# Question #22fd7

Dec 15, 2015

True.

#### Explanation:

For a given equilibrium reaction, the equilibrium constant tells you what the ratio between the equilibrium concentrations of the products and the equilibrium concentrations of the reactants, all raised to the power of their respective stoichiometric coefficients.

For a general equilibrium reaction

$a \text{A" + b"B" rightleftharpoons c"C" + d"D}$

the equilibrium constant takes the form

$\textcolor{b l u e}{{K}_{c} = \left({\left[\text{C"]^c * ["D"]^d)/(["A"]^a * ["B}\right]}^{b}\right)}$

The important thing to keep in mind here is that all those concentrations are equilibrium concentrations.

Now, the reaction quotient, ${Q}_{c}$ takes the same form as the equilibrium constant, with the important distinction that the concentrations of the chemical species used in its expression are not equilibrium concentrations.

You can think of the reaction quotient as being a snapshot of how the equilibrium reaction proceeds at a given moment in time, let's say $t$.

$\textcolor{b l u e}{{Q}_{c} = \left({\left[\text{C"]_t^c * ["D"]_t^d)/(["A"]_t^a * ["B}\right]}_{t}^{b}\right)}$

The concentrations used here are specific to the products and reactants at a moment $t$.

When ${Q}_{c} = {K}_{c}$, the system is at equilibrium and no net reaction takes place, i.e. the forward and reverse reactions occur at the same rate.

Now, when ${Q}_{c} > {K}_{c}$, the ratio between products and reactants is too high in favor of the products.

This means that in order for equilibrium to be established, the reverse reaction must be favored. Simply put, the equilibrium will shift to the left, favoring the conversion of products into reactants.

Similarly, when ${Q}_{c} < {K}_{c}$, the equilibrium will shift to the right, favoring the conversion of reactants to products.

So, for a given equilibrium reaction, the reaction quotient will tell in which direction that equilibrium will proceed from the moment you calculated ${Q}_{c}$.