JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Law Of Mass Action And Equilibrium Constant

Law Of Mass Action And Equilibrium Constant

Category : JEE Main & Advanced

On the basis of observations of many equilibrium reactions, two Norwegian chemists Goldberg and Waage suggested (1864) a quantitative relationship between the rates of reactions and the concentration of the reacting substances. This relationship is known as law of mass action. It states that

The rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants at a constant temperature at any given time.

The molar concentration i.e. number of moles per litre is also called active mass. It is expressed by enclosing the symbols of formulae of the substance in square brackets. For example, molar concentration of A is expressed as [A].

Consider a simple reversible reaction

$aA+bB$? $cC+dD$          (At a certain temperature)

According to law of mass action

Rate of forward reaction $\propto {{[A]}^{a}}{{[B]}^{b}}={{k}_{f}}{{[A]}^{a}}{{[B]}^{b}}$

Rate of backward reaction $\propto {{[C]}^{c}}{{[D]}^{d}}={{k}_{b}}{{[C]}^{c}}{{[D]}^{d}}$

At equilibrium,

Rate of forward reaction = Rate of backward reaction

${{k}_{f}}{{[A]}^{a}}{{[B]}^{b}}={{k}_{b}}{{[C]}^{c}}{{[D]}^{d}}$

$\frac{{{k}_{f}}}{{{k}_{b}}}={{K}_{c}}=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$

Where, ${{K}_{c}}$ is called equilibrium constant.

In terms of partial pressures, equilibrium constant is denoted by ${{K}_{p}}$ and

${{K}_{p}}=\frac{P_{C}^{c}\ P_{D}^{d}}{P_{A}^{a}\ P_{B}^{b}}$

In terms of  mole fraction, equilibrium constant is denoted by ${{K}_{x}}$ and

${{K}_{x}}=\frac{{{({{X}_{C}})}^{c}}\ {{({{X}_{D}})}^{d}}}{{{({{X}_{A}})}^{a}}\ {{({{X}_{B}})}^{b}}}$

Relation between Kp, Kc and Kx

${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$

${{K}_{p}}={{K}_{x}}{{(P)}^{\Delta n}}$

Dn = number of moles of gaseous products – number of moles of gaseous reactants in chemical equation.

As a general rule, the concentration of pure solids and pure liquids are not included when writing an equilibrium equation.

 Value of Dn Relation between Kp and Kc Units of Kp Units of Kc 0 Kp = Kc No unit No unit >0 Kp > Kc (atm)Dn (mole l–1)Dn <0 Kp < Kc (atm)Dn (mole l–1)Dn

Characteristics of equilibrium constant

(1) The value of equilibrium constant is independent of the original concentration of reactants.

(2) The equilibrium constant has a definite value for every reaction at a particular temperature. However, it varies with change in temperature.

(3) For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward reaction.

In general,  ${{K}_{\text{forward reaction}}}=\frac{1}{{{{{K}'}}_{\text{backward reaction}}}}$

(4) The value of an equilibrium constant tells the extent to which a reaction proceeds in the forward or reverse direction.

(5) The equilibrium constant is independent of the presence of catalyst.

(6) 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

(i) 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.

(ii) When DH = +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.

(iii) When DH = – 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.

(7) The value of the equilibrium constant depends upon the stoichiometry of the chemical equation.

For the reaction

$2S{{O}_{3}}(g)$ ? $2S{{O}_{2}}(g)+{{O}_{2}}(g)$ 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 ${K}'=\frac{[S{{O}_{2}}]{{[{{O}_{2}}]}^{1/2}}}{[S{{O}_{3}}]}$

${K}'=\sqrt{K}\text{ or }{{(K)}^{1/2}}$

(i) 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., ${K}'={{K}^{2}}$

(ii) 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 $K={{K}_{1}}\times {{K}_{2}}$

Applications of equilibrium constant

(1) Judging the extent of reaction

(i) If ${{K}_{c}}>{{10}^{3}}$, products predominate over reactants. If ${{K}_{c}}$ is very large, the reaction proceeds almost all the way to completion.

(ii) If ${{K}_{c}}<{{10}^{-3}}$, reactants predominate over products. If ${{K}_{c}}$ is very small, the reaction proceeds hardly at all.

(iii) If ${{K}_{c}}$ is in the range ${{10}^{-3}}$ to ${{10}^{3}}$, appreaciable concentration of both reactants and products are present.

(2) Predicting the direction of reaction : The concentration ratio, i.e., ratio of the product of concentrations of products to that of reactants is also known as concentration quotient and is denoted by Q. Concentration quotient, $Q=\frac{[X][Y]}{[A][B]}$.

It may be noted that Q becomes equal to equilibrium constant (K) when the reaction is at the equilibrium state. At equilibrium, $Q=K={{K}_{c}}={{K}_{p}}$. Thus,

(i) If Q > K, the reaction will proceed in the direction of reactants (reverse reaction).

(ii) If Q < K, the reaction will proceed in the direction of the products (forward reaction).

(iii) If Q = K, the reaction mixture is already at equilibrium. Thus, a reaction has a tendency to form products if Q < K and to form reactants if Q > K.

Other Topics

Notes - Law of mass action

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