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Chemistry: Activation Energy (Arrhenius equation)

by lonelyassassin
Tags: activation, arrhenius, chemistry, energy, equation
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lonelyassassin
#1
Feb9-07, 02:57 PM
P: 3
1. The problem statement, all variables and given/known data

A certain reaction, with an activation energe of 64.5 kJ/mole, is run at 25.0'C (degree Celsius) and its rate is measured. How many times faster would this reaction be if it were run at 50.0'C (degree Celsius)?


2. Relevant equations

I think: (Arrhenius equation)
ln(k1/k2) = Ea/R (1/T2 - 1/T1)

R = 8.314 J/mole*K


3. The attempt at a solution

I don't know where to start. Only Ea = 64.5 kJ/mole, 1 temperature = 25.0'C and R (8.314 J/mole*K) are given. I don't know where to get the other numbers to plug in the equation.
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dmoravec
#2
Feb9-07, 03:04 PM
P: 147
remember, you also have a T2 (50degC). You don't necessarily need to solve for the individual k values, just figure out a relationship between them.
qwedsa
#3
Feb11-07, 01:36 PM
P: 16
i just did this problem on my homework

we have ln( k2/k1 ) = (Ea/R) * (1/T1 - 1/T2)

where k is the reaction constant, Ea is activation energy, R is gas constant, T is temperature

what are we solving for? well we want to see how k changes. we rearrange the equation:

[antilog of both sides]
k2/k1 = antilog[ (Ea/R) * (1/T1 - 1/T2) ]
[multiple by k1 on each side]
k2 = k1 * antilog[ (Ea/R) * (1/T1 - 1/T2) ]
[antilog(x) just means e^(x)]
k2 = k1 * e^[ (Ea/R) * (1/T1 - 1/T2) ]

here we go! we wanted to see how k changes right? the new k (k2) value equals the old k (k1) value times e^[ (Ea/R) * (1/T1 - 1/T2) ]

the reason it was tricky is because we weren't solving for a variable, we just wanted to see how k would change when the temperature changed

so the answer is: when the temperature is 50C, it is ((( e^[ (Ea/R) * (1/T1 - 1/T2) ] ))) times faster than when it is at 25C

just plug and chug:

Ea = 64.5kj/mol = 64500 j/mol
T1 = 25*C = 298K
T2 = 50*C = 323K
R = 8.314 J/mole*K


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