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Homework Help: Chemistry: Activation Energy (Arrhenius equation)

  1. Feb 9, 2007 #1
    1. The problem statement, all variables and given/known data

    A certain reaction, with an activation energe of 64.5 kJ/mole, is run at 25.0'C (degree Celsius) and its rate is measured. How many times faster would this reaction be if it were run at 50.0'C (degree Celsius)?

    2. Relevant equations

    I think: (Arrhenius equation)
    ln(k1/k2) = Ea/R (1/T2 - 1/T1)

    R = 8.314 J/mole*K

    3. The attempt at a solution

    I don't know where to start. Only Ea = 64.5 kJ/mole, 1 temperature = 25.0'C and R (8.314 J/mole*K) are given. I don't know where to get the other numbers to plug in the equation.
  2. jcsd
  3. Feb 9, 2007 #2
    remember, you also have a T2 (50degC). You don't necessarily need to solve for the individual k values, just figure out a relationship between them.
  4. Feb 11, 2007 #3
    i just did this problem on my homework

    we have ln( k2/k1 ) = (Ea/R) * (1/T1 - 1/T2)

    where k is the reaction constant, Ea is activation energy, R is gas constant, T is temperature

    what are we solving for? well we want to see how k changes. we rearrange the equation:

    [antilog of both sides]
    k2/k1 = antilog[ (Ea/R) * (1/T1 - 1/T2) ]
    [multiple by k1 on each side]
    k2 = k1 * antilog[ (Ea/R) * (1/T1 - 1/T2) ]
    [antilog(x) just means e^(x)]
    k2 = k1 * e^[ (Ea/R) * (1/T1 - 1/T2) ]

    here we go! we wanted to see how k changes right? the new k (k2) value equals the old k (k1) value times e^[ (Ea/R) * (1/T1 - 1/T2) ]

    the reason it was tricky is because we weren't solving for a variable, we just wanted to see how k would change when the temperature changed

    so the answer is: when the temperature is 50C, it is ((( e^[ (Ea/R) * (1/T1 - 1/T2) ] ))) times faster than when it is at 25C

    just plug and chug:

    Ea = 64.5kj/mol = 64500 j/mol
    T1 = 25*C = 298K
    T2 = 50*C = 323K
    R = 8.314 J/mole*K
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