Equilibrium constant

The equilibrium constant of a chemical reaction describes the relationship between the concentrations of the reactants and the concentrations of the products of the reaction at dynamic equilibrium.

Chemists found that when a reversible reaction reaches dynamic equilibrium, the product of the concentrations (or partial pressures) of the reaction products raised to the power of their stoichiometric coefficients divided by the product of the concentrations (or partial pressures) of the reactants raised to the power of their stoichiometric coefficients, has a constant value at a particular temperature (see this advanced level article for derivation). For the decomposition of PCl₅ at 620K,

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

we have

$K_c=\frac{\left [ PCl_3 \right ]\left [ Cl_2 \right ]}{\left [ PCl_5 \right ]}$

where [i] is the concentration in mol dm^-3 of the species i.

We called this constant, K_c, the equilibrium constant. Since the species in this reaction are in the gaseous state and that the partial pressure of gas is proportional to its concentration, we can also express the equilibrium constant of this reaction in terms of partial pressures of the species:

$K_p=\frac{P_{PCl_3}P_{Cl_2}}{P_{PCl_5}}$

where p_i is the partial pressure of gas i and K_p is the equilibrium constant in terms of partial pressures. Note that K_c may or may not be equal to K_p for a particular reaction (see here for details).

Another example is the reaction:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

where

$K_c=\frac{\left [ NH_3 \right ]^2}{\left [ N_2 \right ]\left [ H_2 \right ]^3}\; \; \; or\; \; \; K_p=\frac{P_{NH_3}^{\; \; \; \; \; \; 2}}{P_{N_2}P_{H_2}^{\; \; \; \; 3}}$

In general, for a reaction

$mA+nB+...\rightleftharpoons pC+qD+...$

$K_c=\frac{\left [ C \right ]^p\left [ D \right ]^q...}{\left [ A \right ]^m\left [B \right ]^n...}\; \; \; or\; \; \; K_p=\frac{p_C^{ \; \; \; p}p_D^{ \; \; \;q}...}{p_A^{ \; \; \; m}p_B^{\; \; \; n}...}$

Question

Calculate the equilibrium constant for the dissociation of PCl₅ in a 250 ml vessel if the initial and equilibrium amounts of the reactant are 0.0175 mol and 0.0125 mol respectively (assuming no products formed initially).

Answer

	PCl₅	PCl₃	Cl₂
Initial conc, M	0.0175/0.250	0	0
Equilibrium conc, M	0.0125/0.250	(0.0175-0.0125)/0.250	(0.0175-0.0125)/0.250

$K_c=\frac{\frac{0.0050}{0.250}\frac{0.0050}{0.250}}{\frac{0.0125}{0.250}}=8.0\times 10^{-3}\: M$

Equilibrium constant

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