Chemical Equilibrium PDF

# CHEM-201 Topics Included - Introduction to Chemical Equilibrium - General Equilibrium Expressions - Reaction Quotients - Equilibrium reactions in solid, liquid and gas phases - Extent of reactions and Equilibrium constants - Effect of Temperature and pressure on the equilibrium constant - Gibbs E...

# CHEM-201 Topics Included - Introduction to Chemical Equilibrium - General Equilibrium Expressions - Reaction Quotients - Equilibrium reactions in solid, liquid and gas phases - Extent of reactions and Equilibrium constants - Effect of Temperature and pressure on the equilibrium constant - Gibbs Energy of formation and calculations of Equilibrium Constants - Van't Hoff Equation - Le-Chatalier's principle ## Introduction to chemical Equilibrium **Equilibrium** Life is all about "balance". In Science, the word of Equilibrium is utilized to indicate the balance. Mostly (in Chemistry), we are familiar with forward reactions. A + B → Products However, in reality, most chemical reactions never finish or reach to an end. Some of the reactant concentrations are always left behind. Sometimes, there is another reaction is observed. When sufficient amount of product is acquired in the medium, we get backward reaction. A + B ← Products Why some reactions are reversible? **Answer "Gibbs Free Energy"** For reversible reactions, both forward and backward reactions can happen spontaneously (ΔG < 0). Therefore, they don't require an energy input from us to go backward. They do so accordingly with respect to the ΔG values. For a reversible reaction, When a reaction forward progress (reactants reacting to form products) is perfectly balanced with the reverse process (products changing back into products reactants) then the system is said to be in equilibrium state. In Physics, the condition of a system where neither its state of motion nor its internal energy states tends to change with respect to time. - Static - Dynamic ### Static Equilibrium For a single particle, equilibrium arises if the vector sum of all the forces acting upon the particle is zero. For a rigid body, in addition to the states listed for the particle, the vector sum of all the torque acting on the body equals to be zero. So that its state of rotational motion remains constant. F₁ = F₂ Object - F₁ No linear acceleration No angular acceleration Torque= 0 ### Dynamic Equilibrium Important in Thermodynamics In Thermodynamics, the concept of equilibrium is extended to include possible changes in the internal state of systems, as characterized by its Temperature, Pressure, density or any other quantity needed to specify its state completely. ## Dynamic Both processes are occurring but their rates have become equal. So it will look like nothing is happening but both processes are happening at the same time. ## Chemical Equilibrium Is A Dynamic Equilibrium Based Process - Irreversible reaction does not show chemical Equilibrium. - Reversible reactions are required to understand chemical equilibrium. Chemical equilibrium represents a state of reversible reactions in which a constancy in measurable properties (pressure, concentration, color, density etc.) is observed under a given set of conditions. Reactions which proceed in both directions are called reversible reactions. ## Generalization Let us understand this concept by taking a general reaction taking place in a closed vessel. A + B → C + D conc. of reactant = A&B. conc. of products = C&D. 1) At the start of reaction, conc. of reactants = maximum, and that of products = 0 = maximum. That of backward reaction is zero. 2) With the passage of time, conc. of A & B and therefore rate of forward reaction also decreases. ↑Rate of backward reaction & conc. of products ↑ ↑ Rate of backward reaction & conc. of products ↑ 1) At the start of reaction, conc. of reactants = maximum, and that of products = 0 = maximum. That of backward reaction is zero. 2) With the passage of time, conc. of A & B and therefore rate of forward reaction also decreases. ↑Rate of backward reaction & conc. of products ↑ 3) Ultimately, a stage is achieved when the rate of forward reaction become equal to rate of backward direction reaction. At this stage; conc. of A, B, C & D become constant. This state is called state of chemical equilibrium. ### Characteristics - It can be attained from either side i.e reactants or products. - There is no apparent change in measurable properties with time. - It is dynamic in nature (R_f = R_b) and spontaneously equilibrium is reached again when the disturbing forces are removed (Le-Chatelier's Principle). - It is independent of the direction in which the reaction takes place. - Chemical equilibrium involving gases are only attained in a closed vessel. - A catalyst has no effect on the state of equilibrium. Both forward and backward reactions are enhanced when dealing with the equilibrium state & catalyst. The equilibrium state is achieved in lesser time during the presence of catalyst... Thermodynamically speaking, a system in equilibrium state has maximum stability and entropy as well as minimum free energy. (ΔG)_{T, P} = 0 (maximum) (ΔG)_{T, P} = 0 (minimum) Capability to do non-PV work. ## Some Trick Questions N₂ + 3H₂ → 2NH₃ If the system is in equilibrium; is there any reaction happening? N₂ + 3H₂ → 2NH₃ (Forward reaction) N₂ + 3H₂ → 2NH₃ (backward reaction) Both reactions are happening, they are just happening at same speed. The 2 moles of NH₃ which is formed by forward reaction will be broken down to give N₂ + 3H₂ at the same time. So, concentration of NH₃, N₂ and 3H₂ will remain constant but reactions will be happening in the system. Rate of reaction vs. Time (Where Equilibrium is present?) <start_of_image> Vs time (A) Rate of forward Rx. Rate of backward Rx. Time (B) At single point - Equilibrium was attained. Time (C) Equilibrium was never Attained. Time All graph, represent equilibrium, as conc is constant. Initial concentration has no effect on relative concentration at equilibrium at a given temperature. K_eq does not get affected by the initial concentration. Although the absolute amount of reactants and products at equilibrium does. (K_eq = ([Product]_eq / [reactant]_eq No initial conc. involved Measurable parameters become constant. Color, Pressure, density, concentration all become constant. ## Types of Chemical Equilibrium There are two types of laws of Chemical Equilibrium: 1. **Homogeneous Equilibrium** in which all components are in single phase. For example, a system containing only gases or totally miscible liquids. N₂O₄(g) → 2NO₂(g) CH₃COOH(l) + C₂H₅OH(l) → CH₃COOC₂H₅(l) + H₂O(l) 2. **Heterogeneous Equilibrium** in which more than a single phase exists. For example. CO(g) + CO₂(g) → 2CO₂(g). CO(g) + H₂O(l) → CO₂(g) + H₂(g). ## General Equilibrium Expressions: Law of Mass Action This law was first enunciated by C.M. Guldberg and P.W. Waage in 1864. Relationship between conc. of reactants & products with the equilibrium. According to this law: At constant Temperature, the rate of chemical reaction is proportional to the effective concentration of a masses of reacting substances. The term active mass" is the effective concentration of a reactant or product which is related to the concentration. a = fc where a represents activity, f activity co-efficient and c - concentis at low pressure or dilute solutions. Ordinary systems like gases and f=1 concentration = activity. ⇒ activity may be replaced by molar concentrations (chemical kinetics). [] ← Active mass. General Equation: For a 1 molar reactions where solid and liquids are involved. A + B → C + D According to law of mass action: Rate of forward rate (R_f) α [A] [B] R_f = k_f[A][B] Where k_f is the rate constant for forward reaction. Depends upon Temperature. and nature of reactants. Similarly; Rate of backward reaction (R_b) α [C] [D] At Equilibrium, R_b = k_b [C][D] R_f = R_b k_f [A] [B] = k_b [C] [D]^α K_c = k_f / k_b = [C]^c[D]^d / [A]^a [B]^b K_eq = equilibrium constant of reaction in terms of concentration units of mole/dm³ It is the ratio of product of equilibrium concentrations of products to the product of equilibrium concentrations of reactants. Some BOOKS DOES NOT MAKE THIS DISTINCTION K_c = key of concentration / active mass of Product at equilibrium. K_eq = Concentration / active mass of Reactant at equilibrium. If we specify that concentration unit used is moles per dm³ K_c = ([Products]_eq / [Reactants]_eq) = [C][D]^α / [A][B]^α = C^c / A^a. C^d / B^b Concentration is specified by the moles/dm³ ## Stoichiometric Interpretation aA + bB → cC + dD (For solids and liquids). Generally R_f α [A]^a [B]^b R_f = k_f [A]^a [B]^b Similarly, R_b α [C]^c [D]^d R_b = k_b [C]^c [D]^d At equilibrium, R_f = R_b k_f [A]^a [B]^b = k_b [C]^c [D]^d Kc = k_f / k_b = [C]^c[D]^d/ [A]^a [B]^b = C^c / A^a. C^d / B^b Where a,b,c &d represents the stoichiometric co-efficients of reactants and products. K_c is a type of equilibrium constant in terms of the concentration unit of moles/dm³ or molar concentration. In case of reactions involving gases; k_eq can be sometimes represented as K_p which is another type of equilibrium constant. Where the concentration of gases are represented in form of their partial pressures at any given temperature. Kp = (P_c)^c (P_d)^d/ (P_a)^a (P_b)^b If the concentration is represented in terms of mole fraction, we can write K_x, which is the equilibrium constant of reaction using the concentration unit of mole fraction. K_x = (X_c)^c (X_d)^d / (X_a)^a (X_b)^b If no. of moles are utilized then, K_n = (n_c)^c (n_d)^d / (n_a)^a (n_b)^b K_eq can be represented as K_c, K_p, K_x and K_n. ## Relationship Between Equilibrium Constants: (Gaseous) ### I. Relationship between K_p and K_c For an ideal gas, the partial pressure of the ith component in the mixture is given by PV=nRT P_iV=n_iRT Where n_i is the no. of moles of the gas occupying volume V. P_i = (n_i / V)RT No. of moles per unit volume = concentration = moles/dm³ P_i = C_iRT Kp = P_c^cP_d^d/P_a^aP_b^b = (C_cRT)^c(C_dRT)^d/(C_aRT)^a(C_bRT)^b = (C_c)^c(C_d)^d(RT)^c+d / (C_a)^a(C_b)^b(RT)^a+b = (C_c)^c(C_d)^d / (C_a)^a(C_b)^b (RT)^(c+d)-(a+b Kp = K_c. (RT)^(c+d)-(a+b) Kp = K_c (RT)^An Eqy. Where An = (Sum of the products co-efficients) - (Sum of reactants co-efficients) An = difference in the number of moles of reactants and reactants. An may be positive, negative or zero Equations represents the relationship between K_p and K_c. If An = 0. That means; An = (Sum of products moles) - (Sum of reactants moles) 0 = (Sum of products moles) - (Sum of reactants moles) Sum of reactants moles = sum of products moles. Kp = K_c (RT)^An Kp = K_c (RT)⁰ Kp = K_c ### II. Relationship Between K_p and K_x According to Dalton's law of partial pressure, the partial pressure of any ith component is given by P_i = X_iP X_i is the mole fraction of an ith component of a gaseous mixture, P is the total pressure of the gaseous mixture. Kp = P_c^cP_d^d/P_a^aP_b^b = (X_cP)^c(X_dP)^d / (X_aP)^a(X_bP)^b = (X_c)^c(X_d)^d (P)^c+d / (X_a)^a (X_b)^b (P)^a+b = (X_c)^c(X_d)^d / (X_a)^a (X_b)^b (P)^(c+d)-(a+b) Kp = K_x. P^An Kp= K_x (RT)^An For An = 0 Kp= K_x ### III. Relation Between K_p and K_n P_i = n_i / N P Since X_i = n_i / N = no. of moles of ith componend / Total no. of moles. Kp = P_c^cP_d^d/P_a^aP_b^b = (n_c/N P)^c(n_d/N P)^d / (n_a/N P)^a(n_b/N P)^b = (n_c)^c(n_d)^d(P/N)^c+d/ (n_a)^a(n_b)^b (P/N)^a+b = (n_c)^c(n_d)^d / (n_a)^a(n_b)^b (P/N)^(c+d)-(a+b) = K_n (P/N)^(c+d)-(a+b) Kp = K_n. (P/N)^An If An= 0; Kp = K_c(RT)^An = K_x(P)^An = K_n(P/N)^An Kp = K_n So, if An = 0 Kp = K_c = K_x = K_n ## Examples: Application of law of mass action to the homogenous gaseous equilibria. The gaseous reactions are broadly classified into two categories. ### Type-I (An = 0) Some examples are 2HI(g) → H₂(g) + I₂(g) N₂(g) + O₂(g) → 2NO(g) #### Hydrogen Iodide Equilibrium H₂(g) + I₂(g) → 2HI(g) at t = 0 a moles a b moles b 0 0 (no. of moles) (concentration) at t = equilibrium: (a - x) (b - x) (2x) For equilibrium constant: Kc = [HI]² / [H₂] [I₂] = (2x/v)² / (a-x/v). (b-x/v) Kc = 4x² / (a - x). (b - x) If An = 0, then K_c does not depend upon the change in pressure and volume. It is true for all the reactions having An = 0. #### Calculate K_p at t = 0 H₂(g) + I₂(g) → 2HI(g) a b 0 at t = eq (a - x) (b - x) 2x Initial no. of moles. No. of moles at equilibrium. Total no. of moles at equilibrium = (a - x) + (b - x) + 2x = (a - x + b - x +2x) = (a + b + 2x - 2x) = a + b. If P is the total equilibrium pressure of the system, then Partial Pressure of H₂ at equilibrium = X_H₂. P = (No. of moles of H₂ / Total no. of moles) P = (a - x) / (a + b) P. Partial Pressure of I₂ at equilibrium = X_I₂. P = (No. of moles of I₂ / Total no. of moles) P = (b - x) / (a + b) P Partial Pressure of HI at equilibrium = P_HI = X_HI. P = (No. of moles of HI / Total no. of moles) P = 2x / (a + b) P. We know that Kp = (P_HI)² / (P_H₂) (P_I₂)² = (2x/a + b)². P² / ( (a - x/a + b) . P). ((b - x) / a + b). P) Kp = 4x² . P² / (a - x)( b - x) P² = 4x² / (a - x)(b - x) Proof | K_p = K_c when An=0. ### Type-II (An≠0) Gaseous reaction in which no. of moles of reactants and products are different. Some examples are: - PCl₅(g) → PCl₃(g) + Cl₂(g) - N₂O₄(g) → 2NO₂(g) - N₂ + 3H₂(g) → 2NH₃(g) #### Dissociation of PCl₅ PCl₅(g) → PCl₃(g) + Cl₂ at t = 0 1 mole 0 0 (initial concentration) at t = eq 1 - x x x (No. of moles at equilibrium) (concentration) where α is the degree of ionization No. of moles dissociated Total no. of moles take α is derived by using unitary method. Let's suppose you have 3 moles at the start and 3 moles are dissociated. Out of 3 moles, the no. of moles dissociated 3 moles For 1 mole, the no. of moles dissociated = 3 moles / 3 moles = 1 α can be defined as the no. of moles dissociated when 1 mole of a substance is taken. So, if 1 mole of it is taken, upon dissociation, it gives α products. So, remaining reactant would be equal to (1 - α). K_c = [PCl₃] [Cl₂] / [PCl₅] = (x/V) . (x/V) / ((1 - x)/V) = x² / V . (1 - α) For very dilute electrolytes solution or non-dissociating gas 1 - α ≈ 1 K_c = x² / V K_c. V = x² K_c. V = α² α² = K_c. V α ∝ √K_c. V Increase in Volume, increases degree of dissociation. Ostwald's law of dilution. #### Calculate K_p At equilibrium, PCl₅ → PCl₃ + Cl₂ 1 - x x x Total no. of moles at equilibrium = 1 - x + x + x = 1 + x P_PCl₅ = X_PCl₅. P_Total = (1 - x) / (1 + x). P P_PCl₃ = X_PCl₃. P = x / (1 + x) P P_Cl₂ = X_Cl₂. P = x / (1 + x). P Kp = P_PCl₃ .P_Cl₂ / P_PCl₅ = (x/1+xP) . (x/1+xP) / (x/1+xP) = α² / (1 + α). P = α² / (1 + α)(1 + α). P = α² / (1 + α)². P = [ (1)² - (α)²] / (1 + α)² . P = 1 - α² / (1 + α)² . P Kp = 1 - α² / (1 + α)² . P Rearrange it to get the dependance of α on P. Kp (1 - α²) = α² . P Kp - Kp α² = α² . P Kp = Kp α² + α² . P Kp = α² (Kp + P) Kp / (Kp + P) = α² At high values of P, (Kp + P ) ≈ P Kp / P = α² → α = √ (Kp / P) → α ∝ √ (Kp / P) Another way of Ostwald dilution law. ## Homogeneous Equilibrium In Liquid System Similar K_c values as observed for gaseous systems, in case of non-pure liquids for An = 0. Example Acetic acid esterification. CH₃COOH(l) + C₂H₅OH(l) → CH₃COOC₂H₅(l) + H₂O(l) at t = 0 a b 0 0 moles moles concentration at t = eq (a - x) (b - x) x x Kc = [Products]² / [Reactants]² = [CH₃COOC₂H₅] [H₂O] / [CH₃COOH] [C₂H₅OH]= x² / (a - x)(b - x) Kc = x² / (a - x)(b - x) Kc is not affected by the volume change. ## Equilibrium In Heterogeneous System Heterogeneous reactions involving two or more phases, one of which is a solid or liquid treats the K_c a bit differently. According to law of mass action. The value of activity is always taken as unity at all temperatures. CaCO₃(s) → CaO(s) + CO₂(g) Taking activemasses as concentration unit Keq = ([CaO]_s [CO₂]_g / [CaCO₃(_s] = A_caO . A_CO₂ / A_CaCO3 = Kc Pure solids and pure liquids have activity values = 1 (concentration). Something is dissolved in something else. Pure solids and liquids are not dissolved. So, (a = 1) as concentration. Concept cannot be applied on them. If we consider the molecules of solvent or gas surrounding these phases to calculate a concentration, you come to conclusion that this concentration is constant in a practical way. Because very large no. of extra interaction is rendering some molecular molecules not active in participating in the molecular interaction. Vapour pressure of pure liquids and solids is constant at a given temperature. Consequently, their concentration is constant in the gaseous phase.. It still remains same. So, if you dissolve it in solvent it is not considered K_eq. V. P of the water at The Pacific ocean or a pool in your house have some atmospheric pressure. With the application of the above condition. A_cao = A_cacos = A_co₂ = 1 (K_eq) = K_a = A_co₂ = 1. For an ideal gas, activity = partial pressure of the gas at its equilibrium stage. (K_eq) = K_a = P_co₂. The K_eq is independent of the equilibrium active masses of the pure liquids and solids in heterogeneous system. ## Reaction Quotient Reactions Irreversible Reversible Physical Reversible Reaction Chemical Biological → Environmental Reactants Products Reaction Quotient) Q = [Products] / [Reactants] Ratio of molar concentration of products to reactants at any given time eq. (Equilibrium) Q = [Products]_eq / [Reactants]_eq Eventually, Q reaches K_eq at the equilibrium stage. Time Reactants Products 0 100 0 50 58 42 100 42 58 75 Q = 0.72 Q = 1.4 Q = 2.2 150 41 59 K = 1.9 conc. 200 31 69 25 250 34 66 0 0 300 37 63 350 35 65 50 100 150 200 250 300 350 Time (s) Kc = [Products]_eq / [Reactants]_eq = 65 / 35 = 1.9 Q = Ratio of product and reactant molar concentrations at any time 't'. At t = 0 sec. Q = 0/100 = 0 At 50 sec. 42/58 = 0.72 At 100s; Q = 58 / 42 = 1.4 At 150 sec. Q = 59 / 41 = 1.4 At 200 sec. Q = 69 / 31 = 2.2 At 250 sec. Q = 66 / 34 = 1.9 At 300 sec. Q = 63 / 37 = 1.7 At 350 sec. Q = 65 / 35 = 1.9 Q is what we have along the way, while K is the value what we get in The end. Now, let's suppose you are performing a chemical reaction. What is the significance of these K and Q values. Relationship between K and Q will give information about the direction in which your reaction should go as well as how much products or reactant are there in the equilibrium. K = [Products]_eq / [Reactants]_eq ↑ K α [Product]_eq ↑ Higher values of K represents the fact that more product is present in the medium. K α 1 / [Reactants]_eq ↑ If more reactants are present in the medium; [Product] / [Reactants] ↑ Higher. So, overall ratio will be lower. [Product]_eq / [Reactant]_eq As, a result K will decrease as well. If in the medium, you have: Then K = 1 = [Product]_eq / [Reactant]_eq It would mean you have equal concentration of product and reactant at a given temperature for the system in the equilibrium state. K = 1 → [Product]_eq = [Reactant]_eq K >>> 1 or K ≫ 1: It means our numerator is bigger than denominator So, [Product]_eq > [Reactant]_eq K<<<1 or K ≪ 1: Opposite is true; and the denominator value is bigger So, [Product]_eq < [Reactant]_eq K value is fixed. K = 1.9. (From the previous example) Q → K If we have a system where Q value is less than K value, then Q < K. 1.4 → 1.9 Reactants → Products. ↑ Q = [Product]_eq / [Reactant]_eq ↑ Forward reaction should dominate. At this certain point, enough product should be present in the medium to have the value of Q equal to K. If we have a system where Q value is greater than K. Q → K 2.2 → 1.9 Your Q value wants to approach K_c. For that Q value has to go lower. ↓ Q = [Product]_eq / [Reactant]_eq ↓ Reactants → Product So more reactants can be produced. backward. If Q = K. Then there is no distinction between reaction quotient and equilibrium constant. Q<K Reaction moves in forward direction. Q>K Reaction will move in backward direction. Q = K Then, equilibrium is achieved. ## Le-Chatelier's Principle N₂(g) + 3H₂(g) → 2NH₃(g) Equilibrium Concentration Time H₂ NH₃ N₂ When the equilibrium is reached, Rate of forward reaction becomes equal to rate of backward direction reaction. N₂ + 3H₂ → 2NH₃ at the same time. N₂ + 3H₂ → 2NH₃ No. of moles over here = constant. - No. of moles over here = desirable. We want to have the products or as much NH₃ as possible. So, reversible reaction here spoils that. (Equilibrium) (Equilibrium disturbances) → Out of Equilibrium But Adjust accordingly with respect to Le-Chatelier's principle. New Equilibrium By creating these equilibrium changes; maximum product can be attained. Henry Le Chatelier (1884) studied the effect of concentration, temperature, pressure on equilibrium systems and sum up his findings in the form of generalization. If a system at equilibrium is disturbed by applying some sort of stress, the system will adjust itself in order to nullify the effect of that stress. Disturbance Equilibrium A Q = [Product]_t / [Reactants]_t (Equilibrium constant = K = [Products]_eq / [Reactants]_eq Q = [Product]_t / [Reactants]_t Disturbance N₂(g) + 3H₂(g) → 2NH₃(g) H₂ 5 H₂ 4 NH₃ NH₃ 3 2 N₂ N₂ 1 0 New Equilibrium position is established after the system. System adjusted itself to nullify the change. Stress is applied. K_eq before disturbance K = [NH₃]² / [N₂] [H₂]³ = (3)² / (1)(4)³ = 9 / 64 = 0.14. Apply disturbance, systems is now out of equilibrium, K ≠ Q. Q = [NH₃]² / [N₂] [H₂]³ = (3)² / (1) (6)³ = 9 / 216 = 0.042. Disturbance here is the increase in the H₂ conc. from 4 → 6 Q ≤ K → forward At equilibrium (new): K = [Products]_eq / [Reactants]_eq = (4)² / (0.9)(5)³ = 16 / 112.5 = 0.14 Important Injections from Experiment: 1) Q < K, so the reaction has to move in the forward reaction to gain equilibrium again. Therefore, NH₃ concentration during adjusting. Before disturbance = 3 Adjustment → 4. That meant more consumption of reactants. Similarly, increase in NH3 meant their concentration. N₂ 1 → 0.9 5 → 6 ii) K_eq = 0.14 Before disturbance = New Equilibrium k_eq = 0.14 So, by changing the variable in the system K_eq does not change. However, equilibrium position does change. iii) Although K_eq is constant but the value of individual concentration does change. The change happens in such a way that the overall [Product]_eq / [Reactants]_eq remain constant. K_eq = (3)² / (1) (4)^³ = 0.14 After shifting the H₂ at 6, (K_eq)_new = (4)<sup>