# Reaction Rate and Pressure

absolutezer0es
The question is:

The rate equation for the reaction A2 + B2 -> 2C (all gases) is rate = k[A2]2[B2]. If the gaseous reaction misture is compressed to half its original volume, by what factor will the reaction rate change. Assume temperature is constant.

My thoughts are that the reaction rate wouldn't change. I know the rate changes if the concentrations of any of the reactants do, which is quite simple to calculate, but I'm not totally sure on pressure's effect on reaction rate. Any pushes in the right direction would be great.

Mentor
Imagine having 1 mole of an ideal gas squeezed into 1 L volume. What is its concentration?

Apply twice the pressure. What is the volume now? What is the concentration now?

absolutezer0es
Ah, I see. The concentration of both would be 2M, so substituting would give me:

rate = 8k[A2]2][B2]

Thanks for the help!

Homework Helper
Gold Member
Ah, I see. The concentration of both would be 2M, so substituting would give me:

rate = 8k[A2]2][B2]

Thanks for the help!

No! The rate has increased by a factor of 8 but it's still k[A2]2][B2] , just each of the things inside the brackets has doubled.

absolutezer0es
What do you mean by "it's"? The reaction rate doesn't increase by a factor of 8?

Homework Helper
Gold Member
What do you mean by "it's"? The reaction rate doesn't increase by a factor of 8?

Yes I said the reaction rate increases 8X. "It's" means "it is".

Consider [A2], [B2] to represent variables, k a constant.

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Mentor
In other words, rate is in both cases k[A2]2[B2], but rate'=8*rate.