Posted by Donna on .
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 * (T1T2/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 
DrBob222,
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(300X/300X)
?? = k(300X/300X)
300X*??= k(300X)
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 
PIERRE AMEDUITE,
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/T21/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*104 s1
K2= 8.80*104s1
Ea=104Kj/mol
R=8.314j/k.mol because of Ea, R=8.80*103kj/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
644273=371
T2= 371¡ãC
Thank you. PIERRE 
Chemistry, Arrhenius law 
PIERRE AMEDUITE,
There is some add of some messing letters from the system. Please, move them away to get the correct result. Thank you !!!