How does pressure affect the rate of a gaseous reaction?

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In the discussion about the reaction A2 + B2 -> 2C, participants analyze how compressing the gaseous reaction mixture to half its original volume affects the reaction rate. The rate equation is given as rate = k[A2]²[B2]. When the volume is halved, the concentration of both reactants doubles, leading to an increase in the reaction rate. Specifically, the new rate becomes rate' = 8k[A2]²[B2], indicating that the reaction rate increases by a factor of 8 due to the squared dependence on the concentration of A2. The conversation clarifies that while the form of the rate equation remains the same, the increased concentrations significantly boost the overall reaction rate.
absolutezer0es
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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.
 
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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?
 
Ah, I see. The concentration of both would be 2M, so substituting would give me:

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

Thanks for the help!
 
absolutezer0es said:
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.
 
What do you mean by "it's"? The reaction rate doesn't increase by a factor of 8?
 
absolutezer0es said:
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". :oldsmile:

Consider [A2], [B2] to represent variables, k a constant.
 
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In other words, rate is in both cases k[A2]2[B2], but rate'=8*rate.
 
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