Chemical Equilibrium The Equilibrium Condition ● The majority of chemical reactions are reversible. ● Equilibrium is the state where the rate of the forward reaction is equal to the rate of the reverse reaction and results in no net observable change. At equilibrium, concentrations of all reactants and products remain constant at constant temperature. ● Equilibrium is dynamic. ⇄ or ↔ indicates that the reaction is proceeding forward and in reverse. ○ Once equilibrium is established, the rate of each direction is equal. ○ This keeps the concentration of reactants constant ○ This keeps the concentration of products constant ○ This does NOT mean that concentration of products and concentration of reactants are the same or equal to each other. ● Any chemical reactions carried out in a closed vessel will reach equilibrium. ● Concept Check 3H2(g) + 2N2(g) ↔ 2NH3(g) H2 gas and N2 gas were placed in a rigid vessel and allowed to reach equilibrium in the presence of a catalyst according to the equation above. The diagram below shows how the concentrations of H2, N2, and NH3 in this system changed over time. Based on the graph, which of these claims is most likely correct? powder we a) The reaction reached equilibrium at t1, because the concentrations of H2 and NH3 were the same. b) The reaction reached equilibrium at t2, because the concentrations of N2, H2, and NH3 stopped changing. c) The reaction reached equilibrium at t3, because the reaction stopped proceeding. d) The reaction did not reach equilibrium, because not all of the N2 and H2 was consumed at time t3. The Equilibrium Position ● Whether the reaction favors products or favors reactants depends on three main factors: ○ Initial concentrations (more collisions--faster reaction) ○ Relative energies of reactants and products (nature goes to minimum energy) ○ Degree of organization of reactants and products (nature goes to maximum disorder) ● If the rate of the forward reaction is greater than the rate of the reverse reaction, then there is a net conversion of reactants to products and the forward reaction is favored. ● If the rate of the forward reaction is less than the rate of the reverse reaction, then there is a net conversion of products to reactants and the reverse reaction is favored. ● At equilibrium, the rates are equal and neither direction is favored. ● Concept Check ○ A reversible reaction is represented by the equation above. The amount of reactants and products at time 1 and time 2 are shown. Based on the diagram, what can we infer about the rates of the forward and reverse reactions over time? a) Over time, the forward reaction is favored, because the concentration of AB increases b) Over time, the reverse reaction is favored, because the concentration of A decreases c) Over time, the system reaches equilibrium, because the total number of A and B atoms does not change d) Over time, the system reaches equilibrium, because the reaction stops ○ The reaction, N2(g) + 3H2(g) ↔ 2NH3(g), is allowed to go to completion. What sequence of the particle diagrams shown below best represents the progress of the reaction at time t1, t2, and t3? The Reaction Quotient ● For a reversible reaction, we can represent the relative quantities of reactants and products at any given point in time as a ratio of products to reactants. This value is called the reaction quotient, Q. ● For the general reaction: dort equilibrium aA + bB ↔ cC + dD w ○ ○ ○ ○ ○ Qc = [C]c [D]d [A]a [B]b Qc for concentration [ ] means molarity QP for pressure - can only use with gases Qc =/ QP Solids and liquids are not included because do not have concentrations or pressure ● Value of Q will tell us whether the reaction will need to proceed in the forward direction or the reverse direction in order to reach equilibrium. ● Concept Check ○ Write the reaction quotient, Qc, for the reaction 2HF(aq) ↔ H2(aq) + F2(aq) ○ Write the reaction quotient, Qc, for the reaction CaO(s) + CO2(g) ↔ CaCO3(s) The Equilibrium Constant ● When the system is at equilibrium, no longer use reaction quotient, use equilibrium constant, K ● The equilibrium constant expression for the general reaction: r f Equilibrium cow aA + bB ↔ cC + dD Kc = [C]c [D]d [A]a [B]b ● Temperature dependent ● Equilibrium constant at a given temperature will remain the same, but the concentrations will not always be the same. ● The value of K will tell us whether there are more products or more reactants at equilibrium ● Concept Check ○ Write the equilibrium expression, KP, for the following reaction: 4NH3(g) + 7O2(g) ↔ 4NO2(g) + 6H2O(g) ○ The graph below shows the relationship between concentration and time for a reversible reaction involving reactants A(g) and B(g) and product C(g) a) Write a balanced chemical equation that could be represented by the graph b) From the balanced equation in (a), write the equilibrium constant expression, Kc, for the reaction at equilibrium. During what time range is this expression valid? c) From the balanced equation in (a), write the reaction quotient expression, Qc, for the reaction. During what time range would it be most appropriate to use Qc? Calculating the Equilibrium Constant ○ The following equilibrium concentrations were observed for the Haber process for the synthesis of ammonia at 127°C: N2(g) + 3H2(g) ↔ 2NH3 -2 [NH3] = 3.1 x 10 mol/L; [N2] = 8.5 x 10-1 mol/L; [H2] = 3.1 x 10-3 mol/L Calculate the equilibrium constant at 127°C for this reaction. ○ For the reaction X(g) + Y2(g) ↔ XY2(g), a system at 298 K has the following partial pressures at equilibrium in three separate experiments: Experiment Px PY2 Pxy2 1 2.0 atm 5.0 atm 1.0 atm 2 2.0 atm 10.0 atm 2.0 atm 3 1.0 atm 5.0 atm 0.5 atm Calculate the value of KP at 298 K for each of the three experiments. K Protects Magnitude of the Equilibrium Constant reactants ● Can be any value from 1 x 10100 to 1 x 10-100 and beyond! ● Products are directly proportional to K and reactants are inversely proportional to K reactions forward ● K > 1 means that the reaction favors the products at equilibrium ● ● ● ● ○ A very large K means the reaction essentially goes to completion. K < 1 means that the reaction favors the reactants at equilibrium reversereaction ○ A very small K means the given reaction does not occur to any significant extent. K = 1 means the reactants = products and neither is favored notcommon 17 12 related. The size of K and the time required to reach equilibrium are not directly ○ The time required to achieve equilibrium depends on the reaction rate, which is determined by the size of the activation energy. ○ The size of K is determined by thermodynamic factors such as the difference in energy between products and reactants. Knowing the equilibrium constant allows you to predict: ○ The tendency of a reaction to occur (not the speed of the reaction) ○ Whether a given set of concentrations represents an equilibrium condition ○ The equilibrium position that will be achieved from a given set of initial concentrations Properties of the Equilibrium Constant Change Made Effect on Keq Stoichiometry coefficients in a balanced equation are changed by a factor of “n” K new = (K old )n Reverse equation is used Several balanced equations are added to get an overall equation. (K1, K2, K3 are the equilibrium constants for the balanced equations.) ● Concept Check ○ Recall the earlier problem of K new = 1 K old K new = K 1 K 2 K 3 … N2(g) + 3H2(g) ↔ 2NH3 [NH3] = 3.1 x 10-2 mol/L 10 [N2] = 8.5 x 10-1 mol/L [H2] = 3.1 x 10-3 mol/L ■ Calculate the equilibrium constant at 127°C for the reverse reaction: 2NH3 ↔ N2(g) + 3H2(g) take recip K 3.8 k 4 2600 o4 ■ Calculate the equilibrium constant at 127°C for the reaction: 1 N (g) + 32 H2(g) ↔ NH3 2 2 raise to power of change 4 2 3.8 10 Calculating Equilibrium Concentrations 4 Reaction Initial concentration (pressure) Change in concentration (pressure) Equilibrium concentration (pressure) CO2(g) + H2(g) ←→ CO(g) + H2O(g) Suppose that Kc is 0.64 and that we start with 0.100 M CO2 and H2. Calculate the equilibrium concentrations of the products and the reactants. cinnamon R airman vii 7 initial consume amor orange I C Equilibrium CO2(g) + 0.100 0.100M E 100 X 64 ←→ CO(g) H2O(g) 0 Odetta 100044 56 u coz CHz 8 xxx X TF g X X H2O 100 x C100 look at coefficient X 100 X O + O M X X µ 64 H2(g) assume not if zero mentioned x 08 080 x ClodfHz0 I 100X box X 1.80 X 044 i 044N A vessel initially has a partial pressure of NO equal to 0.526 atm and a partial pressure of Br2 equal to 0.329 atm. At equilibrium the partial pressure of Br2 is 0.203 atm. Calculate Kp for the reaction 2NO(g) + Br2(g) ←→ 2NOBr(g) R 2NO(g) I 526 atm C E + Br2(g) ←→ 2NOBr(g) 329atm 2x X 526 2x 329 X O t ZX 203 X 126 526 21.126 2x zsz 274 Kp PNob Tho p Kp 25232 Kpi 4.17 C 27474.203 Representations of Equilibrium ● Can count particles for concentration. Using the information provided, determine which vessels are at equilibrium. x rparticles OU µ Z K FYI g 14 12 3 172 4 Q'Czyz 2 1.2 Concentration or At what time does the system below reach equilibrium.? Justify your answer. partialpressure Le Châtelier’s Principle ● If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. ● Change is called stress ● Shifts occur to reestablish equilibrium positions. aka stress ● Change in Concentration (constant T and P or constant T and V) ○ Add reactant ■ Tries to use up the reactant ■ Shifts right, towards products ■ Fe3+(aq) + SCN-(aq) ↔ Fe(SCN)2+(aq) Add in more SCN- Left right ○ Add product ■ Tries to use up the product ■ Shifts left, towards reactants ○ Diluting ■ All molarity values decrease ■ Shifts in the direction that will form the greater amount of particles. 2 particles 1 particle no charge 2 Azt 0252420 ● Change in Pressure (only works for problems with gases) ○ Three ways to change the pressure of a system ■ Add or remove a gaseous reactant or product (Change in concentration) ■ Add an inert gas not involved in reaction (While it increases the total pressure, has no effect on the concentrations or partial pressures of the reactants or products) ■ Change the volume of the container ○ Increase pressure ■ Volume decreases ■ → more collisions ■ Moves towards the side with fewer collisions ■ Go to the side with smaller number of moles of gas (stoichiometry) ○ Decrease pressure ■ Volume increases ■ → lower number of collisions ■ Moves towards the side with more collisions ■ Go to the side with the greater number of moles of gas (stoichiometry) ● Change in Temperature ○ The value of K changes with temperature but can use Le Châtelier’s Principle to predict the direction of the change. ○ Treat heat as a reactant or product ■ Exothermic → heat is a product ■ Endothermic → heat is a reactant ○ Direction of shift is then the same as in a change of concentration ● Catalyst ○ NO EFFECT on K ○ Establishes equilibrium faster ● Solid ○ NO EFFECT on K ● Concept Check ○ For the system N2O3(g) ←→ NO(g) + NO2(g) ΔH = +39.7 kJ, predict the effect that each of the following changes will have on the position of equilibrium (i.e. will it shift to the right, left, or unchanged). Change decrease the container size adding NO lower the temperature decrease the external pressure increase the temperature add Helium gas Shift [N2O3] [NO] [NO2] Value of Keq Reaction Quotient and Le Châtelier’s Principle ● If Q < K, the system is not at equilibrium ○ Products are too small and the reactants are too big. ○ Shifts to the right towards equilibrium to make products and reduce reactants ○ Reactants → products to make Q = K at equilibrium. ● If Q = K, the system is at equilibrium ● If Q > K, the system is not at equilibrium ○ Reactants are too small and products are too big ○ Shifts to the left towards equilibrium to make reactants and reduce products ○ Reactants ← Products to make Q = K at equilibrium ● Predicts what will happen given any set of data. EASY METHOD: Put on a number line and then draw a line from Q to K. The line is the direction that the reaction will have to shift to establish equilibrium. Q<K Q>K ● Concept Check ○ For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0x10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases: ■ [NH3] = 1.0 x 10-3; [N2] = 1.0 x 10-5; [H2] = 2.0 x 10-3 ■ [NH3] = 2.0 x 10-4; [N2] = 1.50 x 10-5; [H2] = 3.54 x 10-1 ■ [NH3] = 1.0 x 10-4; [N2] = 5.0; [H2] = 1.0 x 10-2 CNHDZ Q NDC Hz k a O s 3 i 0 10 3 5 1.0 10 l Q 2 2.0 10 3 Equilibrium positioneff 1.1 10 J 2 0 10 1.5 4 10 5 2 3.54 10 1 3 i 060 no shift
