Chemistry, Arrhenius law
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 * (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?
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*?? = 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.
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
T2= (Ea*T1)/(Ln(K1/(K2 ))R*T1+Ea)
K1= 4.60*10-4 s-1
R=8.314j/k.mol because of Ea, R=8.80*10-3kj/k.mol
Thank you. PIERRE
There is some add of some messing letters from the system. Please, move them away to get the correct result. Thank you !!!