Posted by **Donna** on Wednesday, May 28, 2008 at 12:23am.

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 -
**DrBob222**, Wednesday, May 28, 2008 at 12:59am
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 -
**PIERRE AMEDUITE**, Sunday, April 20, 2014 at 2:47pm
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

Ea=104Kj/mol

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

644-273=371

T2= 371¡ãC

Thank you. PIERRE

- Chemistry, Arrhenius law -
**PIERRE AMEDUITE**, Sunday, April 20, 2014 at 3:26pm
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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