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Radagast

Equilibrium Problem!

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I got a question in which I am having considerable difficulty with:

 

The following reaction takes place in a 1.00L vessel at 500C.

 

2HI(g) <---> H2(g) + I2(g)

 

Equilibrium concentrations were found to be 1.76 moles/L HI, 0.20 moles/L of H2, and 0.20 moles/L of I2. If an additional 0.5 moles of hydrogen iodide gas is introduced, what are the concentrations of all gases once equilibrium has again been reached?

 

I have tried it several times, but my Chemistry teacher said that my answers were wrong. It's due for Monday (Nov.19), so if anyone can answer it before then it would be greatly appreciated :).

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I got a question in which I am having considerable difficulty with:

 

The following reaction takes place in a 1.00L vessel at 500C.

 

2HI(g) <---> H2(g) + I2(g)

 

Equilibrium concentrations were found to be 1.76 moles/L HI, 0.20 moles/L of H2, and 0.20 moles/L of I2. If an additional 0.5 moles of hydrogen iodide gas is introduced, what are the concentrations of all gases once equilibrium has again been reached?

 

I have tried it several times, but my Chemistry teacher said that my answers were wrong. It's due for Monday (Nov.19), so if anyone can answer it before then it would be greatly appreciated :).

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"It's due for Monday (Nov.19)"

 

You don't happen to mean Monday Nov 22nd? :) Since 19th was Friday, the magnificent day when I finished Half-Life 2. :>

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"It's due for Monday (Nov.19)"

 

You don't happen to mean Monday Nov 22nd? :) Since 19th was Friday, the magnificent day when I finished Half-Life 2. :>

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Okay. I think I can help you here. The first thing you need to do is calculate the Keq for the reaction. This is done via the following equation: Keq = (([product1]^n)([product2]^n))/([reactant]^n). So for your problem, you would have:

 

 

([i2][H2])/([HI]^2) = Keq.

 

Now that you have the equillibrium constant, you can just add in the extra .5 mol/L of HI and calculate the new concentrations with some minimal algebra. Once you've come up with a solution, post it here and we'll let you know if you're right. :D

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Okay. I think I can help you here. The first thing you need to do is calculate the Keq for the reaction. This is done via the following equation: Keq = (([product1]^n)([product2]^n))/([reactant]^n). So for your problem, you would have:

 

 

([i2][H2])/([HI]^2) = Keq.

 

Now that you have the equillibrium constant, you can just add in the extra .5 mol/L of HI and calculate the new concentrations with some minimal algebra. Once you've come up with a solution, post it here and we'll let you know if you're right. :D

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Alright, I'll try it out. :)

 

Keq = [i2] x [H2] / [HI]^2

= [0.20] x [0.20] / [1.76]^2

= [0.04] / [3.0976]

= 0.0129

 

Now for the adding of the new concentration, that would make it:

Keq = [i2] x [H2] / [HI]^2

.0129 = [0.20+x] x [0.20 + x] / [(1.76+0.50)-2x]^2

(Taking the perfect square of that..)

0.1136 = (0.20+x) / (2.26-2x)

0.1136 (2.26-2x) = (0.20+x)

0.257 - 0.227x = 0.20 + x

0.057 = 1.227x

x = 0.0465

 

Therefore:

[i2] = 0.20 + .0465 = 0.2465 mole/L

[H2] = 0.20 + .0465 = 0.2465 mole/L

[HI] = 2.26 - 2(.0465) = 2.167 mole/L

 

 

Sound right? :confused:

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Alright, I'll try it out. :)

 

Keq = [i2] x [H2] / [HI]^2

= [0.20] x [0.20] / [1.76]^2

= [0.04] / [3.0976]

= 0.0129

 

Now for the adding of the new concentration, that would make it:

Keq = [i2] x [H2] / [HI]^2

.0129 = [0.20+x] x [0.20 + x] / [(1.76+0.50)-2x]^2

(Taking the perfect square of that..)

0.1136 = (0.20+x) / (2.26-2x)

0.1136 (2.26-2x) = (0.20+x)

0.257 - 0.227x = 0.20 + x

0.057 = 1.227x

x = 0.0465

 

Therefore:

[i2] = 0.20 + .0465 = 0.2465 mole/L

[H2] = 0.20 + .0465 = 0.2465 mole/L

[HI] = 2.26 - 2(.0465) = 2.167 mole/L

 

 

Sound right? :confused:

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That sounds pretty good to me. I'll double check on that once I get a chance, but your method sounds proper. :D

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That sounds pretty good to me. I'll double check on that once I get a chance, but your method sounds proper. :D

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What on Earth is "Keq" ? Maybe a) I'm just tired b) I don't know enough English chemistry terms. :|

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What on Earth is "Keq" ? Maybe a) I'm just tired b) I don't know enough English chemistry terms. :|

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What on Earth is "Keq" ? Maybe a) I'm just tired b) I don't know enough English chemistry terms. :|

 

Equilibrium Constant.

 

It's a dimensionless number that relates the concentrations in an equilbrium system and the value of the constant changes with the temperature of the system.

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What on Earth is "Keq" ? Maybe a) I'm just tired b) I don't know enough English chemistry terms. :|

 

Equilibrium Constant.

 

It's a dimensionless number that relates the concentrations in an equilbrium system and the value of the constant changes with the temperature of the system.

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Alright, I just conferred with a friend, and now were both confused lol.

 

Where I had this line in my calculations:

Keq = [i2] x [H2] / [HI]^2

.0129 = [0.20+x] X [0.20 + x] / [(1.76+0.50)-2(0+x)]^2

 

She had:

.0129 = [0.20+x] X [0.20+x] / [(1.176+0.50)-2(.20+x)^2]

 

Now we are both confused and both unsure of our methods. We don't know if once the 0.5 moles of HI is introduced, whether or not the product side will start initially at 0 or start at 0.20.

 

Any help sorting this one out?

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Alright, I just conferred with a friend, and now were both confused lol.

 

Where I had this line in my calculations:

Keq = [i2] x [H2] / [HI]^2

.0129 = [0.20+x] X [0.20 + x] / [(1.76+0.50)-2(0+x)]^2

 

She had:

.0129 = [0.20+x] X [0.20+x] / [(1.176+0.50)-2(.20+x)^2]

 

Now we are both confused and both unsure of our methods. We don't know if once the 0.5 moles of HI is introduced, whether or not the product side will start initially at 0 or start at 0.20.

 

Any help sorting this one out?

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Okay. To start off, we have the Keq for the reaction which is 0.0129. We also have the concentrations of the gasses at equillibrium which will be our starting concentrations. [H2]=0.20, [i2]=0.20, [HI]=(1.76+0.50)=2.26. The initial concentration of HI is 2.26 and those of the hydrogen and iodine are 0.20 since the extra 0.5 moles is added to the system once it has reached equillibrium. Now you can set up an 'I.C.E.' chart which shows the initial concentration, the change in concentration, and the equillibrium concentration.

 

2HI <=> H2 + I2

 

2.26 0.20 0.20 (Initial Concentration)

-2x +x +x (Change in Concentration)

2.26-2x 0.20+x 0.20+x (Equillibrium Concentration)

 

So you plug the Equillibrium concentrations into your Keq equation and solve for x. So you have done it correctly and she has done it incorrectly. :D

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Okay. To start off, we have the Keq for the reaction which is 0.0129. We also have the concentrations of the gasses at equillibrium which will be our starting concentrations. [H2]=0.20, [i2]=0.20, [HI]=(1.76+0.50)=2.26. The initial concentration of HI is 2.26 and those of the hydrogen and iodine are 0.20 since the extra 0.5 moles is added to the system once it has reached equillibrium. Now you can set up an 'I.C.E.' chart which shows the initial concentration, the change in concentration, and the equillibrium concentration.

 

2HI <=> H2 + I2

 

2.26 0.20 0.20 (Initial Concentration)

-2x +x +x (Change in Concentration)

2.26-2x 0.20+x 0.20+x (Equillibrium Concentration)

 

So you plug the Equillibrium concentrations into your Keq equation and solve for x. So you have done it correctly and she has done it incorrectly. :D

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Can't trust females to do a Chemistry problem eh;)

 

Thanks a whole lot though. My gratitude is unlimited.

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Can't trust females to do a Chemistry problem eh;)

 

Thanks a whole lot though. My gratitude is unlimited.

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Can't trust females to do a Chemistry problem eh;)

 

Thanks a whole lot though. My gratitude is unlimited.

 

 

Heh. No problem. I have no problem helping out as long as some effort at completing the problem is shown. I was very happy to see that you were able to figure out with only a little hint. (Too often on chemistry message boards, people post their homework here without ever showing any attempt at solving it themselves. You'll never learn if you go about it that way. But you asked a legitimate question, you showed your work after a little help was given, and you came up with the right answer. THAT is what should go on). :D

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Can't trust females to do a Chemistry problem eh;)

 

Thanks a whole lot though. My gratitude is unlimited.

 

 

Heh. No problem. I have no problem helping out as long as some effort at completing the problem is shown. I was very happy to see that you were able to figure out with only a little hint. (Too often on chemistry message boards, people post their homework here without ever showing any attempt at solving it themselves. You'll never learn if you go about it that way. But you asked a legitimate question, you showed your work after a little help was given, and you came up with the right answer. THAT is what should go on). :D

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I absolutely agree. A similiar question could easily be asked on my upcoming test, and now that I understand the method and have completed the work on my own, I am much more prepared.

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