Question

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)?

Answer 1

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

Answer 2

Cs, graphite + CO2 (g) ⇌ 2 CO (g)

Answer 3

@Abhisar

Answer 4

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

Answer 5

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

Answer 6

the balanced reaction will create 2 moles of CO, you're correct. But you don't have 2 moles of CO at equilibrium

Answer 7

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

Answer 8

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

Answer 9

your initial value of CO in the flask is zero. You need to find the concentrations \(at\) \(equiliubrium\)

\[K_C = \frac{[CO]^2}{[CO_2]}\]

Answer 10

but kc is given

Answer 11

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

Answer 12

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

Answer 13

yes

Answer 14

but 0^2/0.025 is 0...

Answer 15

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

Answer 16

yes

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

wouldn't it be -2x cause Q<Kc

Answer 22

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

Answer 23

the equilibrium equation then looks like this

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

I just thought when Q<Kc the equilibrium shifted left :/

Answer 28

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

Answer 29

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

Answer 30

0.01829034 M = CO at equilibrium?

Answer 31

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...

Answer 32

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

Answer 33

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

Answer 34

i graphed it.

Answer 35

what did you plug in for the coefficients?

Answer 36

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

Answer 37

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

Answer 38

are you good at lab reports?

Answer 39

@JFraser

Answer 40

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