# Question #28ef9

Apr 24, 2017

$\left(a\right) < \left(c\right) < \left(b\right)$

#### Explanation:

The equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the products raised to the power of their respective stoichiometric coefficients and the product of the equilibrium concentrations of the reactants raised to the power of their respective stoichiometric coefficients.

This basically tells you that an equilibrium constant that is smaller than $1$ corresponds to an equilibrium reaction in which the reactants are favored.

Likewise, an equilibrium constant that is greater than $1$ corresponds to an equilibrium reaction in which the products are favored.

You can thus say that the higher the value of ${K}_{c}$, the closer the reaction will be to going to completion, i.e. the forward reaction will be favored.

${K}_{c} = 9.3 \cdot {10}^{- 13}$

${K}_{c} = 3.6 \cdot {10}^{20}$

${K}_{c} = 2.0 \cdot {10}^{9}$

So, you know that the higher the value of ${K}_{c}$, the closer the reaction will be to going to completion. You can thus say that, in order of the increasing tendency to go toward completion, you will have

$9.3 \cdot {10}^{- 13} \text{ "< " "2.0 * 10^9 " " < " } 3.6 \cdot {10}^{20}$

For

${K}_{c} = 9.3 \cdot {10}^{- 13}$

you will have more reactants than products at equilibrium, i.e. the reverse reaction will be favored.

For

${K}_{c} = 2.0 \cdot {10}^{9} \text{ " and " } {K}_{c} = 3.6 \cdot {10}^{20}$

you will have more products than reactants at equilibrium, i.e. the forward reactions will be favored.

The difference in magnitude between the two values implies that for ${K}_{c} = 3.6 \cdot {10}^{20}$, the reaction converts significantly more of the reactants to products than for ${K}_{c} = 2.0 \cdot {10}^{9}$.