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I am trying to solve a question that asks to compare Q vs. K. K = 4.2, and Q = 1.0.

Problem Details:

2.0 mol of Samples of A, B, and C

$\ce{A + B \rightleftharpoons 2C}$

I know that the reaction equilibrium is to the right, but, from this information, how can I tell if it is far to the right or slightly to the right?

Thanks.

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Try writing out the K expression. Remember that in general,

${K = \frac{[product]}{[reactant]}}$

In your case, ${K = \frac{[C]^2}{[A][B]}}$

Or in other words,

$K = \frac{[product]^2}{[reacant][reactant]} = \frac{[product]^2}{[reacant]^2}$

I combined the $[reactant]$ terms because the equation you have, there is 1:1 ratio of both reactants.

We can take the square root of both sides:

$\sqrt{K} = \sqrt\frac{[product]^2}{[reactant]^2} = \frac{[product]}{[reactant]}$

So now just plug in your numbers to determine how far to the right the equilibrium lies.

$\sqrt{K} = \sqrt{4.2} = \frac{[product]}{[reactant]} = 2.0$

Therefore (upon consideration of reaction stoichiometry), there is a 2:1:1 ratio of products to reactants at equilibrium.

We can also plug in the Q (reaction quotient) value; remember that K is just a special Q value (K is the Q value when the system is at equilibrium):

$\sqrt{Q} = \sqrt{1.0} = \frac{[product]}{[reactant]} = 1.0$

The Q value indicates that the system currently lies away from the equilibrium mix of a 2:1:1 ratio between products and reactant gases. At Q = 1.0 we have a 1:1:1 mix of A, B, and C.

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  • $\begingroup$ Thanks for your reply. I have written out the expressions, but am unable to determine if the rx will shift far to the right, or only slightly to the right. Is there any way to tell this from the data? $\endgroup$ – Jake Chasan May 15 '14 at 21:38
  • $\begingroup$ Yes. Sorry, I had not completed my answer. $\endgroup$ – Dissenter May 15 '14 at 21:40
  • $\begingroup$ So is 2:1 a large or small ratio? (meaning, will it be only slightly shifted, or shifted far?) $\endgroup$ – Jake Chasan May 15 '14 at 21:42
  • $\begingroup$ I would say large. 2 is after all twice one. $\endgroup$ – Dissenter May 15 '14 at 21:43
  • $\begingroup$ So why is taking the square root of both sides necessary, if you already have the Q and K values? $\endgroup$ – Jake Chasan May 15 '14 at 21:44

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