JEE Chemistry Equilibrium Notes Set 04

Reversible Reactions and the Equilibrium State

Precipitation, neutralisation and reactions in open vessels that lose a gas go to completion in one direction. A reversible reaction, possible in a closed vessel, runs in both directions under the same conditions. Starting from reactants only, the forward rate falls and the reverse rate rises until they become equal. This is chemical equilibrium: rate forward = rate backward, and all measurable properties (concentration, pressure, colour, density) stay constant.
Features: the state can be reached from either side; it is dynamic, not static (a steady state with one-way flow is not equilibrium); a catalyst speeds up the approach but does not change the equilibrium composition, because it lowers the activation energies of both reactions equally; the free energy change is zero at equilibrium; equilibrium is shifted only by changing concentration, pressure, volume or temperature. Equilibria are homogeneous (one phase, such as \( \ce{H2 + I2 <=> 2HI} \)) or heterogeneous (\( \ce{CaCO3(s) <=> CaO(s) + CO2(g)} \)), and physical (phase change, dissolution) or chemical.

Law of Mass Action, Kc and Kp

The rate of a reaction is proportional to the product of the active masses (molar concentrations) of the reactants raised to their stoichiometric coefficients. For \( \ce{aA + bB <=> cC + dD} \), setting the forward rate equal to the backward rate gives \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \) with equilibrium concentrations; \( K_p \) is the same expression in partial pressures.
Pure solids and pure liquids have constant active mass and are taken as 1, so for \( \ce{CaCO3(s) <=> CaO(s) + CO2(g)} \) we have \( K_c = [CO_2] \) and \( K_p = P_{CO_2} \). In aqueous ionic equilibria the ion concentrations are used, for example \( K_c = 1/([Ag^+][Cl^-]) \) for \( \ce{Ag^+ + Cl^- <=> AgCl(s)} \).
Relation: \( K_p = K_c (RT)^{\Delta n_g} \), where \( \Delta n_g \) = moles of gaseous products - moles of gaseous reactants (it may be zero, negative or fractional). Units of Kp are atm to the power \( \Delta n_g \) and of Kc are (mol per L) to the power \( \Delta n_g \); the thermodynamic constant is unitless.

Reaction Quotient and Extent of Reaction

The reaction quotient Q has the form of K but uses the concentrations (or pressures) at any instant. If Q < K the reaction proceeds forward, if Q > K backward, and if Q = K the system is at equilibrium. Example: for \( \ce{NOBr <=> NO + 1/2 Br2} \) with Kp = 0.15 atm, mixing NOBr, NO and Br2 at 0.5, 0.4 and 0.2 atm gives Qp = 0.2^(1/2) x 0.4 / 0.5 = 0.36, which exceeds Kp, so the reaction goes backward and Br2 is consumed.
A large K (greater than about 1000) means the products dominate; a very small K (below about 0.001) means the reactants dominate. The percentage of products is highest for the largest K, but K says nothing about how fast equilibrium is reached.

Properties of the Equilibrium Constant

K depends only on temperature, not on the concentrations, pressure, volume or the presence of a catalyst. If a reaction is reversed, K becomes 1/K. If the equation is multiplied by n, K becomes \( K^n \). If two reactions are added, their K values multiply: \( K = K_1 K_2 \).
Effect of temperature (van't Hoff equation): \( \log \frac{K_2}{K_1} = \frac{\Delta H^\circ}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \). For an endothermic reaction (positive \( \Delta H \)) K rises with T; for an exothermic reaction K falls with T. Use R and \( \Delta H \) in the same energy unit.

Thermodynamics of Equilibrium

\( \Delta G = \Delta G^\circ + 2.303RT \log Q \). At equilibrium \( \Delta G = 0 \) and Q = K, so \( \Delta G^\circ = -2.303RT \log K \). Using \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \) gives \( \log K = -\frac{\Delta H^\circ}{2.303R}\cdot\frac{1}{T} + \frac{\Delta S^\circ}{2.303R} \). A plot of log K against 1/T is a straight line with slope \( -\Delta H^\circ/2.303R \) and intercept \( \Delta S^\circ/2.303R \). A reaction is spontaneous when \( \Delta G < 0 \).
In the thermodynamic K, pure solids and liquids have activity 1, ideal gases use partial pressure in atm and solutes use molar concentration.

Ester Equilibrium and Heterogeneous Systems

For \( \ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O} \) in the liquid phase Kc = [ester][water]/([acid][alcohol]). If 1 mol each of acid and alcohol with 1 mol water is used and 54.3% of the acid is esterified, then the equilibrium amounts are 0.457, 0.457, 0.543 and 1.543 mol, so Kc = (0.543 x 1.543)/(0.457 x 0.457) = 4.0 (the volume cancels).
For heterogeneous equilibria omit solids and liquids from K. Vapour pressure of a liquid is the Kp of \( \ce{H2O(l) <=> H2O(g)} \) and depends only on temperature. Relative humidity = partial pressure of water vapour / vapour pressure at that temperature.

Degree of Dissociation and Vapour Density

The degree of dissociation \( \alpha \) is the fraction of one mole that dissociates. For \( A_n \rightleftharpoons nA \), if a mol is taken, the total moles at equilibrium are \( a[1 + (n-1)\alpha] \). Hence \( M_{obs} = M_{th}/[1 + (n-1)\alpha] \) and \( \alpha = \frac{D - d}{(n-1)d} = \frac{M_{th} - M_{obs}}{(n-1)M_{obs}} \), where D is the vapour density before dissociation and d the observed value. This is not usable when n = 1 (HI, NO), because the number of moles does not change.
For \( \ce{PCl5 <=> PCl3 + Cl2} \) the equilibrium amounts are a(1 - alpha), a alpha and a alpha, with total a(1 + alpha). This gives \( K_p = \frac{\alpha^2 P}{1 - \alpha^2} \). For \( \ce{N2O4 <=> 2NO2} \) with a mixture vapour density of 38.3, \( M_{mix} = 76.6 \) and \( \alpha = 92/76.6 - 1 = 0.2 \); starting with 100 g (1.087 mol) of \( \ce{N2O4} \), the moles of \( \ce{NO2} \) are 2 x 1.087 x 0.2 = 0.435.

Le Chatelier's Principle

If a stress is applied to a system at equilibrium, the system shifts in the direction that reduces the stress. Adding a reactant or removing a product shifts the reaction forward; adding a product shifts it backward. Adding a solid component changes nothing.
Volume and pressure: increasing the volume (decreasing the pressure) favours the side with more moles of gas. So a rise in volume shifts a reaction forward when \( \Delta n_g > 0 \), backward when \( \Delta n_g < 0 \), and has no effect when \( \Delta n_g = 0 \). Raising the pressure does the opposite. This follows from Q: for \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \) Qc is proportional to 1/V, so increasing V decreases Qc below Kc and the reaction goes forward.
Catalyst: no shift. Inert gas: at constant volume it has no effect; at constant pressure the volume grows, so the reaction behaves as if the volume were increased (it shifts toward more gas moles).
Temperature: heating favours the endothermic direction. For an exothermic reaction K decreases on heating, for an endothermic one K increases. Temperature is the only factor that changes the value of K.

Industrial and Physical Applications

Haber process \( \ce{N2 + 3H2 <=> 2NH3} \) (exothermic, fewer moles of gas): high pressure and low temperature favour ammonia, but very low temperature is too slow, so a moderate temperature with a catalyst is used. Contact process \( \ce{2SO2 + O2 <=> 2SO3} \) (exothermic): higher pressure, lower temperature (about 500 C with a catalyst) and excess oxygen favour \( \ce{SO3} \). Birkeland-Eyde process \( \ce{N2 + O2 <=> 2NO} \) (endothermic, no change in moles): pressure has no effect and high temperature (1200 to 2000 C) favours NO.
Graphite to diamond is endothermic and the volume decreases, so high pressure and high temperature favour diamond. Ice melts with a decrease in volume, so higher pressure favours melting (the melting point falls). Boiling gives a large increase in volume, so raising the pressure raises the boiling point.
Efflorescence example: for \( \ce{CuSO4.5H2O(s) <=> CuSO4.3H2O(s) + 2H2O(g)} \), Kp = 1.084 x 10^-4 atm squared, so the equilibrium water vapour pressure is 1.04 x 10^-2 atm = 8 mm Hg. With a vapour pressure of 24 mm Hg at 25 C, the pentahydrate loses water when the relative humidity falls below 8/24 = 33%.

Simultaneous Equilibria

When two equilibria share a species, the concentration of the common species is the total from all the equilibria. For example if A(s) gives X + Y and B(s) gives Z + Y, with t and u mol per litre of the two decompositions, then \( K_1 = t(t+u) \) and \( K_2 = u(t+u) \), taking the concentration of Y as t + u in both.