## Lena772 one year ago Hydrogen reacts with iodine to form hydrogen iodide as follows: H2 (g) + I2 (g) ⇌ 2 HI (g) At 445 °C, Kc = 50.2. If one starts with 0.100 M H2 (g), 0.100 M I2 (g) and 0.0500 M HI (g) what is the equilibrium concentration of HI?

1. Lena772

@Photon336

2. Photon336

$Kc = \frac{ [HI]^{2} }{ [H_{2}][I_{2}] }$

3. Photon336

I would also calculate K to be sure of what direction our reaction is in. $\frac{ [0.05]^{2} }{ [0.10][0.10] } = 0.25 Q$ Q < K so the reaction is going to go to the right

4. Photon336

I guess we could do an ice table for this

5. Lena772

0.25

6. Photon336

|dw:1442107231301:dw|

7. Photon336

$\frac{ [0.05+2x] }{ [0.10-x][0.10-x] } = 50.2$

8. Photon336

sorry the numerator should be squared

9. Lena772

0.01-0.10x-0.10x+x 0.01-0.20x+x (0.0025+2x^2)/(0.01-0.20x+x) = 50.2

10. Photon336

$\frac{ [0.05+2x]^{2} }{ [0.1-x][0.1-x] } = 50.2$

11. Photon336

I think you can disregard the x in the denominator that's how I was taught it.

12. Photon336

$\frac{ (0.05+2x)^{2} }{ (0.01) } = 50.2$

13. Photon336

$(0.05+2x)^{2} = 0.502$

14. Abhisar

That's a very neat explanation photon c:

15. Photon336

Thanks @abhisar, @Lena772 had another question that I couldn't answer before, I was hoping that she would reopen it, to see what you thought about it

16. Abhisar

Sure, give me the link c:

17. Photon336

This was from wolfram alpha app lol. $(2x+0.05)^{2} = 4x ^{2}+0.2x+0.0025 = 0.502$ so wolfram alpha says the solution to this is x = 0.3296 $\frac{ [(0.05+2(0.3296)]^{2} }{ 0.01 }$ = 50.2

18. Photon336

Thanks to the wolfram alpha app this has made our job(s) so much easier When I put this answer back into the equation it gives me the Kc of 50.2

19. Photon336

So the equilibrium concentration will be $0.05+2(0.3296) = 0.7092$

20. Photon336

$\frac{ 0.502 }{ 0.01 } = 50.2$

21. Photon336

So with ALL that, the equilibrium concentration should be 0.7092 for HI don't know if they want any sig figures. I guessed that I assumed that the concentration for the reactants changed by a small amount, My math skills are okay but not great If the answer happens to be wrong I guess you would have to find a way to solve this for x I disregarded x in the denominator. |dw:1442108700281:dw| that would be set to 50.2

22. Lena772

0.05 +( 2 * 0.3296)=0.7902 HI

23. Lena772

OH YOU ALREADY DID THAT LOL THANK YOU SO MUCH @Photon336 :')

24. Lena772

was incorrect :c

25. Lena772

@Photon336

26. Photon336

It must be, that in our answer we had to account for the change in the reactants which I disregarded

27. Photon336

I think I see the problem, we could not disregard the change in x for the reactants because the concentrations were large enough so that they couldn't be disregarded. not a problem I enter this in on wolfram alpha $\frac{ [0.05+2x]^{2} }{ [(0.1-x)^{2}} = 50.2$ and I get two answers x = 0.0724827 x = 0.149162 Only one of these answers makes sense we have 0.10 moles of each reactant, so clearly 0.149162 cant be the answer otherwise we would have negative reactants. it must be x = 0.073

28. Photon336

@Lena772 my first assumption was to disregard the change -x in reactants but then I realized that I couldn't do this

29. Photon336

Before we go any further let's plug x = 0.073 into what I had set up above

30. Lena772

0.196

31. Lena772

0.05 + (2*0.073) = 0.196

32. Photon336

$0.196^{2} = 0.038416$ $(0.10-0.073)^{2} = 0.000729$ Now $\frac{ 0.038416 }{ 0.000729 } = 52.6~$

33. Photon336

Now my justification to put this all together

34. Lena772

So is the answer 0.000729? im confused

35. Photon336

$[HI] = 0.05M + 2(0.73) = 0.196M$ $I_{2} = 0.10M - 0.073 = 0.027 M$

36. Photon336

$H_{2} = 0.10M - 0.073 = 0.027 M$

37. Photon336

@Lena772 I plugged in the value 0.073 into this remember so plug in 0.073 into this and tell me what you get $\frac{ [2x+0.05]^{2} }{ [0.1-x]^{2} }$ I did this to see if the value was correct and it is because it equals our Kc roughly 50

38. Lena772

52.7

39. Lena772

So 0.196?

40. Photon336

I realized the last answer didn't make sense because we started with 0.10 for each. |dw:1442110682306:dw|

41. Photon336

@lena772 do they mark it wrong if you don't get sig figures right?

42. Lena772

not really, there's always a tolerance. that was correct. Thanks so much!

43. Photon336

OMG YES!!! |dw:1442111149801:dw|