Question

I have a rate law that states
rate=k[A][B]^2

How do I figure out what happens when [A] is tripled while [B] is doubled?

Answer 1

If you think about some equation:\[x=y\]If you double y, x will be doubled also. If you triple y, x will be tripled as well.

If you think about some equation:\[x=y^2\]If you double y, x will be multiplied by 4 (2^2), and if you triple y, x will be multiplied by 9 (3^2).

Does that help?

Answer 2

So if we apply this to the equation at hand:\[Rate=k[A][B]^2\]k is constant, so it doesn't affect how rate will change.\[Rate=3[A]*(2[B])^2=3[A]*4[B]^2=12[A][B]^2\]

Answer 3

Yes, that makes sense! So then if [A] is doubled and [B] is halved, then:

Rate= 2[A] * ((1/2)[B]\[^{2}\]

Answer 4

Whoa. Wonky formatting there...

Rate= 2[A] * ((1/2)[B]^2)

Answer 5

Yep. Which ends up simplifying to:\[Rate=2[A]*\frac{1}{4}[B]^2=\frac{1}{2}[A][B]^2\]The net effect is that the reaction rate is halved.

Answer 6

Excellent! Thank you!

Answer 7

Actually, one more question in regards to this, would you have to raise the 1/2 to the second power? I ask this because if [A] is constant and [B] is doubled, then the rate quadruples, doesn't it since [B] is second order. Mathematically wouldn't that be:

[A] * 2[B]^2 = 4

Answer 8

That's correct:\[[A]*(2[B])^2\]

Answer 9

And if B were halved:\[[A]*(\frac{1}{2}[B])^2=\frac{1^2}{2^2}=\frac{1}{4}\]

Answer 10

It's often a good idea just to throw in some numbers. If you are trying to draw general conclusions, of course they need to be true about any particular case.

So, for example, let k = 1, [A] = [B] = 1.0. What happens when [A] goes to 3.0 and [B] goes to 2.0?