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Lena772
 one year ago
At 1000 K, Kc = 2.11E2. 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
 one year ago
At 1000 K, Kc = 2.11E2. 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)?

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Lena772
 one year ago
Best ResponseYou've already chosen the best response.0I caclulated ad got x=0.021644, but when I plugged that in to find the conc. of CO it was wrong.

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0Cs, graphite + CO2 (g) ⇌ 2 CO (g)

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2if you included the "concentration" of the solid graphite, you shouldn't

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0i did. do i sassume 2 moles of CO as well because it doesn't explicitly say that but its 1to1to2

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the balanced reaction will create 2 moles of CO, you're correct. But you don't have 2 moles of CO at equilibrium

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2remember 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2your initial value of CO in the flask is zero. You need to find the concentrations \(at\) \(equiliubrium\) \[K_C = \frac{[CO]^2}{[CO_2]}\]

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2KC is given, but the concentrations are missing. that's what you're solving for

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the starting concentration of \(CO_2\) is 0.025M, correct?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the \(equilibrium\) concentration of \(CO_2\) will be (0.025M  x)

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the equilibrium concentration of \(CO\) is not zero

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2as the \(CO_2\) is used, some CO is \(formed\)

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2so the equilibrium concentration of \(CO\) is "2x"

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0wouldn't it be 2x cause Q<Kc

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2\(CO\) is the product \(formed\), so it's amount is "2x"

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the equilibrium equation then looks like this

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0would it be 2.11e2=(2x)^2/(0.025+x)?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2\[K_C = \frac{[CO]^2}{[CO_2]} =\frac{[2x]^2}{[0.025x]} = 0.0211\]

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the \(CO_2\) is \(used \space up\) so its equilibrium concentration must be less than what you started with

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0I just thought when Q<Kc the equilibrium shifted left :/

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2it does, but that's not what this question is asking you for

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2it 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
 one year ago
Best ResponseYou've already chosen the best response.00.01829034 M = CO at equilibrium?

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0No 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
 one year ago
Best ResponseYou've already chosen the best response.0Multiplying the first one gives a negative, and we can't have negative concentration

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2did you solve it quadratically, or take the lazy shortcut?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2what did you plug in for the coefficients?

Lena772
 one year ago
Best ResponseYou've already chosen the best response.02.11e2=((2x)^2)/(0.025x)

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2I don't normally solve them that way, but it's probably close enough. I got 0.0229M

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0are you good at lab reports?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2usually, but what things you should put in it really depends on what your teacher has asked for
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