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Chemistry, Arrhenius law

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The rate constant of a first order reaction is 4.60x10^-4/s at 350 degrees Celsius. If the activation energy is 104 kJ/mol, calculate the temp. at which its rate constant is 8.80 x 10^-4/s.

So I think that I can use this version of the equation:

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

Because I am looking for T2.
But now how do I go about solving for T2 when it's in two different places like that?

  • Chemistry, Arrhenius law - ,

    How about trying this? k1, k2, Ea, R are known, so lets just use them as constants and call T1 something like 300.
    ln K = k(300-X/300X)
    ?? = k(300-X/300X)
    300X*??= k(300-X)
    300X*?? = 300k - 300X
    300*??*X + 300X = 300*k
    Which is similar to
    400X+300X = 400
    700X = 400
    X = 400/700
    I know I have not made sense with the values but it shows you how to solve the problem.

  • Chemistry, Arrhenius law - ,

    Lnk1= LnA - Ea/R*T1
    Lnk2= LnA - Ea/R*T2
    lnK1- Lnk2 = LnA ¨CLnA +(- Ea/R*T1+ Ea/RT2)
    Ln(K1/K2)= Ea/R*T2 ¨C Ea/R*T1
    Ln(K1/K2)=(Ea/R)(1/T2-1/T1 )
    1/T2= (Ln(K1/K2))(R/Ea)+1/T1
    T2= (Ea*T1)/(Ln(K1/(K2 ))R*T1+Ea)
    T1= 350¡ãC=350+273=623K
    K1= 4.60*10-4 s-1
    K2= 8.80*10-4s-1
    R=8.314j/k.mol because of Ea, R=8.80*10-3kj/k.mol
    T2=(104*623)/(Ln((4.60*¡¼10¡½^(-4) s^(-1))/(8.80*¡¼10¡½^(-4)*s^(-1)))*8.314*¡¼10¡½^(-3)*623+104)
    T2=643.780K= 644k
    T2= 371¡ãC
    Thank you. PIERRE

  • Chemistry, Arrhenius law - ,

    There is some add of some messing letters from the system. Please, move them away to get the correct result. Thank you !!!

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