Law of equilibrium and Equilibrium constant
Category : NEET
Law of Equilibrium and Equilibrium Constant
Equilibrium constant.
(1) Equilibrium constant in terms of law of mass action: The law of mass action may be applied to a reversible reaction to derive a mathematical expression for equilibrium constant known as law of chemical equilibrium.
Let us consider a simple reversible reaction, \[A+B\rightleftharpoons X+Y\] in which an equilibrium exists between the reactants (A and B) and the products (X and Y). The forward reaction is,
\[A+B\to X+Y\]
According to law of mass action,
Rate of forward reaction \[\propto [A][B]={{k}_{f}}[A][B]\]
Where \[{{k}_{f}}\] is the rate constant for the forward reaction and [A] and [B] are molar concentrations of reactants A and B respectively.
Similarly, the backward reaction is ; \[X+Y\to A+B\]
Rate of backward reaction \[\propto [X][Y]={{k}_{b}}[X][Y]\]
Where \[{{k}_{b}}\] is the rate constant for the backward reaction and [X] and [Y] are molar concentrations of products X and Y respectively.
At equilibrium, the rates of two opposing reactions become equal. Therefore, at equilibrium,
\[{{k}_{f}}[A][B]={{k}_{b}}[X][Y]\]
\[\frac{{{k}_{f}}}{{{k}_{b}}}=\frac{[X][Y]}{[A][B]}\,\,\,\,\,\text{or}\,\,\,\,\ K=\frac{[X][Y]}{[A][B]}\]
The combined constant K, which is equal to \[{{k}_{f}}/{{k}_{b}}\] is called equilibrium constant and has a constant value for a reaction at a given temperature. The above equation is known as law of chemical equilibrium.
For a general reaction of the type : \[aA+bB\rightleftharpoons cC+dD\]
The equilibrium constant may be represented as : \[K=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\]
where the exponents a, b, c and d have the same values as those in the balanced chemical equation. Thus, the equilibrium constant may be defined as,
“The ratio between the products of molar concentrations of the products to that of the molar concentrations of the reactants with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation at a constant temperature.”
(2) Characteristics of equilibrium constant
(i) The value of equilibrium constant is independent of the original concentration of reactants.
For example, the equilibrium constant for the reaction,
\[F{{e}^{3+}}(aq)+SC{{N}^{-}}(aq)=FeSC{{N}^{2+}}(aq)\]; \[K=\frac{[FeSC{{N}^{2+}}]}{[F{{e}^{3+}}][SC{{N}^{-}}]}=138.0\,L\,mo{{l}^{-1}}\] (at 298 K)
Whatever may be the initial concentrations of the reactants, \[F{{e}^{3+}}\] and \[SC{{N}^{-}}\] ions, the value of K comes out to be \[138.0\text{ }L\text{ }mo{{l}^{1}}\] at 298 K.
(ii) The equilibrium constant has a definite value for every reaction at a particular temperature. However, it varies with change in temperature.
For example, the equilibrium constant for the reaction between hydrogen and iodine to form hydrogen iodide is 48 at 717 K.
\[{{H}_{2}}(g)+{{I}_{2}}(g)=2HI(g)\]; \[K=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=48\]
For this reaction, the value of K is fixed as long as the temperature remains constant.
(iii) For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward reaction.
For example, if equilibrium constant, K, for the reaction of combination between hydrogen and iodine at 717 K is 48
\[{{H}_{2}}(g)+{{I}_{2}}(g)~\rightleftharpoons 2HI\left( g \right);\] \[{K}'=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=48\]
Then, the equilibrium constant for the decomposition of hydrogen iodide is the inverse of the above equilibrium constant.
\[2HI\left( g \right)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g);\] \[K=\frac{[{{H}_{2}}][{{I}_{2}}]}{{{[HI]}^{2}}}=\frac{1}{{{K}'}}=\frac{1}{48}=0.02\]
In general, \[{{K}_{\text{forward reaction}}}=\frac{1}{{{{{K}'}}_{\text{backward reaction}}}}\]
(iv) The value of an equilibrium constant tells the extent to which a reaction proceeds in the forward or reverse direction. If value of K is large, the reaction proceeds to a greater extent in the forward direction and if it is small, the reverse reaction proceeds to a large extent and the progress in the forward direction is small.
(v) The equilibrium constant is independent of the presence of catalyst. This is so because the catalyst affects the rates of forward and backward reactions equally.
(vi) The value of equilibrium constant changes with the change of temperature. Thermodynamically, it can be shown that if \[{{K}_{1}}\] and \[{{K}_{2}}\] be the equilibrium constants of a reaction at absolute temperatures \[{{T}_{1}}\] and \[{{T}_{2}}\]. If DH is the heat of reaction at constant volume, then
\[\log {{K}_{2}}-\log {{K}_{1}}=\frac{-\Delta H}{2.303\,R}\left[ \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right]\] (Van’t Hoff equation)
The effect of temperature can be studied in the following three cases
(a) When \[\Delta H=0\] i.e., neither heat is evolved nor absorbed
\[\log {{K}_{2}}-\log {{K}_{1}}=0\] or \[\log {{K}_{2}}=\log {{K}_{1}}\] or \[{{K}_{2}}={{K}_{1}}\]
Thus, equilibrium constant remains the same at all temperatures.
(b) When \[\Delta H=+ve\] i.e., heat is absorbed, the reaction is endothermic. The temperature \[{{T}_{2}}\] is higher than \[{{T}_{1}}\].
\[\log {{K}_{2}}-\log {{K}_{1}}=+ve\] or \[\log {{K}_{2}}>\log {{K}_{1}}\] or \[{{K}_{2}}>{{K}_{1}}\]
The value of equilibrium constant is higher at higher temperature in case of endothermic reactions.
(c) When \[\Delta H=-ve\] i.e., heat is evolved, the reaction is exothermic. The temperature \[{{T}_{2}}\] is higher than \[{{T}_{1}}\].
\[\log {{K}_{2}}-\log {{K}_{1}}=-ve\] or \[\log {{K}_{1}}>\log {{K}_{2}}\] or \[{{K}_{1}}>{{K}_{2}}\]
The value of equilibrium constant is lower at higher temperature in the case of exothermic reactions.
(vii) The value of the equilibrium constant depends upon the stoichiometry of the chemical equation.
Examples:
(a) If the equation (having equilibrium constant K) is divided by 2, then the equilibrium constant for the new equation is the square root of K i.e. \[\sqrt{K}\]. For example, the thermal dissociation of \[S{{O}_{3}}\] can be represented in two ways as follows,
\[2S{{O}_{3}}(g)\] ? \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\] and \[S{{O}_{3}}(g)\] ? \[S{{O}_{2}}(g)+1/2{{O}_{2}}(g)\]
\[K=\frac{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}{{{[S{{O}_{3}}]}^{2}}}\] and \[{K}'=\frac{[S{{O}_{2}}]{{[{{O}_{2}}]}^{1/2}}}{[S{{O}_{3}}]}\]; \[{K}'=\sqrt{K}\text{ or }{{(K)}^{1/2}}\]
(b) Similarly, if a particular equation is multiplied by 2, the equilibrium constant for the new reaction (K¢) will be the square of the equilibrium constant (K) for the original reaction i.e., \[{K}'={{K}^{2}}\]
(c) If the chemical equation for a particular reaction is written in two steps having equilibrium constants \[{{K}_{1}}\] and \[{{K}_{2}}\], then the equilibrium constants are related as \[K={{K}_{1}}\times {{K}_{2}}\]
For example, the reaction \[{{N}_{2}}(g)+2{{O}_{2}}(g)\] ? \[2N{{O}_{2}}(g)\] with equilibrium constant (K) can be written in two steps :
\[{{N}_{2}}(g)+{{O}_{2}}(g)\] ? 2NO (g) ; (Equilibrium constant = \[{{K}_{1}}\])
\[2NO(g)+{{O}_{2}}(g)\] ? \[2N{{O}_{2}}(g)\] ; (Equilibrium constant = \[{{K}_{2}}\])
Now, \[{{K}_{1}}=\frac{{{[NO]}^{2}}}{[{{N}_{2}}][{{O}_{2}}]}\] and \[{{K}_{2}}=\frac{{{[N{{O}_{2}}]}^{2}}}{{{[NO]}^{2}}[{{O}_{2}}]}\]
Therefore, \[{{K}_{1}}\times {{K}_{2}}=\frac{{{[NO]}^{2}}}{[{{N}_{2}}][{{O}_{2}}]}\times \frac{{{[N{{O}_{2}}]}^{2}}}{{{[NO]}^{2}}[{{O}_{2}}]}=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}]{{[{{O}_{2}}]}^{2}}}=K\]
(3) Types of equilibrium constant: Generally two types of equilibrium constants are used,
(i) \[{{\mathbf{K}}_{\mathbf{c}}}\to \] It is used when the various species are generally expressed in terms of moles/litre or in terms of molar concentrations.
(ii) \[{{\mathbf{K}}_{p}}\to \] It is used when in gaseous reactions, the concentration of gases expressed in terms of their partial pressures.
\[{{\mathbf{K}}_{p}}\] is not always equal to \[{{\mathbf{K}}_{\mathbf{c}}}.{{K}_{p}}\]and \[{{K}_{c}}\]are related by the following expression, \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\]
where, R = Gas constant = 0.0831 \[bar\,d{{m}^{3}}\,mo{{l}^{-1}}\,{{k}^{-1}}\]; T = Temperature in Kelvin
\[\Delta n\]= number of moles of gaseous products – number of moles of gaseous reactants in chemical equation
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