Arrhenius Equation for two temperatures (14.59)

Arrhenius Equation:

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LeilaBushweller1F
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Arrhenius Equation for two temperatures (14.59)

Postby LeilaBushweller1F » Wed Feb 18, 2015 8:55 pm

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
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Re: Arrhenius Equation for two temperatures (14.59)

Postby Neil DSilva 1L » Wed Feb 18, 2015 9:48 pm

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 are either given or canceled out, you can use the resulting difference equation to solve for .

Sonia Kumar 2A
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Re: Arrhenius Equation for two temperatures (14.59)

Postby Sonia Kumar 2A » Thu Feb 19, 2015 12:08 pm

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

Chem_Mod
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Re: Arrhenius Equation for two temperatures (14.59)

Postby Chem_Mod » Thu Feb 19, 2015 1:28 pm

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)


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