## Lena772 one year ago 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)?

1. Lena772

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

2. Lena772

Cs, graphite + CO2 (g) ⇌ 2 CO (g)

3. Lena772

@Abhisar

4. JFraser

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

5. Lena772

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

6. JFraser

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

7. 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

8. 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

9. JFraser

your initial value of CO in the flask is zero. You need to find the concentrations $$at$$ $$equiliubrium$$ $K_C = \frac{[CO]^2}{[CO_2]}$

10. Lena772

but kc is given

11. JFraser

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

12. JFraser

the starting concentration of $$CO_2$$ is 0.025M, correct?

13. Lena772

yes

14. Lena772

but 0^2/0.025 is 0...

15. JFraser

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

16. Lena772

yes

17. JFraser

the equilibrium concentration of $$CO$$ is not zero

18. JFraser

as the $$CO_2$$ is used, some CO is $$formed$$

19. 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

20. JFraser

so the equilibrium concentration of $$CO$$ is "2x"

21. Lena772

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

22. JFraser

$$CO$$ is the product $$formed$$, so it's amount is "2x"

23. JFraser

the equilibrium equation then looks like this

24. Lena772

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

25. JFraser

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

26. JFraser

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

27. Lena772

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

28. JFraser

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

29. 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

30. Lena772

0.01829034 M = CO at equilibrium?

31. 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...

32. Lena772

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

33. JFraser

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

34. Lena772

i graphed it.

35. JFraser

what did you plug in for the coefficients?

36. Lena772

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

37. JFraser

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

38. Lena772

are you good at lab reports?

39. Lena772

@JFraser

40. JFraser

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