Law of Chemical Equilibrium and Equilibrium Consant

LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

What factors can be exploited to alter the composition of an equilibrium mixture? The last question in particular is important when choosing conditions for synthesis of industrial chemicals such as H₂, NH₃, CaO etc.

To answer these questions, let us consider a general reversible reaction:

A+BC+D

where A and B are the reactants, C and D are the products in the balanced chemical equation. On the basis of experimental studies of many reversible reactions, the Norwegian chemists Cato Maximillian Guldberg and Peter Waage proposed in 1864 that the concentrations in an equilibrium mixture are related by the following equilibrium equation,

\({K_c} = \,\frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}}\)(7.1) where K_c is the equilibrium constant and the expression on the right side is called the equilibrium constant expression.

The equilibrium equation is also known as the law of mass action because in the early days of chemistry, concentration was called “active mass”. In order to appreciate their work better, let us consider reaction between gaseous H₂ and I₂ carried out in a sealed vessel

at 731K.

H₂(g)+I₂(g)2HI(g)

1 mol    1 mol      2 mol

Experiment Analysis

Six sets of experiments with varying initial conditions were performed, starting with only gaseous H₂ and I₂ in a sealed reaction vessel in first four experiments (1, 2, 3 and 4) and only HI in other two experiments (5 and 6). Experiment 1, 2, 3 and 4 were performed taking different concentrations of H₂ and / or I₂, and with time it was observed that intensity of the purple colour remained constant and equilibrium was attained. Similarly, for experiments 5 and 6, the equilibrium was attained from the opposite direction.Data obtained from all six sets of experiments are given in Table 7.2.

Table 7.2 Initial and Equilibrium Concentrations of H₂, I₂ and HI

It is evident from the experiments 1, 2, 3 and 4 that number of moles of dihydrogen reacted = number of moles of iodine reacted =

½ (number of moles of HI formed). Also,experiments 5 and 6 indicate that,

[H₂(g)]_eq = [I₂(g)]_eq

Knowing the above facts, in order to establish a relationship between concentrations of the reactants and products, several combinations can be tried. Let us consider the simple expression,

[HI(g)]_eq / [H₂(g)]_eq [I₂(g)]_eq

It can be seen from Table 7.3 that if we put the equilibrium concentrations of the reactants and products, the above expression

Table 7.3 Expression Involving the Equilibrium Concentration of Reactants H₂(g)+I₂(g)2HI(g) is far from constant. However, if we consider the expression,[HI(g)]_eq / [H₂(g)]_eq [I₂(g)]_eq

we find that this expression gives constant value (as shown in Table 7.3) in all the six cases. It can be seen that in this expression the power of the concentration for reactants and products are actually the stoichiometric coefficients in the equation for the chemical reaction. Thus, for the reaction H₂(g) +I₂(g)2HI(g), following equation 7.1, the equilibrium constant K_c is written as,

\({K_c} = \,\frac{{\left[ {HI\left( g \right)} \right]_{eq}^2}}{{{{\left[ {{H_2}(g)} \right]}_{eq}}{{\left[ {{I_2}(g)} \right]}_{eq}}}}\)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBa % aaleaacaWGJbaabeaakiabg2da9maadmaabaGaamisaiaadMeacaGG % OaGaam4zaiaacMcaaiaawUfacaGLDbaadaqhaaWcbaGaamyzaiaadg % haaeaacaaIYaaaaOGaai4lamaadmaabaGaamisamaaBaaaleaacaaI % YaaabeaakiaacIcacaWGNbGaaiykaaGaay5waiaaw2faamaaBaaale % aacaWGLbGaamyCaaqabaGcdaWadaqaaiaadMeadaWgaaWcbaGaaGOm % aaqabaGccaGGOaGaam4zaiaacMcaaiaawUfacaGLDbaadaWgaaWcba % Gaamyzaiaadghaaeqaaaaa!5251! {K_c} = \left[ {HI(g)} \right]_{eq}^2/{\left[ {{H_2}(g)} \right]_{eq}}{\left[ {{I_2}(g)} \right]_{eq}}\)(7.2)

Generally the subscript ‘eq’ (used for equilibrium) is omitted from the concentration terms. It is taken for granted that the concentrations in the expression for K_c are equilibrium values. We, therefore, write,

K_c = [HI(g)]² /[H₂ (g)] [I₂ (g)](7.3)

The subscript ‘c’ indicates that K_c is expressed in concentrations of mol L^–1.

At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.

The equilibrium constant for a general reaction,

a A + b Bc C + d D is expressed as,

K_c= [C]^c[D]^d / [A]^a[B]^b     (7.4)

where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products.

Equilibrium constant for the reaction,

4NH₃(g) + 5O₂(g)4NO(g) + 6H₂O(g) is written as

K_c = [NO]⁴[H₂ O]⁶ / [N₃H ]⁴ [O₂ ]⁵

Molar concentration of different species is indicated by enclosing these in square bracket and, as mentioned above, it is implied that these are equilibrium concentrations. While writing expression for equilibrium constant, symbol for phases (s, l, g) are generally ignored.

Solved Examples

PROBLEM 1

The following concentrations were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium at 500K.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaaca % WGobWaaSbaaSqaaiaaikdaaeqaaaGccaGLBbGaayzxaaGaeyypa0Ja % aGymaiaac6cacaaI1aGaey41aqRaaGymaiaaicdadaahaaWcbeqaai % abgkHiTiaaikdaaaGccaWGnbGaaiOlamaadmaabaGaamisamaaBaaa % leaacaaIYaaabeaaaOGaay5waiaaw2faaiabg2da9iaaiodacaGGUa % GaaGimaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI % YaaaaOGaamytaaaa!50ED! \left[ {{N_2}} \right] = 1.5 \times {10^{ - 2}}M.\left[ {{H_2}} \right] = 3.0 \times {10^{ - 2}}M\) and \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaaca % WGobGaamisamaaBaaaleaacaaIZaaabeaaaOGaay5waiaaw2faaiab % g2da9iaaigdacaGGUaGaaGOmaiabgEna0kaaigdacaaIWaWaaWbaaS % qabeaacqGHsislcaaIYaaaaOGaamytaiaac6caaaa!449A! \left[ {N{H_3}} \right] = 1.2 \times {10^{ - 2}}M.\) Calculate equilibrium constant.

SOLUTION

The equilibrium constant for the reaction,

N₂(g) + 3H₂(g)   2NH₃(g) can be written as,

\({K_c} = \,\frac{{{{\left[ {N{H_3}(g)} \right]}^2}}}{{\left[ {{N_2}(g)} \right]{{\left[ {{H_2}(g)} \right]}^3}}}\)

\({K_c} = \,\frac{{{{\left[ {1.2x{{10}^{ - 2}}} \right]}^2}}}{{\left[ {1.5x{{10}^{ - 2}}} \right]{{\left[ {3.0x{{10}^{ - 2}}} \right]}^3}}}\)

= 0.106 x 10⁴ = 1.06 x 10³

PROBLEM 2

At equilibrium, the concentrations of N₂ =3.0 × 10^–3M, O₂ = 4.2 × 10^–3M and NO= 2.8 × 10^–3M in a sealed vessel at 800K. What will be K_c for the reaction N₂(g) + O₂(g)2NO(g)

SOLUTION

For the reaction equilibrium constant \({K_c}\) can be written as

\({K_c} = \,\frac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{O_2}} \right]}^3}}}\)

\({K_c} = \,\frac{{{{\left[ {2.8x{{10}^{ - 3}}M} \right]}^2}}}{{\left[ {3x{{10}^{ - 3}}M} \right]\left[ {4.2x{{10}^{ - 2}}M} \right]}}\)

= 0.622