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.
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.
