rate constant

[Oh_Gee]

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i thought rate constants don't change (since it's a constant) but according to this question it does. can someone explain this to me? I know Keq can change if temp changes but I thought that was only for Keq (and Ksp as well?)

 

[The Brown Knight]

Look up the Arrhenius Equation to understand what the rate constant depends on. I believe it goes something like this:
k = pze^(-Ea/RT),
where p is collision orientation, z is collision frequency, and Ea is activation energy.
Just from this you can see why increasing activation energy decreases k and increasing temperature increases k.
The rate constant is pretty much an empirically observed term for a given reaction with its temperature and associated catalyst. As long as the temperature and catalyst are the same, changing reactant amounts won't affect its value.

 

[Oh_Gee]

Look up the Arrhenius Equation to understand what the rate constant depends on. I believe it goes something like this:
k = pze^(-Ea/RT),
where p is collision orientation, z is collision frequency, and Ea is activation energy.
Just from this you can see why increasing activation energy decreases k and increasing temperature increases k.
The rate constant is pretty much an empirically observed term for a given reaction with its temperature and associated catalyst. As long as the temperature and catalyst are the same, changing reactant amounts won't affect its value.

oh wow totally forgot about this equation lol. on the MCAT, would they expect you remember this?

 

[Oh_Gee]

Look up the Arrhenius Equation to understand what the rate constant depends on. I believe it goes something like this:
k = pze^(-Ea/RT),
where p is collision orientation, z is collision frequency, and Ea is activation energy.
Just from this you can see why increasing activation energy decreases k and increasing temperature increases k.
The rate constant is pretty much an empirically observed term for a given reaction with its temperature and associated catalyst. As long as the temperature and catalyst are the same, changing reactant amounts won't affect its value.

how would a higher T affect this equation. i'm looking at a graph of e^-x but can't figure it out. intuitively though, i can see how a higher temp would make the reaction go faster which would mean k would go up. is that logical? i feel like it's flawed reasoning

 

[The Brown Knight]

You probably wouldn't need to know the eqn for the MCAT, but I guess it would help to understand it.
Rewrite the equation as k = pz/e^(Ea/RT)
So if T increases the term in the exponent will decrease; if the exponent term decreases the term in the denominator decreases as well and thereby increasing k.
Think of it as a double fraction: T is in the denominator of an exponential term that is itself in the denominator.

 

[Oh_Gee]

You probably wouldn't need to know the eqn for the MCAT, but I guess it would help to understand it.
Rewrite the equation as k = pz/e^(Ea/RT)
So if T increases the term in the exponent will decrease; if the exponent term decreases the term in the denominator decreases as well and thereby increasing k.
Think of it as a double fraction: T is in the denominator of an exponential term that is itself in the denominator.

oh wow flipping the e made sense. forgot you could do that lol

 

[Oh_Gee]

You probably wouldn't need to know the eqn for the MCAT, but I guess it would help to understand it.
Rewrite the equation as k = pz/e^(Ea/RT)
So if T increases the term in the exponent will decrease; if the exponent term decreases the term in the denominator decreases as well and thereby increasing k.
Think of it as a double fraction: T is in the denominator of an exponential term that is itself in the denominator.

statement III is wrong because otherwise how would you calculate k in the first place? right? also concentrations aren't in the arhenius equation

 

[The Brown Knight]

statement III is wrong because otherwise how would you calculate k in the first place? right? also concentrations aren't in the arhenius equation

Yeah, that's right

 

[NextStepTutor_1]

I don't think the MCAT will ask a question that necessitates that you have memorized that equation in order to answer correctly.
It's easy to remember that k (rate constant) is a constant and unable to change except when the temperature changes, much like K (the equilibrium constant).
And if you remember that association we still don't need to know the equation in order to see how the 2 variables (temperature and k) are related. It's intuitive to think that as the temperature increases the reaction proceeds at a faster rate whereas if the temperature decreases the reaction rate decreases.