Consider the following equilibrium: $\ce{A2\ _{(g)} + 2B_{(g)} <=> 2AB_{(g)}}$

A chemist puts 8.0 moles of $\ce{A2}$ and 10.0 moles of $\ce{B}$ into an evacuated 1.0 L container. The system reaches equilibrium at a certain temp according to the process outlined above. When the system is at equilibrium the chemist determines $\ce{[AB]}$ is 8.0 M. What is the $K_c$ for this process at this temperature (278 K). I set up a nice chart and everything and I cant figure this out. My answer choices are 10, 3.0, 1.0, 4.0, and .50

Initially I thought that The Kc was [AB]^2/([A2][B^2]) but I put in the values and didn't get any of the answers...

Then i set up an ice chart

It's hard to show on here but I then ended up with [8+2x]^2/([8-x][10-2x]^2) but that didn't really get me anywhere

  • $\begingroup$ It would probably be helpful to include the chart in your question so that those answering can point out where you were incorrect. $\endgroup$ – jonsca Jan 3 '14 at 22:41

Well, I haven't done that in a while, but we seem to agree that the equilibrium constant is expressed as

$$ K_c = \frac{[AB]^2}{[A_2][B]^2}$$

What else do we know?

  1. Initially, $[A_2]$ = 8.0M, $[B]$ = 10.0M

  2. In equilibrium, $[AB]$ = 8.0M

  3. For every mole of $B$, one mole of $AB$ is produced.
  4. For every mole of $A_2$, two moles of $AB$ are produced.

With just one cup of coffee so far, I get

$$ K_c = \frac{8^2}{(8 - \frac{8}{2})(10-8)^2} = 4$$

  • $\begingroup$ OK! I didn't know you had to subtract 8/2 so there's my problem. Thank you! $\endgroup$ – sloth1111 Jan 4 '14 at 16:36
  • 2
    $\begingroup$ You're welcome! If the formation of AB would consume A2, this wouldn't be chemistry but magic ;) $\endgroup$ – Klaus-Dieter Warzecha Jan 4 '14 at 17:21

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