chemistry question

[sshaltoni]

ok, about the arrhenius equation: i read that the reaction rate is directly proportional to the activation energy, so if i infer correctly, a high activation energy corresponds to a high reaction rate??

this seems a bit counterintuitive to moi, por que, IF a reaction had a high Ea, that would need quite a bit of input. doesn't the fact that having a large energy barrier to begin with preclude the likelihood of its occurence, thus making its reaction rate hard to attain and rare? ok, now, somehow i think that this doesn't matter anymore. regardless of the chances of the reaction proceeding due to its high barrier, its rate will be high, nonetheless??

i may have answered my question by that musing, but it could be better phrased. another breakdown would be appreciated. anyone??

---

Members don't see this ad.

 

[liverotcod]

Arrhenius Equation:
k=A*exp(-Ea/R*T)
So the rate is inversely proportional to the activation energy, not directly. Because of the negative sign in front of Ea.

 

[RayhanS1282]

From my understanding, the arh. equation (I did't memorize it) is used to describe rxn. rate, which decreases with increasing activation energy and low temperature. The rxn. rate can be increased by applying the opposite of the perviously mentioned conditions.

k = T (temp.)
k = 1/Ea (activation energy)

 

[BisMuOxo]

Don't confuse the rate with the rate constant. The term "k" is the rate constant, not the rate. The arrhenius equation describes how the rate CONSTANT varies with temperature or Ea. The rate CONSTANT is not porportional to Ea or inversely proportional to Ea, there is an exponential involved so a direct proportion is out of the question.

The most accurate way to describe the situation is that if Ea increases, the rate CONSTANT decreases. If Ea decreases, the rate constant increases.

Most often, the arrhenius equation is used to describe how a rate constant varies with temperature since there is no way to change Ea for a reaction.

 

[sshaltoni]

...and the rate constant increases with increasing temp?? and decreases with decreasing temp?

 

[ASDIC]

rate constant is proportional to the temperature and inversely proportional to the activation energy

rate law (rate of the reaction) is proportional to the temperature, proportional to the concentration of reactants and inversely proportional to the activation energy

 

[Assembler]

OK, here goes:

Arrhenius Equation: k=Ae^(-Ea/RT)
A is the Arrhenius factor, Ea is the activation energy, R is the gas constant, T is temperature (in Kelvins). k is not the reaction rate, but it is the reaction constant.

Ex. aA + bB -> cC + Dd. The reaction rate R = k[A]exp(x)exp👍 (also called rate law expression). The reaction constant "k," and the orders of the reaction (x,y), including the total order of the reaction (x+y) can only be experimentally determined. Reaction rate is proportional to reaction constant, but remember, it is still a constant (it will never change for a given reaction).

Arrhenius Eqn: ln(k)=ln(A)*ln[e^(-Ea/RT)]
This becomes ln(k)=ln(A)+(-Ea/RT), which then becomes ln(k)=ln(A)-Ea/RT

So, reaction constant "k" increases when the term (-Ea/RT) is small. This happens only when activation energy is small (ie. catalyst is used), or when temperature is high.

Therefore, increase in "k" is inversely proportional to activation energy, and directly proportional to temperature.

 

[sshaltoni]

ok, so now that we've established all sorts of direct and inverse proportionalities of the reaction rate and its constant... say you've got a second order reaction rate, with 2 reactant concentrations k=[x][y] (this was a question i vaguely remember from april's mcat) ...would doubling [x] double the overall reaction rate?? thanks again.

 

[ifailedmcat]

ok, so now that we've established all sorts of direct and inverse proportionalities of the reaction rate and its constant... say you've got a second order reaction rate, with 2 reactant concentrations k=[x][y] (this was a question i vaguely remember from april's mcat) ...would doubling [x] double the overall reaction rate?? thanks again.

Yes the reaction rate would double.

 

[Assembler]

if R=k[x][y]. we know it's second order, so the exponents are 1 and 1, representatively, and anything to the power of one is the same number itself

So, R=kxy, goes to R=k(2x)^1👍^1=2kxy

Then, yes, it would double.

 

[BisMuOxo]

rate constant is proportional to the temperature and inversely proportional to the activation energy

You guys really need to stop using the word "proportional". It is absolutely not correct in this case. The rate constant cannot be proportional to temperature or inversely proportional to Ea. There is an exponential involved. A proportion would be, for example, if the rate constant doubled when you double the temperature..... The equation k=AExp(-Ea/RT) does not permit this sort of relationship.

All you can say is that k increases with increasing temperature and decreases with increasing Ea.