Tricky, tricky RATE LAW Chemical Kinetics Question

[Doc412]

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:
-Does the rate increase by 4 times?
-Or does it increase by 16 times?

I don't know the answer and was wondering if anyone else did.

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[LittleRocker]

I believe the answer is there's no change in rate. There will be a shift in equilibrium concentrations depending on the # of moles of gas on each side of the balanced equation. But if temp is constant, then there shouldnt be a change in rate. Anyone else have an idea?

Edit: yeah, rate depends on conc of reactants, catalyst, surface area of reactant/catalyst, and temp. not volume.

 

[Doc412]

On wikipedia, it says that rate varies with pressure. And it says that an increase in pressure is like an increase in concentration.

that is why i figured the rate does change.

http://en.wikipedia.org/wiki/Reaction_rate

 

[LittleRocker]

yeah youre right, i forgot to think of it in terms of moles/liter. Ok so if you increase the conc of each reactant by 4 times, then r should increase by 16 (4x4). please disregard my earlier statement

 

[bigserve99]

On wikipedia, it says that rate varies with pressure. And it says that an increase in pressure is like an increase in concentration.

that is why i figured the rate does change.

http://en.wikipedia.org/wiki/Reaction_rate

Wiki isn't giving you the full story. Just to note:

As already noted in this thread, the change in rate is due the change in concentration of the gases (or whatever A and B are, we'll call them gases..)

However, this change is not directly connected to total pressure as Wiki would "suggest" though did not explicitly make clear.

My point is that it is a concentration change.....independent of pressure change. For example, I could flood the chamber with an inert gas such as Argon thereby increasing the total pressure; however, since the partial pressures of the reactants have NOT changed, there is no shift in the equilibrium and thereby no increase in the forward reaction rate.

 

[Doc412]

good point

 

[Dencology]

The rate of the rxn depends on temp, catalyst, pH, medium. but the Rate Constant, K depends on the REACTANT.

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:
-Does the rate increase by 4 times?
-Or does it increase by 16 times?

I don't know the answer and was wondering if anyone else did.