JEE Chemistry Equilibrium Notes Set 01

Dynamic Nature of Equilibrium

1. A reversible reaction in a closed vessel reaches a state where the forward and backward rates are equal.
2. From then on the concentrations (and so the observable properties such as colour, pressure and density) stay constant, but both reactions keep going. This is why it is called dynamic equilibrium.
3. Equilibrium can be approached from either side and is reached only in a closed system at constant temperature.
4. A catalyst speeds up both reactions equally, so it shortens the time needed but does not change the composition at equilibrium or the value of K.

Law of Mass Action and Equilibrium Constant

For \( aA + bB \rightleftharpoons cC + dD \) at equilibrium:

\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}, \qquad K_p = \frac{p_C^{\,c}\,p_D^{\,d}}{p_A^{\,a}\,p_B^{\,b}} \]

1. Concentrations are raised to the power of their coefficients. Pure solids and pure liquids are left out because their active mass is constant.
2. For an elementary reaction \( K_c = k_f/k_b \), the ratio of the forward and backward rate constants.
3. K depends only on the reaction (as written) and on temperature. It does not depend on initial concentrations, pressure, volume or a catalyst.

Relation Between Kp and Kc

\[ K_p = K_c\,(RT)^{\Delta n_g} \]

Here \( \Delta n_g \) is (gas moles of products) minus (gas moles of reactants). When \( \Delta n_g = 0 \) the two constants are equal and the volume cancels from the expression. Use R = 0.0821 L atm mol⁻¹ K⁻¹ when pressures are in atm and concentrations in mol L⁻¹.

Manipulating Equilibrium Constants

Change to the equation	Effect on K
Reverse the reaction	K becomes 1/K
Multiply all coefficients by n	K becomes Kn
Add two reactions	Multiply their K values
Subtract one reaction from another	Divide their K values

Degree of Dissociation

The degree of dissociation \( \alpha \) is the fraction of the initial amount that has dissociated. Start with 1 mol and use a mole table (initial, change, equilibrium).

Dissociation	Total moles at equilibrium	Kp in terms of α and P
\( A \rightleftharpoons B + C \) (for example \( PCl_5 \))	\( 1+\alpha \)	\( \dfrac{\alpha^2 P}{1-\alpha^2} \)
\( A \rightleftharpoons 2B \) (for example \( N_2O_4 \))	\( 1+\alpha \)	\( \dfrac{4\alpha^2 P}{1-\alpha^2} \)
\( 2NH_3 \rightleftharpoons N_2 + 3H_2 \)	\( 1+\alpha \)	complex; use the mole table

For small α the expressions simplify, and α varies inversely with the square root of the pressure for \( A \rightleftharpoons B + C \). Vapour density method: \( \alpha = \dfrac{D - d}{(n-1)\,d} \), where D is the theoretical (undissociated) vapour density, d the observed one and n the number of moles produced from 1 mole.

Reaction Quotient

Q has the same form as K but uses the concentrations present at any moment.

Comparison	Meaning
Q < K	Net reaction goes forward
Q = K	System is at equilibrium
Q > K	Net reaction goes backward

Le Chatelier's Principle

If a stress is applied to a system at equilibrium, the system shifts in the direction that reduces the stress.

Change	Shift	Effect on K
Add reactant or remove product	Forward	None
Add product or remove reactant	Backward	None
Increase pressure (decrease volume)	Toward fewer gas moles; none if Δn = 0	None
Increase temperature	Toward the endothermic direction	K changes
Decrease temperature	Toward the exothermic direction	K changes
Inert gas at constant volume	No shift	None
Inert gas at constant pressure	Toward more gas moles (like a volume increase)	None
Catalyst	No shift	None

Effect of Temperature on K

\[ \log\frac{K_2}{K_1} = \frac{\Delta H}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \]

For an endothermic reaction (\( \Delta H > 0 \)) K increases when T increases. For an exothermic reaction K decreases when T increases.

Industrial Examples

1. Haber process \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) (exothermic, 4 to 2 moles): favoured by high pressure, moderate-low temperature and an iron catalyst.
2. Contact process \( 2SO_2 + O_2 \rightleftharpoons 2SO_3 \) (exothermic, 3 to 2 moles): favoured by high pressure, low temperature and a catalyst.
3. Formation of NO \( N_2 + O_2 \rightleftharpoons 2NO \) (endothermic, \( \Delta n = 0 \)): favoured by high temperature; pressure has no effect.

Worked Examples

Example 1. For \( H_2 + I_2 \rightleftharpoons 2HI \), 1 mol each of \( H_2 \) and \( I_2 \) are heated and 1.6 mol of HI is found at equilibrium. Then \( H_2 = I_2 = 1 - 0.8 = 0.2 \) mol and \( K_c = \dfrac{1.6^2}{0.2 \times 0.2} = 64 \).

Example 2. \( PCl_5 \) is 40% dissociated at a total pressure of 2 atm. Then \( K_p = \dfrac{\alpha^2 P}{1-\alpha^2} = \dfrac{0.16 \times 2}{0.84} = 0.38 \) atm.

Example 3. If \( K_p = 0.04 \) atm for \( N_2O_4 \rightleftharpoons 2NO_2 \) at 300 K, then \( K_c = K_p/RT = 0.04/(0.0821 \times 300) = 1.6 \times 10^{-3} \) mol L⁻¹.