## Arrhenius Equation for two temperatures (14.59)

Arrhenius Equation: $\ln k = - \frac{E_{a}}{RT} + \ln A$

Moderators: Chem_Mod, Chem_Admin

LeilaBushweller1F
Posts: 12
Joined: Fri Sep 26, 2014 2:02 pm

### Arrhenius Equation for two temperatures (14.59)

Hi. Could someone explain how we got the equation ln(k'/k)=(Ea/R)((1/T)-(1/T')) that is given in the solutions manual?

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Arrhenius Equation for two temperatures (14.59)

The derivation for that equation can be found on page 589 of the textbook. Essentially, you're just subtracting two Arrhenius equations for different T and k values (in the same way we were able to subtract one Van't Hoff equation from another). T and k correspond to one pair of values and T' and k' corresponds to another pair of values; since all variables except $E_{a}$ are either given or canceled out, you can use the resulting difference equation to solve for $E_{a}$.

Sonia Kumar 2A
Posts: 6
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Arrhenius Equation for two temperatures (14.59)

It is also one of the equations on the back of the lamented periodic table that was given with the course reader.

Chem_Mod
Posts: 19491
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 878 times

### Re: Arrhenius Equation for two temperatures (14.59)

The Arrhenius equation is k=Ae^(-E/RT) so lnK = lnA - E/RT

This is like a linear equation y=mx+b when written as lnK = (-E/R) (1/T) + lnA

Recall from basic algebra that a relation y=mx+b can be written in a "point-slope form" (y2-y1) = m(x2-x1), so doing this we instantly find

lnK2 - lnK1 = (-E/R) (1/T2 - 1/T1) or
ln(K2/K1) = (E/R) (1/T1 - 1/T2)

Return to “Arrhenius Equation, Activation Energies, Catalysts”

### Who is online

Users browsing this forum: No registered users and 1 guest