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$$\ce{CO (g) + Cl2(g)<=>COCl2}(g)$$ We have 2 mols of $\ce{CO}$, 1 Mol of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$ contained in 1 L. The system is currently in equilibrium.

We then decrease the volume by half to 0.5 L

What will be the new concentration in equilibrium of CO?

I tried calculating at first the constant of concentration;

$Kc=\frac{[\ce{CoCl2}]}{[\ce{CO}]\times[\ce{Cl2}]}$

I found it to be $4.8825$

then I calculated the concentration with the new Volume:

$[\ce{CO}]= \pu{2mol}/\pu{0.5L}= \pu{4 M}$

$[\ce{Cl2}]= \pu{1mol}/\pu{0.5L}= \pu{2 M}$

$[\ce{CoCl2}]= \pu{9.765mol}/\pu{0.5L}= \pu{19.53 M}$

obviously this wasn't in equilibrium so I went to do this:

$\ce{CO + Cl2<=>COCl2}$

$4M-x$, $2M-x$, $19.53+x$

(as the number of mols in the products is smaller and the system wants to fix the increase in pressure)

When I calculated all of this in the Kc formula it came to me that the Concentration of CO increased, and is now larger than 2 M , and in the textbook answers it is stated that it is supposed to be less than 2M, how is it possible?

I would be very glad for help with calculations and technique of solving as well!

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    $\begingroup$ What have you tried so far? Do you have thoughts on how to approach the problem? You should edit the question to include these so it doesn't get closed as homework. @RonAvraham $\endgroup$ – Tyberius Mar 11 at 15:00
  • $\begingroup$ I've edited the question to use some math and chemistry formatting, let me know if I messed anything up. @RonAvraham $\endgroup$ – Tyberius Mar 11 at 15:33
  • $\begingroup$ HINTS - Is the system a gas or an aqueous solution? How should the equilibrium change and why? $\endgroup$ – MaxW Mar 11 at 17:33
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    $\begingroup$ the system is gas, and the equilibrium is supposed to move to the products as they have less mols. $\endgroup$ – Ron Avraham Mar 11 at 17:47
  • $\begingroup$ After getting to values 4M−x, 2M−x, and 19.53+x, I assume you solved the quadratic equation? So long as x is a positive real number, then the equilibrium will shift to favor the product and [CO]<4 $\endgroup$ – MaxW Mar 11 at 19:29
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I solved the quadratic equation by Wolfram Alpha

enter image description here

The solution indicates that $\ce{[Cl2]}=\pu{1.26926M}<\pu{2M}$,but$[\ce{CO}]=\pu{1.26926M}<\pu{2M}$, if we have 1 mol of $\ce{CO}$, 2 Mols of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$ contained in 1 L.

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  • $\begingroup$ @MaxW I appreciate to provide me feedback about the answer $\endgroup$ – Adnan AL-Amleh Mar 12 at 14:54

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