Chemical Equilibrium-Reversible Reactions, Chemistry tutorial

Introduction:

Matter is constantly changing. Such changes are as an outcome of physical and chemical methods. The burning of fuel, melting of ice, burning of magnesium, digestion of food, respiration, making of soap or detergent, dissolving of sugar in water and numerous lab preparations are all illustrations of methods that outcome in changes in matter.

Most of the chemical reactions can only go in one direction. Take for illustration the reaction of magnesium having dilute hydrochloric acid.

Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

There is no manner the reverse reaction can take place. Hydrogen will not react by magnesium chloride, under any condition to generate magnesium and hydrochloric acid. This is an illustration of an irreversible reaction.

A few reactions, though, are reversible. A reversible reaction is a reaction which can go in either direction based on the conditions of the reaction.  There are a number of common illustrations of reversible reactions.

Reversible Reactions:

Reversible reactions are reactions which don't proceed to completion and the transformation of the products to reactants is as well possible beneath a set of reaction conditions. They are represented through ↔ between the reactants and the products. Reactions in which the reactants go virtually to give products and there is no tendency for products to revert to the reactants are irreversible reactions. They are symbolized by → between the reactants and the products.

The reaction of magnesium by dilute hydrochloric acid, the precipitation of silver chloride whenever aqueous solutions of silver nitrate and sodium chloride are mixed and the reaction of aqueous solutions of acid and alkali are illustrations of irreversible reactions.

Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (aq)

General illustrations of reversible reactions:

i) Heating of hydrated copper (II) tetraoxosulphate (vi) crystals:

CuSO4. 5H2O (s) ↔ CuSO4(s) + 5H2O (l)

If the hydrated salt is heated, water vapor is given off. The blue color of the hydrated salt changes to white and the crystals turn to powder. If a few drops of water are added to the cold white powder, the blue color returns as the hydrated salt is made up of.

ii) Heating ammonium chloride crystals:

NH4Cl (s) ↔ NH3 (g) + HCl (g)

If ammonium chloride crystals are heated the salt dissociates to ammonia gas and hydrogen chloride gas. As the gases cool, they join to form the solid. One test for ammonia gas is to hold the stopper of a concentrated hydrochloric acid close to the source of ammonia. A dense white fume of ammonium chloride is made.

iii) Formation of calcium hydrogen trioxocarbonate (iv):

Whenever surplus carbon (iv) oxide is bubbled via lime water Ca (OH)2 (aq) an initial white precipitate turns clear as calcium hydrogen trioxocarbonate (iv) is made. If this solution is heated, the precipitate returns as the insoluble calcium trioxocarbonate (iv) forms.

CaCO3 (s) + H2O (l) + CO2 ↔ Ca(HCO3)2 (aq)

iv) Reaction of iron by steam:

If steam is passed over heated iron a slow reaction takes place generating hydrogen gas and an oxide of iron. The reverse reaction as well takes place if hydrogen is passed over the heated oxide.

3Fe (s) + 4 H2O ↔ Fe3O4 (s) + 4H2 (g)

Reversible reactions and the equilibrium state:

If a chemical reaction takes place spontaneously, it carries on till a state of dynamic equilibrium is reached in which both the forward and reverse reactions are occurring at similar rate. The concentrations of the reactants and products no longer change by time. The reaction is stated to be at equilibrium or that a state of equilibrium has been reached. This equilibrium can merely be established in a sealed system, where no chemicals can enter, or leave the system. Such a system is known as a closed system. The equilibrium state remains unless it is disturbed. The equilibrium state can be approached from the reactant or product direction. The equilibrium state is the dynamic state in which there is no total change in properties like the density or concentration of the different parts of the system.

At equilibrium state, the free energy change is equivalent to zero (ΔG = 0)

Remember, ΔG = ΔH - TΔS

If ΔG = 0

ΔH = Teq ΔS Teq equilibrium temperature

ΔS = ΔH/Teq

The entropy change which accompanies the reversible reaction can be computed if the enthalpy change is acknowledged.

The reversible reaction and the equilibrium constant:

There is a fixed relationship, at a particular temperature, between the molar concentration of the products and the reactants in the equilibrium mixture. The above is the equilibrium law.

For a reaction of the form:

A (g) + B (g) ↔ C (g) + D (g)   KC - Concentration equilibrium constant

KC = ([C] [D])/([A] [B])

[] molar concentration

From the above, we see that the equilibrium constant (KC) is an index of how far the reaction goes in the direction of C(g) and D(g). Very high value of KC means that the reaction goes virtually to completion.

Extremely small values of KC means that the reaction doesn't go far in favor of C(g) and D(g).

The equilibrium constant for the reverse reaction.

C (g) + D (g) ↔ A (g) + B (g)

KC = ([A(g)] [B(g)])/([C(g)] [D(g)])

The equilibrium constant of the reverse reaction is the reciprocal of the re-equilibrium constant of the forward reaction.

Let consider real reactions.

i) H2 (g) + I2 (g) ↔ 2 HI (g)

KC = [HI(g)]2/([I2(g)] [H2(g)])

ii) N2(g) + 3H2(g) ↔ 2 NH3(g)

KC = [NH3(g)]2/[N2(g)] [H2(g)]3

Note that the molar concentrations are increased to the power of the reactant coefficient in the equation. A properly balanced equation is needed to allow you write the right equation for KC.

Now, represent the equilibrium constant equation for the reaction.

2SO2 (g) + O2 (g) ↔ 2SO3 (g)

For the gas reactions we can as well write the equilibrium constant expression by employing the partial pressures of the gases.

For illustration:

H2 (g) + I2 (g) ↔ 2HI (g)

KP = [PHI (g)]2/[PH2 (g)] [PI2 (g)]

P = Partial pressure

KP and KC are not equivalent however are related. Remember that for a gas, pressure is directly proportional to the concentration. The units of KC and KP based on the specific reaction that is considered.

For illustration:

H2 (g) + I2 (g) ↔ 2HI (g), KC and KP have no unit as the number of molecules is similar on reactant and product side.

For the decomposition reaction:

N2O4 (g) ↔ 2NO2 (g)

KC and KP will encompass units of mol dm-3 and Nm-2 or mm Hg (that is, pressure unit)

KC = [NO2 (g)]2/[N2O4(g)]

KP = [PNO2(g)]2/[PN2O4(g)]

The equilibrium constant is associated to the free energy change of the reaction

ΔG = ΔGo + RT ln K    ΔG - free energy standard

At equilibrium ΔG = 0 and

ΔGo = - RT ln k

Factors Affecting an Equilibrium State:

The position of equilibrium can be modified by changing the conditions. The effect on the equilibrium state caused through a change of certain condition of the reaction is predicted via Le Chatelier's principle.

Le Chatelier's principle:

This principle defines that for a system in equilibrium if any change is made to the conditions, the equilibrium will modify so as to oppose the change.

It can as well be defined that if a system in equilibrium is subjected to the constraint, it will adjust to annul or nullify the effect of the constraint.

The above principle manages the behavior of equilibria and the changes in condition which could be modified comprise temperature, pressure (for gases) and concentration of either the products or reactants.

Le Chatelier's principle and temperature effect:

Whenever the temperature is raised the reaction will move in the direction which uses up heat that is, the endothermic reaction is favored. For a decrease in temperature the exothermic reaction is favored. Note here that the reaction is in dynamic equilibrium and the forward and reverse reactions can take place. If the forward reaction is exothermic, the reverse reaction will be endothermic and vice-versa. A change in temperature will thus favor the forward or the reverse reaction. The value of equilibrium constant will rise for the endothermic reaction however reduces for the exothermic reaction.

Example: N2 (g) + 3H2 (g) ↔ 2NH3 (g), ΔH -ve

KC reduces by increasing temperature.

Le Chateliers principle and pressure change:

This applies just to gaseous reactions. Whenever the pressure is increased at constant temperature, the system will move in the direction of decreased volume.

Remember the Boyle's Law: P1V1 = P2V2

The direction of decreased volume is the direction of small number of gas molecules. Keep in mind that Avogadro's law, equivalent volumes of gases at similar temperature have similar number of molecules. The smaller number of molecules will encompass the smaller volume.

Example: 2NO2 (g) ↔ N2O4 (g)

Increase pressure will favor the forward reaction and a diminish pressure will favor the reverse, whenever there is no change in the number of gas molecules pressure will encompass no effect.

Le Chatelier's principle and concentration change:

Whenever the concentration of one substance is raised, the reaction will move in the direction to utilize the added substance. If the added substance is a reactant, then the reaction will move in the product direction and if it is a product, the movement will be in the favor of reverse reaction. If the concentration of one substance is decreased the reaction will move in the direction to form more of the substance eliminated. If the eliminated substance is a reactant, the reverse reaction is favored however if the product is eliminated the forward reaction is favored.

Based on the above, most of the equilibrium reactions have been made to go virtually to completion by the elimination of products as they are made up of.

Effect of Catalyst on Reversible Reactions:

A catalyst decreases the activation energies for both the reverse and forward reactions. The catalyst doesn't change the enthalpy change of the reaction. In the presence of catalyst, the reactants and products are not modified. The catalyst just raises the rate of the reaction. The equilibrium constant is not changed however the time to get to the equilibrium position is decreased in the presence of a catalyst.

Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)

Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology.  Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online chemistry tutoring. Chat with us or submit request at info@tutorsglobe.com