Explain the reaction quotient and how it is used to predict the direction of a reaction.

(N/A) The equilibrium constants $K_{c}$ and $K_{p}$ help in predicting the direction in which a given reaction will proceed at any stage.
For this,we calculate the reaction quotient $Q$ ($Q_{c}$ for molar concentrations and $Q_{p}$ for partial pressures).
It is defined in the same way as $K_{c}$,except that the concentrations in $Q_{c}$ are not necessarily equilibrium values.
For a general reaction: $aA + bB \rightleftharpoons cC + dD$
$Q_{c} = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$
$(i)$ If $Q_{c} < K_{c}$,the reaction will proceed in the forward direction (towards products).
$(ii)$ If $Q_{c} > K_{c}$,the reaction will proceed in the reverse direction (towards reactants).
$(iii)$ If $Q_{c} = K_{c}$,the reaction mixture is at equilibrium.
Example: Consider the gaseous reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,where $K_{c} = 57.0$ at $700 \ K$. If the concentrations at an arbitrary time $t$ are $[H_{2}]_{t} = 0.1 \ M$,$[I_{2}]_{t} = 0.2 \ M$,and $[HI]_{t} = 0.40 \ M$,then:
$Q_{c} = \frac{[HI]^{2}}{[H_{2}][I_{2}]} = \frac{(0.40)^{2}}{(0.1)(0.2)} = \frac{0.16}{0.02} = 8.0$
Since $Q_{c} < K_{c}$,the reaction will proceed in the forward direction to form more $HI$ until $Q_{c} = K_{c}$.