Equilibriums

From MyMCAT

Introduction

When we speak of a reaction, we generally consider the forward case. What happends to the reagents and how do they get converted in products? But more often than not, this reaction is not the only thing that is occurring- if a reaction can proceed, then it must also be possible for the same reaction to go in reverse. This special case, in which reactants are being converted to products and products are being converted to reactants is called an equilibrium.

Consider a reaction in which A is converted to B, and the reverse reaction, B being converted to A (at the same rate). If the two species, A and B, are in a stable equilibrium than the amount of A being converted to B is equal to the amount of B being converted to A. In this case, the concentrations cannot possible change because the amounts are always the same. To an outside observer, it looks as though no reaction is taking place, in fact however, reactions are taking place but in such away that the system as a whole never changes...

Now consider the case where we only have A and we let the reaction begin. Initially, we only have A thus only the forward reaction, in which B is formed, can occur. If we let this proceed for some time, then the amount, or concentration of B, will increase. If any B is present, then the reverse reaction is allowed to occur, forming A again. Because initially there is so much A, more A is converted to B, then B is converted back to A. This process will continue until the two find a stable equilibrium.

The Law of Mass Action

In developing the mathematical expressions required to discuss equilibriums further, one must quantify how the concentrations of all the elements in the reaction act. This relationship can be expressed by the law of mass action. In a system at equilibrium, at a fixed temperature and pressure, the product of the equilibrium concentration of the products divided by the product of the concentrations of the reactants, each being raised to the coefficient of the substance in the equation, must be equal to a constant.

In other words, if any reaction,

$aA + bB + ... \rightleftharpoons cC + dD + ...$

is to be in equilibrium, then the forward and reverse reactions must be occuring at an equal rate. This follows that

$forward rate &=& k_+ \left[ A^a \right]\left[B^b\right]...$ $reverse rate &=& k_- \left[ C^c \right]\left[D^d\right]...$

Since the rates are equal we can combine the two,

$K = \frac{k_+}{k_-} = \frac{ \left[ C^c \right] \left[ D^d \right] ... } {\left[ A^a \right] \left[ B^b \right] ... }$

Keq

The expression above for K is called the equilibrium constant, or K_eq. Every reaction has its own unique K_eq which is always the same, unless temperature or pressure change. If a K_eq value is given along with a reaction, it is assumed to be carried out at standard temperature and pressure unless otherwise stated.

Because this constant is measured as products or reactants, information can be obtained from the number alone. If the number is large, then the equilibrium must be in favour of more products (because a large product over small reactant makes a large K value).

Note, that catalysts, which speed up reactions, do NOT change the K_eq. If the reaction goes faster, then the reverse must go faster too, and thus, the equilibrium does not change so neither does K_eq.

Q

If we try to calculate K when the reactants and products are not in equilibrium, we will get a number which does NOT correspond to K_eq. This new value, which we call Q (reaction quotient) tells us how the system is at this given point in time, but the system is NOT stable, as an equilibrium will eventually be reached altering the concentrations of all the species and thus changing the K value.

If Q is calculated to be larger than K_eq, then it follows that their must be more product (or less reactant) that under equilibrium conditions. Given enough time, the concentrations will level out, and Q will decrease until it becomes equal to K_eq, at which point the system is at equilibrium.

Similarly, if Q is smaller than K_eq, then the current system must have more reactants (or less product) than what should be under equilibrium, and the system will slowly equilibrate.

If we are unsure as to whether the system is in equilibrium, all we need to do is calculate Q, and determine if it is equal to the standard K_eq for that reaction. If it is not, then we can at least determine which side the system is currently on, and how the system will change to reach equilibrium...

Le Chatelier's Principle

Le Catelier's Principle is a common application of understanding equilibrium dynamics. The prinicple is as follows, if a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to partially counter-act the imposed change.

Examples of Le Chatelier's Principle

Concentration

Changing the concentration of an ingredient will shift the equilibrium to the side that would reduce that change. (i.e. If something is added to the right, the equilibrium will shift left, if something is removed from the left the equilibrium will shift to the left, and so on.)

This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.

CO + 2 H₂ ⇌ CH₃OH

Suppose we were to increase the concentration of CO in the system. Using Le Châtelier's principle we can predict that the amount of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to fill the “gap” and favor the side where the species was reduced.

As the concentration of CO is increased, the frequency of collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if a desired product is not thermodynamically favored, the end product can be obtained if it is continuously removed from the solution.

Temperature

Let us take for example the reaction of nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases react to form ammonia:

N₂ + 3 H₂ ⇌ 2 NH₃ ΔH = −92kJ

This is an exothermic reaction when producing ammonia. If we were to lower the temperature, the equilibrium would shift in such a way as to produce heat. Since this reaction is exothermic to the right, it would favor the production of more ammonia. In practice, in the Haber process the temperature is instead increased to speed the reaction rate at the expense of producing less ammonia.

Changes in Pressure

Once again, let us refer to the reaction of nitrogen gas with hydrogen gas to form ammonia:

N₂ + 3 H₂ ⇌ 2 NH₃ ΔH = −92kJ

Note the number of moles of gas on the left hand side, and the number of moles of gas on the right hand side. We know that gases at the same temperature and pressure will occupy the same volume. We can use this fact to predict the change in equilibrium that will occur if we were to change the total pressure.

Suppose we increase total pressure on the system by decreasing the volume: now, by Le Châtelier's principle the equilibrium would move to decrease the pressure. Noting that 4 moles of gas occupy more volume than 2 moles of gas, we can deduce that the reaction will move towards the products if we were to increase the pressure.

Thus, an increase in pressure causes the reaction to shift to the side with the fewer moles. It has no impact on a reaction where the number of moles is the same on each side or when one of the reactants or products is not a gas.

Effect of adding an inert gas

An inert gas (or noble gas) such as helium is one which does not react with other elements or compounds. Adding an inert gas into a closed system at equilibrium may or may not result in a shift. For example, consider adding helium to a container with the following reaction:

N₂ + 3H₂ ⇌ 2NH₃

The main effect of adding an inert gas to a closed system is that it will increase the total pressure or volume (depending on how it is added). An inert gas would not be directly involved in the reaction as it is not part of the Keq equation, but it could still result in a shift through indirect means.

Volume held constant

If volume is held constant, as would be in case in any rigid sealed container, the individual concentrations of the above gases do not change. The partial pressures also do not change, even though we have increased the total pressure by adding helium. This means the reaction quotient does not change, so the system is still be at equilibrium and no shift occurs.

Pressure held constant while volume increases

If while the inert gas is added, the volume is allowed to increase such that the total pressure does not change, the partial pressures then must all decrease (because all the partial pressures must add up to the total pressure and we now have a new partial pressure for helium added).

In the above reaction, because there are more stoichiometric moles on the left-hand side of the equation (4 moles vs. 2 moles), the decrease in partial pressures affects the left-hand side more than the right-hand side. Therefore, the reaction would shift to the left until the system is at equilibrium again.