First vs. Second Order

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Ashley Van Belle 2B
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First vs. Second Order

Postby Ashley Van Belle 2B » Wed Feb 08, 2017 10:44 pm

What is the difference between First Order and Second Order Reactions?

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Re: First vs. Second Order

Postby 204781248 » Wed Feb 08, 2017 10:49 pm

If a reaction is "first order," doubling the concentration of a reactant doubles the rate. If a reaction is "second order," doubling the concentration of the reactant increases the rate by 2^2, or four.

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Re: First vs. Second Order

Postby Armo_Derbarsegian_3K » Fri Feb 10, 2017 2:27 am

How many orders can you have? Does it keep going?

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Re: First vs. Second Order

Postby Aashi_Patel_3B » Thu Feb 16, 2017 4:15 pm

I believe the number of orders is based off the number of substance you have. This would mean a reaction with three chemicals is third order. Still, I am confused too. Please clarify...I found this on the internet

"order of reaction with respect to a given substance (such as reactant, catalyst or product) is defined as the index, or exponent, to which its concentration term in the rate equation is raised"

nikita bhat 2D
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Re: First vs. Second Order

Postby nikita bhat 2D » Thu Feb 16, 2017 8:29 pm

How do you know if it is zero order?

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Re: First vs. Second Order

Postby Mirian_Garcia_2G » Thu Feb 16, 2017 8:59 pm

Can someone explain the difference between first, second, and zero order? I'm having trouble understanding exactly what the difference is- how does the concentration and reaction rate differ in each? What's the relationship between the two?

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Re: First vs. Second Order

Postby Natalie_Boyd_1C » Fri Feb 17, 2017 2:52 am

We've talked about which direction a reaction will go based on the concentrations given (i.e. we've talked about le chatelier's principle), and so now we're exploring how the speed of the reaction changes by changing the concentrations of the reactants. A rate law mathematically describes how the concentrations of the reactants affects the rate of the reaction. The order is just a part of this mathematical description.
aA + bB --> cC + dD

with rate=k[A]^m[B]^n

1. 0 order

If m is 0, that is the order for A is 0, then changing the concentration of A doesn't change how fast the reaction occurs.
Maybe you have a really bad headache and you don't have a car and the hilltop is closed, so you decide to make aspirin in your dorm because you just happen to have some salicylic acid and acetic anhydride lying around (the ingredients, or reactants, that are used to make aspirin).

salicylic acid + acetic anhydride --> aspirin + acetic acid
A + B --> C + D

Well, if the salicylic acid has an order of 0, that means no matter how much salicylic acid you keep adding to the beaker, you won't make aspirin any faster. The rate of the reaction wouldn't depend on the concentration of salicylic acid.
if rate=k[A]^0[B]^n, anything to the 0th power is 1, so the concentration of [A] (or salicylic acid in this case) would drop out of the equation

2. 1st order

If m is 1, that means the rate changes in the same proportion that [A] does. If salicylic acid had an order of 1, then doubling its concentration would double the rate of the reaction. If you tripled it, you'd triple the rate. If you quadrupled it, you'd quadruple the rate. This is because anything raised to the 1st power is itself i.e. 4^1 = 4 5^1 = 5
so if we ignored everything else, and just said rate=[salicylic acid]^1 then if 1 mol/(L*s)=[1Msalicyclic acid]^1, if you added another M of salicylic acid you'd have [2M]^1=2mol/(L*s) (the rate doubled).

3. 2nd order

If m is 2, that means changing the concentration of [A] affects the rate in a "squared fashion". What I mean is if we double the amount of salicylic acid we have, we quadruple the rate. Let's say [1M salicylic acid]^2=1mol/(L*s). Then if [2M salicylic acid]^2=4mol/(L*s) or [3M salicylic acid]^2=9mol/(L*s)

4. You can have higher orders than this, but they're unlikely.

Martha Xuncax 3G
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Re: First vs. Second Order

Postby Martha Xuncax 3G » Sun Feb 19, 2017 9:23 pm

The course reader mentions that the probability of 3 molecules colliding simultaneously is unlikely, so termolecular reactions are less common (to the person asking how many reactions there were).

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