Dr Julie Juliewatty

Chemical Equilibrium Criteria of equilibrium and rate constant Free energy criteria for equilibrium Van’t Hoff equation Equilibrium in open and closed systems

Criteria of equilibrium Chemical equilibrium applies to reactions that can occur in both directions. In a reaction such as: CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) When the net change of the products and reactants is zero the reaction has reached equilibrium *both reactions are still occurring

If the system is NOT at equilibrium, the ratio is different from the equilibrium constant. In such cases, the ratio is called a reaction quotient which is designated as [C ] c [D] d = Q [A] a [B] b A system not at equilibrium tend to become equilibrium, and the changes will cause changes in Q that its value approaches the equilibrium constant, K Q ---  K

Equilibrium/Rate Constant (K) determine the amount of each compound that will be present at equilibrium consider the generic equation: aA + bB cC + dD A, B, C and D=molar concentrations of the reactants and products. A, b, c, and d = coefficients that balance the equation.

Equilibrium/Rate Constant aA + bB cC + dD Kc = Equilibrium constant

Equilibrium Constant Example 1 Using the following equation, calculate the equilibrium constant. N 2 (g) + 3H 2 (g) 2NH 3 (g) A one-liter vessel contains 1.60 moles NH 3, 0.5 moles N 2, and 1.20 moles of H 2. What is the equilibrium constant?

Free energy criteria for equilibrium Under conditions of constant temperature and pressure, chemical change will tend to occur in whatever direction leads to a decrease in the value of the Gibbs free energy. In this lesson we will see how G varies with the composition of the system as reactants change into products. When G falls as far as it can, all net change comes to a stop. The equilibrium composition of the mixture is determined by ΔG° which also defines the equilibrium constant K.

In this lesson we will examine the relation between the Gibbs free energy change for a reaction and the equilibrium constant (K/Kc/Kp).

Lets consider A + B → C + D components are gases at the temperature of interest even if the products have a lower free energy than the reactants, some of the latter will always remain when the process comes to equilibrium.

A + B  C + D In order to understand how equilibrium constants relate to ΔG° values, assume that all of the reactants are gases, so that the free energy of gas A, for example, is given at all times by G A = G A ° + RT ln P A

free energy change for the reaction is sum of the free energies of the products, minus that of the reactants: ΔG = G C + G D – G A – G B ΔG = (G° C + RT ln P C ) + (G° D + RT ln P D ) – (G° B + RT ln P B ) – (G° A + RT ln P A )

We can now express the G° terms collectively as ΔG°, and combine the logarithmic pressure terms into a single fraction: which is more conveniently expressed in terms of the reaction quotient Q P ΔG = ΔG° + RT ln Q

As the reaction approaches equilibrium, ΔG becomes less negative and finally reaches zero. At equilibrium ΔG = 0 and Q = K, so we can write ΔG° = –RT ln K

Example 2 Calculate the equilibrium constant for the reaction H + (aq) + OH – (aq) → H 2 O(l) from the following data:

Answer ΔH° = (∑ ΔH f °,products) – (∑ ΔH f °,reactants) = (–285.8) – (–230) = –55.8 kJ mol –1 ΔS° = (∑ ΔS°,products) – (∑ ΔS°,reactants) = (70.0) – (–10.9) = J K –1 mol –1 The value of ΔG° at 298K is ΔH° – TΔS° = (–55800) – (298)(80.8) = –79900 J mol –1 K = exp(79900/(8.314 × 298) = e 32.2 = 1.01 × 10 14

Equilibrium and temperature ΔG ° = ΔH ° – TΔS° Suppose that the equilibrium constant has the value K 1 at temperature T 1 and we wish to estimate K 2 at temperature T 2: –RT 1 ln K 1 = ΔH ° – T 1 ΔS° and –RT 2 ln K 2 = ΔH ° – T 2 ΔS°

the effect of the temperature on equilibrium: if the reaction is exothermic (ΔH° 0 then increasing T will make the exponent less negative and K will increase.

Van’t Hoff Equation The dependence of equilibrium constant on temperature

When ∆H is negative (exothermic reaction), K decreases with T ∆H is positive (endothermic reaction), K increases with T

Equilibrium in open and closed system In open systems, matter may flow in and out of the system boundaries In a closed system, no mass may be transferred in or out of the system boundaries. The system will always contain the same amount of matter, but heat and work can be exchanged across the boundary of the system. Whether a system can exchange heat, work, or both is dependent on the property of its boundary

Equilibrium in open and closed system In a closed system, changes continue, but eventually there is no NET change over time. Such a state is called an equilibrium state For example, a glass containing water is an open system. Evaporation let water molecules to escape into the air by absorbing energy from the environment until the glass is empty. When covered and insulated it is a closed system. Water vapour in the space above water eventually reaches a equilibrium vapour pressure.