Lena772
  • Lena772
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?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Lena772
  • Lena772
@Photon336
Photon336
  • Photon336
\[Kc = \frac{ [HI]^{2} }{ [H_{2}][I_{2}] }\]
Photon336
  • 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

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Photon336
  • Photon336
I guess we could do an ice table for this
Lena772
  • Lena772
0.25
Photon336
  • Photon336
|dw:1442107231301:dw|
Photon336
  • Photon336
\[\frac{ [0.05+2x] }{ [0.10-x][0.10-x] } = 50.2\]
Photon336
  • Photon336
sorry the numerator should be squared
Lena772
  • Lena772
0.01-0.10x-0.10x+x 0.01-0.20x+x (0.0025+2x^2)/(0.01-0.20x+x) = 50.2
Photon336
  • Photon336
\[\frac{ [0.05+2x]^{2} }{ [0.1-x][0.1-x] } = 50.2\]
Photon336
  • Photon336
I think you can disregard the x in the denominator that's how I was taught it.
Photon336
  • Photon336
\[\frac{ (0.05+2x)^{2} }{ (0.01) } = 50.2\]
Photon336
  • Photon336
\[(0.05+2x)^{2} = 0.502\]
Abhisar
  • Abhisar
That's a very neat explanation photon c:
Photon336
  • 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
Abhisar
  • Abhisar
Sure, give me the link c:
Photon336
  • 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
Photon336
  • 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
Photon336
  • Photon336
So the equilibrium concentration will be \[0.05+2(0.3296) = 0.7092\]
Photon336
  • Photon336
\[\frac{ 0.502 }{ 0.01 } = 50.2\]
Photon336
  • 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
Lena772
  • Lena772
0.05 +( 2 * 0.3296)=0.7902 HI
Lena772
  • Lena772
OH YOU ALREADY DID THAT LOL THANK YOU SO MUCH @Photon336 :')
Lena772
  • Lena772
was incorrect :c
Lena772
  • Lena772
@Photon336
Photon336
  • Photon336
It must be, that in our answer we had to account for the change in the reactants which I disregarded
Photon336
  • 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
Photon336
  • Photon336
@Lena772 my first assumption was to disregard the change -x in reactants but then I realized that I couldn't do this
Photon336
  • Photon336
Before we go any further let's plug x = 0.073 into what I had set up above
Lena772
  • Lena772
0.196
Lena772
  • Lena772
0.05 + (2*0.073) = 0.196
Photon336
  • Photon336
\[0.196^{2} = 0.038416 \] \[(0.10-0.073)^{2} = 0.000729\] Now \[\frac{ 0.038416 }{ 0.000729 } = 52.6~ \]
Photon336
  • Photon336
Now my justification to put this all together
Lena772
  • Lena772
So is the answer 0.000729? im confused
Photon336
  • Photon336
\[[HI] = 0.05M + 2(0.73) = 0.196M\] \[I_{2} = 0.10M - 0.073 = 0.027 M \]
Photon336
  • Photon336
\[H_{2} = 0.10M - 0.073 = 0.027 M \]
Photon336
  • 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
Lena772
  • Lena772
52.7
Lena772
  • Lena772
So 0.196?
Photon336
  • Photon336
I realized the last answer didn't make sense because we started with 0.10 for each. |dw:1442110682306:dw|
Photon336
  • Photon336
@lena772 do they mark it wrong if you don't get sig figures right?
Lena772
  • Lena772
not really, there's always a tolerance. that was correct. Thanks so much!
Photon336
  • Photon336
OMG YES!!! |dw:1442111149801:dw|

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