1/T1 - 1/T2)
    ln(480/760) = (30720/8.314)(1/353.25 - 1/T2)
    First term = ln(480/760) = -0.4595
    30720/8.314 = 3694.97
    1/353.25 = 0.002831
    1/t2 = 1/t2
    -0.4595 = 3694.97(0.002831-1/T2)
    -0.4595 = (10.4599 - 3694.97/T2)

    Check my work. I'll let you finish.

  • Chemistry - ,

    I have -.4595= -3684.51/T2. Would you then just divide -3684.51/-.4595? and how did you het 10.4599 in the last equation?

  • Chemistry - ,

    No. I think your algebra is cockeyed.
    Assuming my work is ok to where I stopped, then multiply through by T2 to get rid of the T2 term in the denominator.
    -0.4595 = (10.4599 - 3694.97/T2)
    -0.4595T2 = 10.4599T2 - 3694.97
    -10.919T2 = -3694.97
    T2 = 338.4 or thereabouts which is approximately 65 C. Check my work but that sounds reasonable to me. Lower pressure means lower boiling point. I can see why 8,000 K didn't sound right to you.

  • Chemistry - ,

    Isn't the formula suppose to be ln(p2/p1)=Dvap/R(1/T2 -1/T1) ? That is what my textbook says.

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