Question

What is the difference between $ Q $ and $ K $ in equilibrium?

Answer 1

Hint : $ Q $ is a quantity that changes as a reaction system approaches equilibrium, whereas $ K $ is the numerical value of $ Q $ at the end of the reaction, when equilibrium is reached. $ Q $ is a concept that is closely related to the equilibrium constant $ K $ .

Complete Step By Step Answer:
The difference between $ Q $ and $ K $ is that, $ K $ is the constant of certain reaction when it is in equilibrium, while $ Q $ is the quotient of activities of products and reactants at any stage of a reaction, therefore, by comparing $ Q $ and $ K $ , we can determine the direction of a reaction.
For a reversible reaction $ aA + bB \rightleftharpoons cC + dD $ , where $ a,b,c $ and $ d $ are the stoichiometric coefficients for the balanced reaction, $ $ We can calculate $ Q $ using the following reaction:
 $ Q = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{[A]}^a}{{[B]}^b}}} $ $ Q $
The value of can be found by raising the products to the power of their coefficients, or stoichiometric factors, divided by the reactants raised to their coefficients. The expression for $ Q $ is very similar to those for $ K $ .
When we set $ Q $ against $ K $ , there are five possible relationships:
$ \bullet Q = K $

$ \bullet Q = 0 $

 $ \bullet Q < K $

$ \bullet Q = \infty $

$ \bullet Q > K$ Situation $ 1 $ : $ Q = K $
When $ Q = K $ , the system is at equilibrium and there is no shift to either left or the right. Take, for example, the reversible reaction shown below:
 $ CO(g) + 2{H_2}(g) \rightleftharpoons C{H_3}OH(g) $
Situation $ 2 $ : $ Q < K $
When $ Q < K $ , there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.
Considered again:
 $ CO(g) + 2{H_2}(g) \rightleftharpoons C{H_3}OH(g) $
For $ Q < K $ :
 $ CO(g) + 2{H_2}(g) \to C{H_3}OH(g) $
So that equilibrium may be established.
 $ Q $ equals zero
If $ Q = 0 $ , then $ Q $ is less than $ K $ . Therefore, when $ Q = 0 $ , the reaction will shift to the right (forward).
 $ CO(g) + 2{H_2}(g) \to C{H_3}OH(g) $
Situation $ 3 $ : $ Q > K $
When $ Q > K $ , there are more products than reactants, to decrease the amount of products, the reaction will shift to the left and produce more reactants.
For $ Q > K $ :
 $ CO(g) + 2{H_2}(g) \leftarrow C{H_3}OH(g) $
 $ Q $ Equals infinity
When $ Q = \infty $ ,the reaction shifts to the left (backward). This is a variation of when $ Q > > > K $
 $ CO(g) + 2{H_2}(g) \leftarrow C{H_3}OH(g) $

Note :
Sometimes it is necessary to determine in which direction a reaction will progress based on initial activities or concentrations. In these situations, the relationship between the reaction quotient, $ {Q_c} $ and the equilibrium constant $ {K_c} $ is essential in solving for the net change.