Equilibrium

Equilibrium

1. Chemical Equilibrium: the state of dynamic equilibrium that occurs in a closed system when the forward and backward reactions of a r eversible reaction occur at the same time. Solubility Eq: most substances dissolve in a solvent to a certain extent, and then the dissolving appears to stop. If the solution is in a closed system so no substance can enter or leave, then observable properties become constant and are said to be in Eq. In chemical reactions it is possible to see when forward and backward reactions are occurring at equal rates.

   • Characteristics of the Eq state: Eq is dynamic, the reaction has not stopped but both forward and backward reactions are still occurring at the same rate. Eq is achieved in a closed system, as it prevents exchange of matter with the surroundings, so Eq is achieved where both reactants and products can react and recombine with each other. Concentrations of products and reactants remain constant as they are being produced and destroyed at equal rates. Macroscopic properties are constant, observable properties such as colour and density change as they depend on the concentrations of the components of the mixture. Eq can be reached no matter where you start, the same mixture will result under the same conditions, no matter whether the reaction is started with all reactants, all products, or a mixture of both.

2. The Eq constant, k, helps determine the position of the equilibrium for a particular reversible reactions. For the general reaction: aA + bB --> cC + dD, where a, b, c, and d are the amount of each substance in moles, Kc = ({C}c{D}d)/({A}a{B}b). Kc is constant for a given T but if the T changes, then the value of Kc will also change. Kc depends on the way the equation is written. The units of Kc vary from one Eq to another. Comparing Kc values can help us determine the extent of a reaction from the magnitude of Kc. Kc > 10^10, the reaction almost goes to completion. Kc = 100, rate at Eq is the same but the reaction has shifted in favour of the products. {products} > {reactants} the plateau is after the Eq. Graph 1. When Kc = 1, which is very rare, {products} = {reactants}. Graph 2. When Kc = 0.01, {reactants} > {products}. Graph 3. When Kc < 10^-10, the reaction barely proceeds.

   • The value of Kc does not change because of the addition of more reactants or products. If that happened, the reaction would readjust itself. Some of the extra products or reactant will react with the other product or reactant to form more of the t=other side and the concentration adjust to establish a new Eq mixture so that the overall value for the expression equal Kc again. Note: only gases are included in the Eq expression.
     • Kc values only vary with T (not with changes in C or in P). Le Chatelier's Principle allows us to predict the direction (forward or reverse) in which the position of the Eq will change if the conditions are altered: When a system in Eq is disturbed, the Eq position will shift in the direction which tends to minimise, or counteract, the effects of the disturbance. The system at Eq is disturbed and responds to counteract the change by responding int he opposite way.
     • The reaction quotient, Q, allows us to predict the direction of the reaction. Q is the same as Kc but the values of the C of reactants and products at any point during the reaction. Q changes during the reaction and allows us to predict its direction. If Q<Kc then the reaction proceeds to the right, in favour of the product. If Q>Kc then the reaction moves to the left in favour of the reactants. If Q = Kc, the reaction is at Eq.
     • Inversing the reaction will inverse the Eq expression and the effect of Kc will be: 1/Kc. Doubling the reaction coefficients will square the expression and lead to Kc^2. Tripling the reaction coefficients triples the expressions and leads to Kc^3. Halving the reaction coefficients leads to square rooting the expressions, (square root)Kc. Adding together two reactions will multiply both expressions, so Kc1 x Kc2.
     • Calculating Kc from initial Eq C. Homogenous C : Eq reactions where all reactants and products are in the same phase/state. A. Eq C provided to find Kc: just plug them in the Kc value that you find using the formula. B. Initial C and one Eq C to find Kc: check to see if the amounts are expressed in moles per liter (C since Kc is for C). Create an ICE table that expresses the initial C, the change in C, and the Eq C for each species in the reaction. Determine the changes in the concentrations of each species and the Eq C. Write the Eq constant for the reaction. C. Calculating Eq C using the Kc and some other value. D. Calculating Eq C when Kc is very small (<10^3). Say that since Kc is very small, C of the things on the denominator will will be approximately equal to the value without the substraction, to the C of the thing on the denominator, therefore (example) 0.10 - 2x = 0.10.

3. Factors altering Eq: Concentration does not change the value of the Kc, it will only cause the Eq to shift in the direction opposite the change. Changing the P for a gas reaction. P only has an affect on gas Eq. Gas phase reactions will only be affected by a change in pressure if the reaction involves a change in the number of molecules on the reactant and product sides. Since molecules cause pressure, the more molecules, the higher the P. Increased P: Eq shifts to the side will less moles.

   • Changes in T. T affects the rate of a reaction in the Eq system. All chemical reactions proceed faster at higher T and vice-versa. However, in an Eq system, we are concerned with which of the reaction (forward or reverse) is affected most by a change in T. A general exothermic reaction can be written as: A + B <--> C + D+ Q, with a negative enthalpy change, whereas a general endothermic reaction can be written as: Q + A + B <--> C + D, with a positive enthalpy change. While both reactions will be speeded up, the endothermic reaction will be affected more at higher T. If the T was reduced, the system would respond by producing Q and favouring the exothermic reaction, a new Eq would be achieved and the value of Kc would be increased. If the T was increased, the system would respond by absorbing Q and favouring the endothermic reaction, a new Eq would be achieved and the value of Kc would be decreased.
     • Summary: Increased T --> Increased Kc for an endothermic reaction. Decreased T --> Decreased Kc for an exothermic reaction.

4. Effect of a catalyst on an Eq reaction. A catalyst increases the rate of a chemical reaction by providing an alternative pathway without undergoing chemical change itself. a catalyst will lower the Ea of both the forward and reverse reactions by the same amount, as both pass through the same transition state. A catalyst will have mo effect on the position of a chemical Eq, the Kc, or the yield. It will simply allow Eq to be reached faster by increasing both rates of reactions equally.

5. Industrial Applications: 1. Haber Process: Industrial production of ammonia (NH3), used to make fertilizers to increase crop yields. A high P will favour the formation of the products, P of 250 atm are commonly used. Although higher P will result in higher yield, it is very expensive to operate with high P. A low T would favour the production of products, but a high 450C is actually used to increase the rate of reaction. An iron catalyst is used to increase the rate of reaction. 2. Contact Process: Industrial production of sulfuric acid to be used for fertilizers, dyes, detergents. Pressure should be very high to favour the products as there are less mole on that side, but 2 atm is used because it is very expensive. Temperature should be low as the forward reaction is exothermic. A very high T is also used to increase the rate of reaction. Catalyst is V2O5 (vanadium oxide).

   • 3. Production of methanol, CH3OH: important in the chemical industry. A high P will favour the products, 5-10x10^6 P is used. A low T would favour the yield but 250C are used to increase the rate of reaction. The catalyst is Cu-ZnO-Al2O3.

6. Free energy and Eq: different reactions have different Kc values, which give information on the extent of the reaction. (example: a Kc = 10^-18 has an Eq that lies to the left and the reaction barely proceeds. ΔG<0 is a spontaneous reaction, which proceeds in the forward reaction. ΔG>0 is a non spontaneous reaction that proceeds in the reverse reaction. ΔG = 0 is when the reaction is at equilibrium and Gproducts = Greactants. Eq occurs when a reaction mixture is at the minimum value of Gibbs free energy. The position of the Eq corresponds to the value of S and a minimum in the value of the Gibbs free energy.

   • Spontaneous Reaction: In a spontaneous reaction starting with reactants, G decreases as the mixture moves towards the Eq. At Eq, G is minimum and ΔG = 0. In a spontaneous reaction, the Eq mixture consists mostly of products.
     • ΔG < 0 and large: appears spontaneous and its Eq mixture has a high proportion of products.
   • In a non spontaneous reaction starting with products, G decreases as the mixture moves towards the Eq, at the Eq, G is minimum and ΔG = 0. In a non spontaneous reaction, the Eq mixture consists mostly of reactants.
     • ΔG > 0 and large: appears non spontaneous and its Eq mixture has a high proportion of reactants.
   • Both ΔG and Kc can be used to measure the position of an Eq: ΔG = -RTlnKc. ΔG can be used to measure the position of an Eq, just like Kc. The Kc value gives info on the extent of the reaction, the Kc values are determined by the sign and value of the ΔG. Kc>1, + lnkc, - ΔG --> Mostly products in the Eq mixture, and spontaneous. Kc < 1, - lnkc, + ΔG --> Mostly reactants and non spontaneous. Kc = 1, lnkc = 0, ΔG = 0 --> appreciable quantities of both (not really equal).

7. Kc gives information about the extent of the Eq but not about its rate. Considering: A + B <--> C + D, rate of forward reaction is Kf{A}{B} and rate of backward reaction is Kb{C}{D}. At Eq, Kf{A}{B} = Kb{C}{D}, or Kf/Kb = {C}{D}/{A}{B} and Kc = Kf/Kb. so, if Kf>>Kb, Kc is large and the reaction proceeds to completion. If kf<<kb, Kc is small and the reaction barely proceeds.

   • Concentration: Increasing {reactant}, increases the rate of reaction, therefore Eq shift to the right, Kc remains constant because change in C does not change Kf or Ob. Catalyst increases the value of both rate constants Kf and Kb, but their Eas are lowered by the same factor so Kc remains constant as Kf/Kb. Temperature: from the Arrhenius equation, when T increases, k also increases. Exothermic reaction s have a lower Ea than endothermic ones, so they will be less affected by an increased T. (Kf is less increased than Kb). As Kf is less increased than Kb, Kc decreases.

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