Writing Expression for Equilibrium Constant

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In this article, we shall stuudy to write expression for equilibrium constant.

Steps Involved in Writing Expression for Equilibrium Constant of a Reaction:

• Write the balanced chemical equation for the reaction.
• Write the products of equilibrium concentrations of the products in the numerator. Omit pure solids, pure liquids and the solvents in dilute solutions.
• Write the products of equilibrium concentrations of the reactants in the denominator. Omit pure solids, pure liquids and the solvents in dilute solutions.
• Raise each concentration term to the power equal to stoichiometric coefficients of the species in the equation.

Homogeneous Equilibrium:

The equilibrium in which all the substances involved exist in a single homogeneous phase is called homogeneous equilibrium.

Examples:

N_2(g) +  3 H_2(g)  ⇌     2NH_3(g)

H_2(g) +  I_2(g) ⇌     2HI_(g)

2SO_2(g) +  O_2(g)  ⇌   2SO_3(g)

NH_3(aq) + H₂O_(l) ⇌     2NH₄+_(aq) +  OH-_(aq)

2N₂O_(g) ⇌ 2N_2(g)  +   O_2(g)

CH₃COOC₂H_5(aq) + H2O_(l) ⇌ CH₃COOH_(aq) +   C2H5OH_(aq)

Heterogeneous Equilibrium:

The equilibrium in which the substance involved are present in different phases is called heterogeneous equilibrium.

Examples:

CaCO_3(s) ⇌  CaO_(s) + . CO_2(g)

CaCO_3(s)+H₂O(l)+CO₂(g) ⇌ Ca²⁺_(aq)+ 2HCO₃^–_(aq)

(NH₄)₂CO_3(s) ⇌ 2NH_3(g) + CO_2(g)+  H₂O_(g)

2Mg_(s) + O2_(g)  ⇌    2MgO_(s)

Note: Pure liquids and solids are ignored while writing the equilibrium constant expression.

Now, for pure solids and pure liquids, the molar mass M and density ρ are constant. Hence the concentration remains constant.

• When pure solids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all solids involved in equilibrium are taken as unity.
• When pure liquids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all liquids involved in equilibrium are taken as unity.
• For equilibria in the aqueous medium, the concentration of solvent (water) will not change appreciably because it is present in large excess. Hence by convention, the concentration of solvent (water) is taken as unity.
• Hence in general, pure liquids, pure solids, and solvents can be ignored while writing the equilibrium constant expression.

Writing the Expression for K_c and K_p:

BaCO_3(s)  ⇌  BaO_(s) + CO_2(g)

4NH_3(g) + 5O_2(g) ⇌   4NO_(g)+ 6H₂O_(g)

NH_3(g) + HCl_(g)  ⇌  NH₄Cl_(s)

3Cl_2(g) + 2NO_2(g)  ⇌  2NO₂Cl_3(g)

Fe³⁺_(aq) + SCN^–_(aq) ⇌ FeNCS²⁺_(aq)

CH_4(g) + 2H₂S_(g)  ⇌  CS_2(g)+ 4H_2(g)

MgCO_3(s)  ⇌  MgO_(s)  + CO_2(g)

AgBr_(s) ⇌ Ag⁺_(aq)+ Br^–_(aq)

K_c = [Ag⁺][Br ^–]

CH₃COCH_3(l) ⇌CH₃COCH_3(g)

K_c = [CH₃COCH_3(g)]

CH_4(g) + 2O_2(g) ⇌ CO_2(g)+ 2H₂O_(g)

Al_(s) + 3H⁺_(aq) ⇌ Al³⁺_(aq)+ 3/2H_2(g)

HPO₄^2-_(aq) + H₂O_(l) ⇌ H₃O⁺_(aq)+ PO₄^3-_(aq)

Ag₂O_(s) + 2HNO_3(aq) ⇌  2AgNO_3(aq)+ H₂O_(l)

Ni_(s) + 4CO_(g) ⇌  Ni(CO)_4(g)

CuO_(s) + H_2(g) ⇌  Cu_(s)+ H₂O_(g)

N_2(g) + 3H_2(g) ⇌  2NH_3(g)

Fe³⁺_(aq) + 3OH^–_(aq) ⇌  Fe(OH)_3(s)

2N₂O_(g) ⇌ 2N_2(g) + O_2(g)

C_(s)+  CO_2(g) ⇌  2CO_(g)

Consider a hypothetical reversible reaction involving homogeneous gaseous phase

aA(g) + bB(g) ⇌  cC(g) + dD(g)

The equilibrium constant in terms of the partial pressure is given by

This is the relation between K_c and K_p.

Where Δn = (Sum of the exponents in the numerator of concentration quotient) – (Sum of the exponents in the denominator of concentration quotient)

Steps Involved in Finding Relation Between K_c and K_p :

• Write the balanced chemical equation for the reaction.
• Find the change in the number of moles of gaseous species by the following formula.

• Now, use the relation

Examples to Find Relation between Between K_c and K_p :

Example – 01:

Example – 02:

Example – 03:

Example – 04: 

For the reaction N_2(g) + 3H_2(g) ⇌  2NH_3(g), the value of the equilibrium constant K_p is 3.6 x 10^-2 at 500 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 05:

For the reaction 2SO_2(g)) + O_2(g) ⇌ 2SO_3(g), the value of the equilibrium constant K_p is 2.0 X 10¹⁰ bar^-1 at 450 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 06:

For the reaction 2NOCl_(g) ⇌ 2NO_(g)+ Cl_2(g), the value of the equilibrium constant K_p is1.8 X 10^-2 at 500 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 07:

For the reaction CaCO_3(s)  ⇌ CaO_(s)+ CO_2(g), the value of equilibrium constant K_p is 167 at 1073 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 08:

Find the ratio of K_p/K_c for the reaction, CO(g) + 1/2O_2(g)  ⇌  2CO_2(g), at 500 K. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 09:

Find the ratio of K_p/K_c  for the reaction, N_2(g) + O_2(g) ⇌   2NO_(g)

Example – 10:

The equilibrium constant K_p for the reaction H_2(g)  + I_2(g)  ⇌  2HI_(g) is 130 at 510 K. Calculate K_c for following reactions a) 2HI_(g)  ⇌ H_2(g)  + I_2(g),  b) HI_(g)  ⇌ 1/2H_2(g)  + 1/2I_2(g), c) 1/2H_2(g)  + 1/2I_2(g)  ⇌  HI_(g).

∴  K_c = 130

For the reaction 2HI_(g)  ⇌ H_2(g)  + I_2(g),

For the reaction, HI_(g)  ⇌ 1/2H_2(g)  + 1/2I_2(g)

For reaction, 1/2H_2(g)  + 1/2I_2(g)  ⇌  HI_(g).

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