Activation Energy

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

smurphy1D
Posts: 51
Joined: Mon Nov 18, 2019 12:18 am

Activation Energy

In the Arrhenious Equation, the Ea being on the top, if it is larger or in absence of a catalyst, would A be larger or smaller than one with a catalyst?

PranaviKolla2B
Posts: 114
Joined: Fri Aug 30, 2019 12:17 am

Re: Activation Energy

Also, can someone explain what each term means in the Arrhenius equation, please?

Vincent Leong 2B
Posts: 207
Joined: Fri Aug 09, 2019 12:15 am

Re: Activation Energy

Ea = activation energy (how much energy is necessary for the R-->P ). R = constant (used to cancel out units) T= temperature (temp at which the reaction happens). A (a pre exponential factor that is used to calculate k, also corresponds to the angular dependence that reactant molecules need to have the reaction start and form products) k = rate constant (some number that is used to cancel out units and determine the ultimate rate (M/s); I like to think of it similar to K (eq constant) because it doesn't change unless T changes)

Vincent Leong 2B
Posts: 207
Joined: Fri Aug 09, 2019 12:15 am

Re: Activation Energy

I'm not too sure about A but I THINK the concept sort of goes in hand with Temperature. If you increase temp, you're increasing the movement of molecules and if you increase the movement of molecules, your reactants are colliding more or have a higher change of colliding WITH THE CORRECT POSITIONAL/ANGULAR spot. With this in mind, A should increase as well? Again, we do not go that in depth with A jsut because it's more of a number to us. k, Ea, and T matter more to us relative to this class because we can determine so much more about a reaction and use it in everyday life. So I would not be too worried about understanding A except to know how to calculate it if you know the rest of the variables.

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