A community for students.
Here's the question you clicked on:
 0 viewing
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?
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?

This Question is Closed

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[Kc = \frac{ [HI]^{2} }{ [H_{2}][I_{2}] }\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3I 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

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3I guess we could do an ice table for this

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3dw:1442107231301:dw

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{ [0.05+2x] }{ [0.10x][0.10x] } = 50.2\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3sorry the numerator should be squared

Lena772
 one year ago
Best ResponseYou've already chosen the best response.00.010.10x0.10x+x 0.010.20x+x (0.0025+2x^2)/(0.010.20x+x) = 50.2

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{ [0.05+2x]^{2} }{ [0.1x][0.1x] } = 50.2\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3I think you can disregard the x in the denominator that's how I was taught it.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{ (0.05+2x)^{2} }{ (0.01) } = 50.2\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[(0.05+2x)^{2} = 0.502\]

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0That's a very neat explanation photon c:

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3Thanks @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

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Sure, give me the link c:

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3This 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

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3Thanks 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

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3So the equilibrium concentration will be \[0.05+2(0.3296) = 0.7092\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{ 0.502 }{ 0.01 } = 50.2\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3So 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

Lena772
 one year ago
Best ResponseYou've already chosen the best response.00.05 +( 2 * 0.3296)=0.7902 HI

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0OH YOU ALREADY DID THAT LOL THANK YOU SO MUCH @Photon336 :')

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3It must be, that in our answer we had to account for the change in the reactants which I disregarded

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3I 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 \[\frac{ [0.05+2x]^{2} }{ [(0.1x)^{2}} = 50.2\] 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

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3@Lena772 my first assumption was to disregard the change x in reactants but then I realized that I couldn't do this

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3Before we go any further let's plug x = 0.073 into what I had set up above

Lena772
 one year ago
Best ResponseYou've already chosen the best response.00.05 + (2*0.073) = 0.196

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[0.196^{2} = 0.038416 \] \[(0.100.073)^{2} = 0.000729\] Now \[\frac{ 0.038416 }{ 0.000729 } = 52.6~ \]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3Now my justification to put this all together

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0So is the answer 0.000729? im confused

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[[HI] = 0.05M + 2(0.73) = 0.196M\] \[I_{2} = 0.10M  0.073 = 0.027 M \]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3\[H_{2} = 0.10M  0.073 = 0.027 M \]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3@Lena772 I plugged in the value 0.073 into this remember so plug in 0.073 into this and tell me what you get \[\frac{ [2x+0.05]^{2} }{ [0.1x]^{2} }\] I did this to see if the value was correct and it is because it equals our Kc roughly 50

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3I realized the last answer didn't make sense because we started with 0.10 for each. dw:1442110682306:dw

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3@lena772 do they mark it wrong if you don't get sig figures right?

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0not really, there's always a tolerance. that was correct. Thanks so much!

Photon336
 one year ago
Best ResponseYou've already chosen the best response.3OMG YES!!! dw:1442111149801:dw
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.