Equilibrium

EquilibriumBrief OutlineWhat is reversible reaction?Examples of reversible reactionDynamic EquilibriumLe Chateliers PrincipleThe Haber Process

2What is reversible reaction?A reversible reaction is a chemical reaction in which the products can be converted back to reactants under suitable conditions. A reversible reaction is shown by the sign ( ) a half-arrow to the right (forward reaction) and a half-arrow to the left (reverse reaction).

3HgOHgOHgOHgOHgOHgOHgOHgO2HgHgHgOHgHgOO2HgO2 HgO (s) 2 Hg (l) + O2 (g) 2 Hg (l) + O2 (g) 2HgO (g)Reversibe Reactions Mercury and oxygen combine to form mercury oxide just as fast as mercury oxide decomposes into mercury and oxygen Upon heating, mercury (II) oxide decomposes to mercury (Hg) and oxygen (O2) [ Equation 1]:Under the same conditions, mercury (Hg) and oxygen (O2) recombine form mercury (II) oxide again [Equation 2]:Mercury and oxygen combine to form mercury oxide just as fast as mercury oxide decomposes into mercury and oxygen

4 2 HgO (s) 2 Hg (l) + O2 (g) Both reactions continue to occur, but there is no net change in the composition of the system.The amounts of mercury(II)oxide, mercury(Hg), and oxygen (O2) remain constant. There is a state of equilibrium between two chemical reactions.

6Dynamic EquilibriumEvaporation(No Equilibrium)EvaporationLiquid Gas(No Equilibrium)(No Equilibrium)Liquid GasLiquid Gas(Equilibrium)Closed SystemOpen SystemExampleFor the decomposition,

CaCO3 CaO + CO2

Would the reaction be the same when temperature is kept constantly high in an open and closed system?

Dynamic EquilibriumAg + + Cl - AgCl (s)ChemicalEquilibriumAg + Cl -Cl - Ag +AgCl (s)Rate of Precipitation = Rate of DissolvingHC2H3O2 (aq) H + + C2H3O2 -Rate of Dissociation (ionization) = Rate of AssociationHC2H3O2H + C2H3O2 -H +HC2H3O2C2H3O2 -Dynamic equilibrium is the state in which the forward and reverse reactions are occurring at the same rate. The forward and reverse reactions both continue. The concentration of all reactants and products remains constant)Dynamic equilibrium vs Static Equilibrium

Dynamic Equilibrium

Static Equilibrium 9Characteristics of Dynamic EquilibriumThe concentrations of the reactants and products (macroscopic properties) remain the same but the reactions don't stop!The reactants are still reacting to form the product and the product is still being converted back to the reactants (microscopic processes).

10

These two sets represent the same chemical reaction system, but with the reactions occurring in opposite directions. Horizontal part : the concentrations of reactants and products remain constant when the system reaches equilibrium. 2 HI H2 + I2 Dissociation of Hydrogen Iodide H2 + I2 2 HI Synthesis of Hydrogen Iodide11The equilibrium state is independent of the direction from which it is approached. Whether we start with an equimolar mixture of H2 and I2 (left) or a pure sample of hydrogen iodide (shown on the right, using twice the initial concentration of HI to keep the number of atoms the same), the composition after equilibrium is attained (shaded regions on the right) will be the same. H2 + I2 2HIFor more information, Click Here12

The adjacent graph shows the changes in the reaction rates of the forward and backward reactions:A + B C + DInitially (t = 0), [A] and [B] were maximum, while [C] and [D] were zero. The rate of the forward reaction decreases as A and B are used up.The rate of the reverse reaction increases as C and D are formed. Equilibrium is attained when the two rates become equal

[A], [B], [C], and [D] remain constant at equilibrium.13Chemical Equilibrium

If NO2 is reddish brown and N2O4 is colorless:What is happening here? What properties are changing?What is happening over time? After a long time?Consider this reaction:The Equilibrium Constant (homogeneous reactions)The equilibrium expression for this reaction would beKc = [C]c[D]d[A]a[B]baA + bBcC + dD

For a reaction of the type

the following is a CONSTANT (at a given T) Kc provides information about how far a reaction proceeds at a particular temperature. (can predict the conc. of reactants & products) Only write the Kc for homogeneous reversible reactions (all in the same physical state)Kc : equilibrium constant[C] and [D] : concentration of products[A] and [B] : concentration of reactants15Your Turn

.

Equilibrium mix

What is the equilibrium expression for the reaction?

Kc ~ 92/(1 x 1) = 81Your TurnUA GenChemYour Turn.

.

Equilibrium mix

What is the equilibrium expression for the reaction?Tier 1Your Turn.

Equilibrium mixKc ~ 22/(6 x 4) = 0.17What Does the Value of K Mean?If K >> 1, the reaction proceeds almost totally towards the products.

If K > 1, the conc of the product is greater than the conc of the reactants at equilibrium.If K Kc , the reaction will move to the left to generate more reactants to reach equilibrium.

The value of Qc in relation to Kc indicates the direction in which the net reaction must proceed as the system moves towards its equilibrium.

Le Chateliers Principleif a system at equiibrium is subjected to some changes, the position of equilibrium will shift to minimise the change.The possible changes in conditions includeChanges in concentration o f either the reactants or products.Changes in pressure for gas phase reactionsChanges in temperaturePresence of a cataystHas industrial significance as it allows chemists to alter the reaction conditions to produce an increased amount of product, hence increase the profitability of the chemical process.

ConcentrationPCl5 (g) PCl3(g) + Cl 2(g)

Kc = Qc If we double the conc. of PCl5 , what happens?

[PCl3][Cl2] [PCl5]Kc = ConcentrationIf one of the products is removed from the mixture, the equilibrium will shift to the right to replace the substance removed.In the reaction below, what happens when water is removed / added ?ethanol , C2H5OH is added / removed ?

CH3CO2H(l) + C2H5OH(l) CH3CO2C2H5(l) + H2O(l)

More carboxylic acid and alcohol would react to replace it producing more ester.This can be done by adding more conc sulfuric acid that provides H+ which acts as the catalyst for the reaction but also as a dehydrating agent removing water from the mixture.The addition of acid favours the prouction of ester since by removing water, it shifts the equilibrium to the right.36ConcentrationConsider the reaction below:2[CrO4]2-(aq) + 2H+(aq) [Cr2O7]2-(aq) + H2O(l)chromium(VI)ion from acid dichromate(VI) ion

When acid (H+ ) is added, acording to Le Cheteliers principle, equilibrium shifts to the right to use up the acid. The solution changes to orange. When alkali (OH- ) is added, the equilibrium shifts to the left because the OH- reacts with H+ to form water. This reduces the concentration of H+ ions. The equilibrium shifts to the left to reduce the effect by replacing the H+ ions.The solution changes to yellow. PressureExampleAn increase in pressure will shift the position of the equilibrium towards the side that involves fewer molecules.In the reaction below, what happens when pressure is (i) increased ? (ii) decreased ? N2O4(g) 2NO2(g) colourless brownAs the pressure is increased, the colour is initially darker as the same number of molecules are squeezed into a smaller space the conc is increased . The mixture then becomes more colourless . According to Le Chateliers principle, the equilibrium shifts to the left , the side with fewer molecules. The colour lightens to almost colourless.

38TemperatureAn increase in temperature will always favour the endothermic process and will shift the equilibrium towards that direction. Lowering the temperature will favour the exothermic process.A change in temperature will alter the Kc.ExampleIn the reaction below, what happens when the temperature is(i) increased ? (ii) decreased ? N2O4(g) 2NO2(g) H = +57kJmol-1 colourless brownIncrease in temp (putting in a hot water bath) cause the colour to be darker as the equilibrium shifts to the right that is endothermic in order to reduce the effect of the higher temperature.Decrease in temp (putting in ice bath) will cause the colour to be lighter.

39The final state depends on: The chemical nature of the reactants and products The conditions of the system (temperature, pressure, volume).Get time progressionCheck silberberg

N2O4 2NO2Low T High TEffect of pressure on KcAt 400K, N2O4 2NO2 2.00M 1.71M

At equilibrium , Kc = 1.46 moldm-3 at 400K. With pressure doubled,conc is doubled, [N2O4]= 4.00M , [NO2 ]=3.42M

The system is not in equilibrium.NO2 must react to produce more N2O4 until the Kc becomes equal.At higher pressure, the percentage of N2O4 is greater => the position of the equilibrium will shift to the left ( the side with fewer molecules of gas) K is not affected by a change in pressure.

Kc = [1.71]2[2.00]Kc = =1.46 moldm-3 Kc = [3.42]2[4.00]Kc = =2.92 moldm-3 Effect of concentration on Kc CH3COOH + C2H5OH CH3COOC2H5 + H2O At equilibrium, 2.88M 2.88M 5.76M 5.76M

With water added, conc. of water increases but conc. of other species will drop as the total volume increased. 2.61M 2.61M 5.22M 10.4M

The system is not in equilibrium.Ethyl ethanoate must react with water to use up the excess water. At higher concentration, the position of the equilibrium will shift to the left in order to maintain the same KcK is not affected by a change in concentration.

Kc = [5.76][5.76][2.88][2.88]Kc = =4 Kc = [5.22][10.4][2.61][2.61]Kc = =8 The equilibrium will shift to the side so as to restore the value of K.

Effect of temperature on KcConsider the reaction CO(g) + 2H2(g) CH3OH(g) H = -90kJmol-1

In an exothermic reaction, according to Le Chateliers principle, the equilibirum will shift to the side that is endothermic, that is to the left. the concentration of the top of Kc will derease ad the concentration of the bottom will increase.Kc decreases.Kc = [CH3OH(g) ][CO(g)][H2(g)]2CO(g) + 2H2(g) CH3OH(g) H = -90kJmol-1

Temperature /KKc / mol-2dm62981.7 x 10175001.1 x 101110002.1 x 106Kc = [CH3OH(g) ][CO(g)][H2(g)]2For exothermic reaction, the increase in temp decreases the Kc value.Exothermic ReactionEndothermic reaction N2 (g) + O2(g) 2NO(g) H = +181kJmol-1

Temperature /KKc / mol-2dm62984.3 x 10-315002.7 x 10-1810007.5 x 10-920004.0 x 10-430000.015Kc = [NO(g) ] 2[N2(g)][O2(g)]For endothermic reaction, the increase in temp increases the Kc value.CatalystA catalyst has no effect on the position of an equilibrium at a particular temperature as it increases the rate of the forward and the reverse reaction equaly.It reduces the time it takes the system to reach equilibrium.

"A catalyst provides an alternative route for the reaction with a lower activation energy."It does not "lower the activation energy of the reaction". There is a subtle difference between the two statements that is easily illustrated with a simple analogy.Suppose you have a mountain between two valleys so that the only way for people to get from one valley to the other is over the mountain. Only the most active people will manage to get from one valley to the other.Now suppose a tunnel is cut through the mountain. Many more people will now manage to get from one valley to the other by this easier route. You could say that the tunnel route has a lower activation energy than going over the mountain.But you haven't lowered the mountain! The tunnel has provided an alternative route but hasn't lowered the original one. The original mountain is still there, and some people will still choose to climb it.

46The equilibrium constant and rateKc provides information about the extent of a reactionIt does not provide information on the rate of reaction. Page 287 Qn 5ExampleDraw a table showing how the position of equilibrium in the reaction A,B and C would be affected by the followng changes:Increased temperatureIncreased pressureReaction A : interconversion of oxygen and ozone3O2(g) 2O3(g) H = +284 kJmol-1Reaction B : the reaction between sulfur dioxide and oxygen in the presence of a platinium/rhodium catalyst.SO2(g) + 2O3(g) SO3(g) H = -197 kJmol-1Reaction C : the reaction between hydrogen and carbon dioxide.CO2(g) + 2H2(g) CO(g) + H2O(g) H = +41 kJmol-1

Conditions affecting the position of EquilibriumAn equilibrium system opposes changesHaber Process

N (g) + 3H (g) 2NH (g)

from airfrom natural gas, CH4 ConditionsPressure: 20000kPa (200 atm)Temperature: about 500CCatalyst: IronAccording to Le Chateliers principle, at higher pressure, there should be greater yield of ammonia, however, there will be an increased cost in building a plant that can operate at a much higher pressure, and other safety and maintenance issues.Hence, a pressure of 200 atm with about 55.8% yield of ammonia is ideal.Haber Process

N (g) + 3H (g) 2NH (g) H = -92 kJmol-1

With a higher temperature at constant pressure, there will bea drop in the yield of ammonia.

Ideally, the reaction should be carried outat a lower temperature. However, the rate ofreaction will be slower at a lower temperature.Hence, a compromise temperature of about 500C is used.

Ammonia and the Haber ProcessExample 1Given that the equilibrium constant for the reactionH2 (g) + I2 (g) 2HI (g) at 700K is 54 and the conc of H2 and I2 at equilibrium are 0.25 moldm-3 and 0.50 moldm-3 what is the equilibrium conc. of HI?

0.12 moldm-3Example 25.00mol H2 and 3.00mol I2 are mixed together in a vessel of volume 10.0dm and allowed to come to equilibrium at 1100K. At equilibrium there were 0.43 mol I2 present in the reaction mixture. Calculate the value of of the equilibrium constant.

25.3Example 33.00mol NO2 and 1.00mol N2O4 are mixed together in a vessel of volume 1.0dm and allowed to come to equilibrium at 398K. At equilibrium there were 1.74 mol N2O4 present in the reaction mixture. Calculate the value of of the equilibrium constant.The reaction is 2 NO2(g) N2O4(g)

0.753 mol-1dm3