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)?

- Lena772

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- Lena772

I caclulated ad got x=0.021644, but when I plugged that in to find the conc. of CO it was wrong.

- Lena772

Cs, graphite + CO2 (g) ⇌ 2 CO (g)

- Lena772

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## More answers

- JFraser

if you included the "concentration" of the solid graphite, you shouldn't

- Lena772

i did. do i sassume 2 moles of CO as well because it doesn't explicitly say that but its 1to1to2

- 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

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

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

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

but kc is given

- JFraser

KC is given, but the concentrations are missing. that's what you're solving for

- JFraser

the starting concentration of \(CO_2\) is 0.025M, correct?

- Lena772

yes

- Lena772

but 0^2/0.025 is 0...

- JFraser

the \(equilibrium\) concentration of \(CO_2\) will be (0.025M - x)

- Lena772

yes

- JFraser

the equilibrium concentration of \(CO\) is not zero

- JFraser

as the \(CO_2\) is used, some CO is \(formed\)

- 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

so the equilibrium concentration of \(CO\) is "2x"

- Lena772

wouldn't it be -2x cause Q

- JFraser

\(CO\) is the product \(formed\), so it's amount is "2x"

- JFraser

the equilibrium equation then looks like this

- Lena772

would it be 2.11e-2=(2x)^2/(0.025+x)?

- JFraser

\[K_C = \frac{[CO]^2}{[CO_2]} =\frac{[2x]^2}{[0.025-x]} = 0.0211\]

- JFraser

the \(CO_2\) is \(used \space up\) so its equilibrium concentration must be less than what you started with

- Lena772

I just thought when Q

- JFraser

it does, but that's not what this question is asking you for

- 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

0.01829034 M = CO at equilibrium?

- 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

Multiplying the first one gives a negative, and we can't have negative concentration

- JFraser

did you solve it quadratically, or take the lazy shortcut?

- Lena772

i graphed it.

- JFraser

what did you plug in for the coefficients?

- Lena772

2.11e-2=((2x)^2)/(0.025-x)

- JFraser

I don't normally solve them that way, but it's probably close enough. I got 0.0229M

- Lena772

are you good at lab reports?

- Lena772

- JFraser

usually, but what things you should put in it really depends on what your teacher has asked for

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