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Rate of reactions with eggs boiling

  1. Mar 4, 2014 #1
    At a high altitude camp in the Rockies, water boils at 95.4°C instead of 100.0°C. A visitor has requested a
    soft-boiled egg (usually boiled for 3.00 minutes at 100.0°C). The activation energy for the reaction in question is 453 kJ/mol.
    6 marks
    egg protein (l) egg protein (s) How long will it take to cook his egg at 95.4°C?

    The answers say 18.6min. However, they use this formula:

    ln(t2)=ln(t1)+(1/T2-1/T1)x(Ea/R)

    I believe the formula should be:
    ln(t2)=ln(t1)-(1/T2-1/T1)x(Ea/R)

    Am I right?
     
  2. jcsd
  3. Mar 4, 2014 #2
    The rates of the reaction at the two temperatures are:

    [tex]r_1=k\exp{\left(-\frac{E}{RT_1}\right)}[/tex]
    [tex]r_2=k\exp{\left(-\frac{E}{RT_2}\right)}[/tex]
    We need to have r1t1=r2t2

    So,[tex]k\exp{\left(-\frac{E}{RT_1}\right)}t_1=k\exp{\left(-\frac{E}{RT_2}\right)}t_2[/tex]
    or[tex]t_2=\exp{\left(\frac{E}{R}(\frac{1}{T_2}-\frac{1}{T_1})\right)}t_1[/tex]
    This confirms their formula.

    Chet
     
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