Jan 13, 2010
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What is true when comparing CH2F2 to CH2Br2 under identical conditions?

I. The diffusion rate of CH2F2 is greater than the diffusion rate of CH2Br2.

II. The collision impulse of CH2F2 is greater than the collision impulse of CH2Br2.

III. The average kinetic energy of CH2F2 is greater than the average kinetic energy of CH2Br2.

Why would II not work? Momentum and, therefore, impulse, is defined by the speed of the molecules, right? A smaller mass (CH2F2) would exhibit greater speed, which means that the impulse is greater…But CH2Br2 has a larger mass. How does the inverse relationship between mass and velocity come into play?
 

BerkReviewTeach

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Impulse is equal to the change in momentum (mv). For two gases at the same temperature we know the following:

1) They will have the same kinetic energy (1/2mv^2)
2) The lighter gas will have a higher average velocity.

We don't usually examine momentum so it's tricky. But if the two gases have the same kinetic energy, then we can derive from that which has the higher momentum:

---CH2F2 : CH2Br2
1/2mv^2 = 1/2mv^2
1/2v(mv) = 1/2v(mv)

For the rearranged equality to hold true, CH2F2 will have to have a smaller momentum (mv) since it has the higher velocity (v).

This is tough to explain so I hope this helps.
Excellent explanation chemnerd. This is one of those questions that always came up during office hours. You really have to see the math as you've explained it to get it.