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.