effect on concentration if Volume decreases by 4?

[Dencology]

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If the rate law for a reaction is second order as so:


rate=k[A]

What happens to the rate when the VOLUME of the system is DECREASED by 4 times? (at constant temperature)


I figured that the Pressure of the system would INCREASE by 4 times.

But what happens to the rate:
I beleivce the rate would increase by the following:
1. temperature
2. reactant concentration. so if the volume decrease there would be more collision b/w the reactant molecules and the rxn rate would increase. according to Kaplan, "For rxns occorung in the gaseous state, the partial pressures of the reactants can serve as a measure of concentration pg. 846
4. catalyst
5. medium
-Does the rate increase by 4 times?
-Or does it increase by 16 times?

 

[Sea of ASH]

i think there would be no change. since the rate is only changed by the stuff you mentioned.
your thinking as if its a gas but did the question state that it is? plus if its a gas the ideal gas law allows us to assume that it occupies no signifacant volume.

 

[Dencology]

what about the rate constant? it changes by reactant only. right? i am sure of this. but double checking.

 

[Sea of ASH]

the only thing that k is dependent on is temp

 

[Dencology]

nobody can still answer this question,?
my answer is reactant. but i hear Temp and other stuff.anyone out there that can put this question to rest?

 

[EdCLS]

Here's how I'd approach it:

Rate=k[A]^1^1 for a second-order reaction. The rate constant, k, only changes with temperature. The only other thing the rate can be dependent on is reactant concentration. If we double the concentration of one reactant...

k[2A]^1^1=2(rate)

...the rate doubles. If we double the concentration of both reactants, the rate quadruples.

Now, let's say we have reactants A & B present in concentrations of 2M (2 mols reactant/1 liter) each.

k[2]^1[2]^1=4k(rate)

What happens if we reduce the volume of the container to one quarter of what it was originally?

2 mols reactant/0.25L=8M concentration. If we plug this into our rate equation:

k[8]^1[8}^1=64k(rate)

If we take our rate after changing the container size and divide by the initial rate:

64/4=16

So by quartering the volume, we'd increase the reaction rate by 16 times.

That's how I'd attack it. I hope this helps!