Chemistry: Chemical equilibrium

Chemical equilibrium

Most chemical reaction is irreversible like neutralization between strong base and acid, they never go back to the reactant side, but some are reversible and we observe the product sometimes react to give the reactants. We write the equation as ⇌ to show that the reaction goes both sides.

Static vs dynamic: In a static equilibrium nothing in the system changes, like the equilibrium achieved in mechanics. However in a chemical equilibrium, a dynamic equilibrium occurs as reaction goes both forward and backward at the same time.

In chemical equilibrium, the concentration of both reactants and products remains unchanged, but reaction still occur and the forward rate = backward rate. Features of equilibrium include:

-          Closed system, otherwise when one chemical species escaped, reaction can't occur.

-          P, T remains constant as they affect concentration and rate of reaction.

-          Reaction never ends, only concentration remains unchanged.

-          It can be affected by the change in condition, i.e., acidity, concentration, P, T, etc.

-          The same equilibrium is achieved independent of the initial concentration. For example consider H₂ + I₂ ⇌ 2HI, the equilibrium [H₂]=[I₂]=0.11mol dm^-3, [HI]=0.78 mol dm^-3 no matter the initial condition is [H₂]=[I₂]=0.5 mol dm^-3, [HI]=0 or [H₂]=[I₂]=0, [HI]=1 mol dm^-3.

-          In a reaction rate-time graph, the forward and backward rate overlaps while the curve of reactants and products in concentration-time graph is zero when equilibrium is achieved.

-          Concentration of reactant decreases and concentration of product increases by time.

Examples

1)       (Haber process) 3H₂ + N₂ ⇌ 2NH₃

2)       (dissociation of weak acid) CH₃COOH ⇌ CH₃COO^- + H⁺

3)       (cobalt(II) chloride paper) CoCl₄^2- (blue, dry) ⇌ Co²⁺ (red), in fact it is about the formation of different complex: [Co(H₂O)₆]²⁺ (hexaaquacobalt(II) ion) + 4Cl^- ⇌ CoCl₄^2- (tetrachlorocobalt(II) ion) + 6H₂O, so upon adding of water it becomes red and in excess HCl it shows blue.

4)       (hydrogen halides) H₂ + I₂ ⇌ 2HI

5)       (chromate(VI) ion) Cr₂O₇^2- (orange) + 2OH^- ⇌ 2CrO₄^2- + H₂O, so dichromate ion exist in acidic condition and chromate exist in alkaline condition. Note that even in orange (dichromate ion) solution, chromate ion still exist, but different acidity gives different equilibrium condition.

6)       (solid involved) CaCO₃(s) ⇌ CaO(s) + CO₂(g)

7)       (esterfication) RCOOH + R'OH ⇌ RCOOR' + H₂O

When one species is removed, the equilibrium is forced to that side lacks chemicals. For example, in esterfication, if water is removed continuously, pure ester is given out.

Equilibrium law and equilibrium constant

Generally for a equation aA + bB ⇌ cC + dD, the equilibrium constant K_c is given by the equilibrium law K_c = [C]^c[D]^d/[A]^a[B]^b, it changes at different temperature so the constant must be given at a certain constant temperature. Sometimes K_a and K_b is used as the equilibrium constant for dissociation of acid and base. The unit of K_c is (mol dm^-3)^c+d-a-b which varies. We put positive power in front of negative power so dm⁶ mol^-2 is more preferred than mol^-2 dm⁶.

For example consider H₂ + I₂ ⇌ 2HI, K_c=[HI]²/[H₂][I₂] which has no unit. Consider the equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g), since the concentration (=moles/V) of solid is fixed, we say K_c=[CO₂] only. Also when one species act as solvent, it must be in large excess and can be ignored in calculation like H₂O is lots of reaction.

When we multiply the equation by n times, new K_c' = K_cⁿ and when reaction is reversed, new K_c' is the reciprocal of the original ones.

Extent of reaction

When K_c is large, then in equilibrium the concentration of product is larger than reactants, and when it's very large (e.g. 10⁴⁰), the reaction goes to complete visually and we usually regard these as irreversible reaction. Similarly when it's small the equilibrium tends to stay at reactants, and when it's very small, reaction is said as hardy proceed.

It also follows that when the reaction is reversed, the K_c is reciprocal of its original value, but the tendency of reaction still remains the same.

Define reaction quotient (Q_c) as the [C]^c[D]^d/[A]^a[B]^b at a certain moment, not necessarily at equilibrium, to determine the net direction of reaction. When Q_c < K_c, it has to be increased, hence the net reaction is to the product side, oppositely if Q_c > K_c it shifts to reactant side. When the two values are equal, equilibrium is achieved.

La Chatelier's principle states that when the condition of system in a equilibrium changed, the position of equilibrium will shift as to oppose the change.

1)       Change in concentration: when products are added to the system, more reactants will be produced as to consume more product to oppose the adding of products. Mathematically when products are added, momentarily Q_c > K_c so it shifts to reactant side. Oppositely when reactant is added, more products are produced. Acidity is also considered as the change of concentration of hydrogen ion or hydroxide ion.

2)       Pressure: Consider the total number of moles, when reactant side has more moles, adding pressure shifts the equilibrium to the product side as to reduce number of moles (PV=nRT) and so as the pressure. (e.g. 3H₂+N₂ ⇌ 2NH₃, when pressure is added, it shifts to the right and for 2NO₂ ⇌ N₂O₄, addition of pressure makes equilibrium shifting to the left.) Note that when the number of moles in both sides is equal, change in pressure gives no effect.

3)       Temperature (The ONLY factor to change K_c): Consider an exothermic reaction, when temperature is increased, reaction shifts to reactant side as to absorb heat and reduce the increase in internal energy, hence K_c decrease.

	Increase in	Shifting of eq.	Kc	Examples
Concentration	Reactants	Products	unchanged	Cr2O72- ⇌ CrO42- in different acidity
	Products	Reaction
Pressure	Increase in pressure may or may not shift the equilibrium depends on number of moles on both sides			3H2 + N2 ⇌ 2NH3, increase in pressure increases the yield of products
Temperature	Exothermic	Reactants	decrease	2NO2 ⇌ N2O4 (exothermic)
	Endothermic	Products	increase
Catalyst has NO effect on shifting equilibrium position, it only change the rate to achieve equilibrium.

Experimental determination of K_c – the approach by determining initial and final concentration

e.g. Esterfication CH₃COOH + CH₃OH ⇌ CH₃COOCH₃ + H₂O

1) Mixing CH₃COOH and CH₃OH in known proportion.

2) Draws 1cc from it and add 25cc of water (for convenience) and titrate against NaOH to find the concentration of CH₃COOH.

3) Add H₂SO₄ (catalyst) to the mixture, draws 1cc again, add water and titrate to find amount of H₂SO₄ added in the mixture.

4) Warm the mixture (reaction starts here) and cool down (for constant temperature), titrate against NaOH to find the final concentration of CH₃COOH, note that the amount of H₂SO₄ is eliminated from calculation.

5) By the given concentration we calculate K_c by [CH₃COOCH₃][H₂O]/[CH₃COOH][CH₃OH]. (water is in taken in account here since it's not the solvent here)

There're also other method to determine the concentration like clock reaction or colorimeter (e.g. the equilibrium Fe³⁺+SCN^- ⇌ [Fe(SCN)]²⁺, iron(III) is yellow while the complex is brown.)

Industrial application on equilibrium

It requires low cost, fast and high yield reactions.

1)       Haber process 3H₂ + N₂ ⇌ 2NH₃. However, this is exothermic that yield is not increased by increasing temperature but by increasing pressure. But in low temperature the rate of reaction is far too slow. Therefore 400 degree Celsius and 200 atm is used to balance the reaction rate and yield (about 30%). Pressure can't be too high to prevent high cost in maintaining high pressure. Note that finely powdered Fe is used as the catalyst.

2)       Contact process 2SO₂ + O₂ ⇌ 2SO₃, it's also exothermic so we can't increase temperature to increase yield. Luckily this reaction undergo steadily in 1atm. Therefore we use 450~600 degree Celsius and 1 atm with V₂O₅ as catalyst by the following equation:

-          2SO₂ + O₂ ⇌ 2SO₃ (The only reversible process; SO₂ commercially available and can also be obtained by burning sulphur.)

-          SO₃(g)+H₂SO₄(l)→H₂S₂O₇ (oleum or fuming sulphuric acid, note that 98% H₂SO₄ is used and it's (l) not (aq))

-          H₂S₂O₇ + H₂O → 2H₂SO₄ (we don't produce H₂SO₄ by SO₃+H₂O→H₂SO₄ because this is largely exothermic.)