Lena772
  • Lena772
At 1000 K, Kc = 2.11E-2. If one starts with 1.00 mole graphite and 1.00 mole CO2 (g) in a 40.0 L vessel, what is the equilibrium concentration of CO (g)?
Chemistry
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SOLVED
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katieb
  • katieb
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Lena772
  • Lena772
I caclulated ad got x=0.021644, but when I plugged that in to find the conc. of CO it was wrong.
Lena772
  • Lena772
Cs, graphite + CO2 (g) ⇌ 2 CO (g)
Lena772
  • Lena772
@Abhisar

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JFraser
  • JFraser
if you included the "concentration" of the solid graphite, you shouldn't
Lena772
  • Lena772
i did. do i sassume 2 moles of CO as well because it doesn't explicitly say that but its 1to1to2
JFraser
  • JFraser
the balanced reaction will create 2 moles of CO, you're correct. But you don't have 2 moles of CO at equilibrium
JFraser
  • JFraser
remember that in equilibrium, only \(some\) of the reactants combine to form \(some\) of the products Use an ICE table approach and set the unknown concentrations equal to the Kc value
Lena772
  • Lena772
I mean is the CO value for moles at initial 2? b/c then you would say 2mol/40.0L of soln to get the initial conc. of CO? Which is 0.05
JFraser
  • JFraser
your initial value of CO in the flask is zero. You need to find the concentrations \(at\) \(equiliubrium\) \[K_C = \frac{[CO]^2}{[CO_2]}\]
Lena772
  • Lena772
but kc is given
JFraser
  • JFraser
KC is given, but the concentrations are missing. that's what you're solving for
JFraser
  • JFraser
the starting concentration of \(CO_2\) is 0.025M, correct?
Lena772
  • Lena772
yes
Lena772
  • Lena772
but 0^2/0.025 is 0...
JFraser
  • JFraser
the \(equilibrium\) concentration of \(CO_2\) will be (0.025M - x)
Lena772
  • Lena772
yes
JFraser
  • JFraser
the equilibrium concentration of \(CO\) is not zero
JFraser
  • JFraser
as the \(CO_2\) is used, some CO is \(formed\)
JFraser
  • JFraser
for every "x" amount of \(CO_2\) that is used, the amount of \(CO\) that is formed is \(twice\) that amount, because of the balanced reaction
JFraser
  • JFraser
so the equilibrium concentration of \(CO\) is "2x"
Lena772
  • Lena772
wouldn't it be -2x cause Q
JFraser
  • JFraser
\(CO\) is the product \(formed\), so it's amount is "2x"
JFraser
  • JFraser
the equilibrium equation then looks like this
Lena772
  • Lena772
would it be 2.11e-2=(2x)^2/(0.025+x)?
JFraser
  • JFraser
\[K_C = \frac{[CO]^2}{[CO_2]} =\frac{[2x]^2}{[0.025-x]} = 0.0211\]
JFraser
  • JFraser
the \(CO_2\) is \(used \space up\) so its equilibrium concentration must be less than what you started with
Lena772
  • Lena772
I just thought when Q
JFraser
  • JFraser
it does, but that's not what this question is asking you for
JFraser
  • JFraser
it tells you that you start with graphite \(C(s)\) and \(CO_2\). there is no \(CO\) formed yet, so the reaction \(must\) proceed forward
Lena772
  • Lena772
0.01829034 M = CO at equilibrium?
Lena772
  • Lena772
No i didn't. I get x=-0.0144202, x=0.00914517. I multiplied the second x value by 2, and that's what I got...
Lena772
  • Lena772
Multiplying the first one gives a negative, and we can't have negative concentration
JFraser
  • JFraser
did you solve it quadratically, or take the lazy shortcut?
Lena772
  • Lena772
i graphed it.
JFraser
  • JFraser
what did you plug in for the coefficients?
Lena772
  • Lena772
2.11e-2=((2x)^2)/(0.025-x)
JFraser
  • JFraser
I don't normally solve them that way, but it's probably close enough. I got 0.0229M
Lena772
  • Lena772
are you good at lab reports?
Lena772
  • Lena772
@JFraser
JFraser
  • JFraser
usually, but what things you should put in it really depends on what your teacher has asked for

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