How does concentration effect dynamic equilibrium?

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How does concentration effect dynamic equilibrium?

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Answer 1 · 2015-06-18T15:41:33

The Law of Mass Action states that when a system (e.g., chemical reaction) is at equilibrium, the ratio of products and reactant concentrations is equal to the equilibrium constant for that reaction. If we change the concentration of any species (e.g., by adding reactant to the mixture), then the reaction will shift toward reactants or products in order to restore the ratio of concentrations to the equilibrium value.

Explanation:

Example:
When $N O_{2}$ $NO_2$ is in equilibrium with its dimer, $N_{2} O_{4}$ $N_2O_4$ , the ratio of concentrations (expressed in mol/L) is equal to 4.7 at room temperature.

$2 N O_{2} \leftrightarrow N_{2} O_{4}$ $2 NO_2 harr N_2O_4$

$K_{c} = \frac{[N_{2} O_{4}]}{{[N O_{2}]}_{2}^{2}} = 4.7$ $K_c=([N_2O_4])/([NO_2]^2)=4.7$

So, if the concentration of $N O_{2}$ $NO_2$ is 0.02M, then we know that the equilibrium concentration of $N_{2} O_{4}$ $N_2O_4$ must be $4.7 {(0.02)}^{2} = 1.88 \times 10^{- 3} M$ $4.7(0.02)^2=1.88times10^-3M$

If the system is in a 1L container at constant volume and temperature and we add 0.03 mol of $N O_{2}$ $NO_2$ , then for a short time the concentration of $N O_{2}$ $NO_2$ is increased to 0.05M. The non-equilibrium ratio of concentrations is
$Q = \frac{[N_{2} O_{4}]}{{[N O_{2}]}_{2}^{2}} = \frac{1.88 \times 10^{- 3}}{{(0.05)}^{2}} = 0.75$ $Q=([N_2O_4])/([NO_2]^2)=(1.88times10^-3)/((0.05)^2)=0.75$
which is smaller than the equilibrium ratio of 4.7. In order to restore equilibrium, the reaction shifts toward products, consuming $N O_{2}$ $NO_2$ and producing $N_{2} O_{4}$ $N_2O_4$ until the equilibrium rations are restored:

$K_{c} = \frac{[N_{2} O_{4}]}{{[N O_{2}]}_{2}^{2}} = \frac{0.00188 + x}{{(0.05 - 2 x)}^{2}} = 4.7$ $K_c=([N_2O_4])/([NO_2]^2)=(0.00188+x)/((0.05-2x)^2)=4.7$

Solving this quadratic equation for the change in concentration, $x$ $x$ , gives $x = 5.37 \times 10^{- 3} M$ $x=5.37times10^-3M$ . The new equilibrium concentrations will be
$[N_{2} O_{4}] = 0.00188 + 0.00537 = 0.00725 M$ $[N_2O_4]=0.00188+0.00537=0.00725M$
$[N O_{2}] = 0.05 - 2 (0.00537) = 0.0392 M$ $[NO_2]=0.05-2(0.00537)=0.0392M$
and the new ratio is
$\frac{[N_{2} O_{4}]}{{[N O_{2}]}_{2}^{2}} = 4.72$ $([N_2O_4])/([NO_2]^2)=4.72$ , which is equal to the equilibrium constant within the rounding error of the significant digits of the calculation.

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How does concentration effect dynamic equilibrium?

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