Chemical Equilibrium Basics Reading: Chapter 16 of Petrucci, Harwood and Herring (8th edition) Problem Set: Chapter 16 questions 25, 27, 31, 33, 35, 43, 71 York University CHEM 1001 3.0 Chemical Equilibrium - 1
Chemical Equilibrium All chemical reactions are reversible; that is, if a reaction can occur in one direction: CO(g) + 2H 2 (g) CH 3 OH(g) then the reverse reaction can also occur: CH 3 OH(g) CO(g) + 2H 2 (g) This applies to both elementary and overall reactions. As a reaction proceeds, the rate of the forward reaction decreases and the reverse reaction increases until they are equal. This balance is called equilibrium. York University CHEM 1001 3.0 Chemical Equilibrium - 2
Approach to Equilibrium - Example Elementary reactions: (1) N 2 O 4 (g) 2NO 2 (g) (forward) (2) 2NO 2 (g) N 2 O 4 (g) (reverse) rate(1) = k 1 [N 2 O 4 ] rate(2) = k 2 [NO 2 ] 2 At t=0 let [N 2 O 4 ] > 0 and [NO 2 ] = 0, then rate(1) > rate(2). As time passes, [N 2 O 4 ] decreases and [NO 2 ] increases. Rate(1) decreases and rate(2) increases until they are equal. When rate(1) = rate(2), we have equilibrium. York University CHEM 1001 3.0 Chemical Equilibrium - 3
Approach to Equilibrium - continued When rate(1) = rate(2), we have equilibrium. Therefore: k 1 [N 2 O 4 ] = k 2 [NO 2 ] 2 = equilibrium constant At equilibrium: Concentrations remain constant. Both reactions continue to occur. This is a dynamic equilibrium. York University CHEM 1001 3.0 Chemical Equilibrium - 4
Constant Conditions In a static system, nothing happens. Example: Neon in the Earth's atmosphere is neither produced nor consumed. At steady-state, the rate of production is balanced by the rate of consumption. Example: Oxygen in the atmosphere is produced by photosynthesis and consumed by respiration. Dynamic equilibrium is a steady-state in which the consumption process is the exact opposite of the production process. Example: N 2 O 4 (g) 2NO 2 (g). People sometimes use equilibrium for all three of these. This is unacceptable in chemistry. York University CHEM 1001 3.0 Chemical Equilibrium - 5
Examples of Equilibrium In a closed container, H 2 O(l) will come into equilibrium with H 2 O(g). (But H 2 O(g) and H 2 O(l) are not in equilibrium in the atmosphere.) (a) I 2 in H 2 O(l) (yellow). (b) I 2 equilibrium between H 2 O(l) (top) and CCl 4 (l) (bottom, purple). Most I 2 is in the CCl 4. York University CHEM 1001 3.0 Chemical Equilibrium - 6
Evidence of Dynamic Equilibrium AgI(s) Ag + (aq) + I - (aq) Make up a saturated solution with excess solid AgI. Add a small amount of radioactive 131 I - (aq) to the solution. The radioactivity will gradually appear in the solid AgI. York University CHEM 1001 3.0 Chemical Equilibrium - 7
The Equilibrium Constant Elementary reaction: aa + bb +... gg + hh +... At equilibrium: k f [A] a [B] b... = k b [G] g [H] h... so = equilibrium constant This applies to every step in a reaction mechanism. For the overall reaction, K C has the same form: products in the numerator, reactants in the denominator multiply concentrations raised to powers powers are the stoichiometric coefficients York University CHEM 1001 3.0 Chemical Equilibrium - 8
Equilibrium Constant vs. Rate Laws These are similar, but different. Both involve products of concentrations. The rate law may or may not include reactants, products, and other species. The equilibrium constant includes all reactants and products, but only reactants and products. In the equilibrium constant, the powers are the stoichiometric coefficients. Not so for rate laws (except elementary reactions). York University CHEM 1001 3.0 Chemical Equilibrium - 9
Equilibrium Constant - Examples CO(g) + 2H 2 (g) CH 3 OH(g) ½CO(g) + H 2 (g) ½CH 3 OH(g) CH 3 OH(g) CO(g) + 2H 2 (g) The correct expression for the equilibrium constant depends on how the reaction is written. York University CHEM 1001 3.0 Chemical Equilibrium - 10
Approach to Equilibrium CO(g) + 2H 2 (g) CH 3 OH(g) Three different initial conditions. Equilibrium reached in each case. Different final conditions. All satisfy the equilibrium condition. York University CHEM 1001 3.0 Chemical Equilibrium - 11
Final State - Equilibrium 14.5 14.4 14.5 K C York University CHEM 1001 3.0 Chemical Equilibrium - 12
Beyond Concentrations In reality, rate laws do not always give the exact dependence of reaction rate on concentration. Also, equilibrium constants using concentrations, K C, are not constant (sometimes not nearly so). The expressions we use here are approximations. The thermodynamic equilibrium constant, K eq, is defined using activities instead of concentrations. We use concentrations as approximations to activities. York University CHEM 1001 3.0 Chemical Equilibrium - 13
Activities Thermodynamic equilibrium constants use activities. Activities are usually defined to be dimensionless. Thermodynamic equilibrium constants are dimensionless. Activities of pure solids and pure liquids are unity (1.0...). For dilute solutions: solute activities approximately equal concentrations, the solvent activity is approximately unity. For ideal gases, activities equal partial pressures. We will treat solutions as dilute and gases as ideal. York University CHEM 1001 3.0 Chemical Equilibrium - 14
Equilibrium Constant - More Examples 2H 2 (g) + O 2 (g) 2H 2 O(l) CaCO 3 (s) CaO(s) + CO 2 (g) 2H 2 O(l) H 3 O + (aq) + OH - (aq) K eq [H 3 O + ][OH - ] HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) York University CHEM 1001 3.0 Chemical Equilibrium - 15
Equilibrium Constants Involving Gases CO(g) + 2H 2 (g) CH 3 OH(g) For this reaction at 483 K, we found that: The thermodynamic equilibrium constant is: These are not the same. York University CHEM 1001 3.0 Chemical Equilibrium - 16
Converting between K C and K P From the ideal gas law: PV = nrt So, for CO: P CO = (n CO /V)RT = [CO]RT So: and K C = 14.5 at T = 483 K. R = 0.08206 L atm mol -1 K -1 So RT = 39.6 L atm mol -1 and K P = 9.23 10-3 York University CHEM 1001 3.0 Chemical Equilibrium - 17
Magnitudes of Equilibrium Constants Equilibrium constants have a very wide range of values. Equilibrium constants can change enormously with changes in temperature. York University CHEM 1001 3.0 Chemical Equilibrium - 18
Significance of Magnitude of K 2H 2 (g) + O 2 (g) 2H 2 O(l) = 1.4 10 83 at 298 K If p O2 = 0.21 bar, what is the equilibrium p H2? P H2 = 5.8 10-42 bar < 1 molecule in the entire atmosphere York University CHEM 1001 3.0 Chemical Equilibrium - 19
Significance of K - continued All reactions are reversible. So there are always some reactants left at equilibrium. A small K means that a reaction can not proceed in the direction written (to the right). It is unfavorable. A reasonably large K means that the reaction may proceed. It is favorable. But it might be very slow. If K is so large that the equilibrium concentration of a reactant is negligible, we say that the reaction "goes to completion". If K is very small, the reverse reaction goes to completion. York University CHEM 1001 3.0 Chemical Equilibrium - 20
Significance of K - example N 2 (g) + O 2 (g) 2NO(g) K P = 4.7 10-31 at 298 K The reaction can not proceed in the direction written. At 298 K, no significant NO can form from air. N 2 (g) + O 2 (g) 2NO(g) K P = 1.3 10-4 at 1800 K At 1800 K, significant NO may form. It turns out that the rate is fast at high temperatures. 2NO(g) N 2 (g) + O 2 (g) K P = 2.1 10 30 at 298 K At 298 K, the NO may convert completely back to N 2 and O 2. But the rate is very slow. York University CHEM 1001 3.0 Chemical Equilibrium - 21
The Equilibrium Constant - Summary All reactions are reversible. When the rates of the forward and reverse reactions are equal, we have equilibrium. Equilibrium is a dynamic state, both forward and reverse reactions continue to occur. The equilibrium constant, K, gives the condition that must obtain at equilibrium. K can be used to determine the direction in which a reaction is allowed to proceed. But the reaction might be very slow. K is expressed in terms of activities. Using concentrations or partial pressures is an approximation. York University CHEM 1001 3.0 Chemical Equilibrium - 22
Predicting the Direction of Change CO(g) + 2H 2 (g) CH 3 OH(g) If we start with only CO and H 2, the reaction will initially proceed to the right. If we start with only CH 3 OH, the reaction will initially proceed to the left. What if we start with a mixture of reactants and products? In what direction will the reaction proceed? The answer depends on where the reaction stops - so it depends on the equilibrium constant. But how? York University CHEM 1001 3.0 Chemical Equilibrium - 23
The Reaction Quotient The reaction quotient, Q, is defined to be the same as the equilibrium constant, but using initial conditions instead of equilibrium conditions. Q is used to decide the direction of change from a specific set of initial conditions. If Q < K, the reaction will proceed to the right. If Q > K, the reaction will proceed to the left. If Q = K, the reaction is at equilibrium. Q C is defined using concentrations; Q P is defined using partial pressures. Must be consistent with K C or K P. York University CHEM 1001 3.0 Chemical Equilibrium - 24
Using the Reaction Quotient CO(g) + 2H 2 (g) CH 3 OH(g) K C = 14.5 at 483 K Initial conditions: [CO] = 1.0 M, [H 2 ] = 0.30 M, [CH 3 OH] = 2.5 M? In which direction will the reaction proceed? Q C > K C The reaction proceeds to the left. York University CHEM 1001 3.0 Chemical Equilibrium - 25
Using the Reaction Quotient - continued What happens as the reaction proceeds? CO(g) + 2H 2 (g) CH 3 OH(g) We have Q C > K C, so the reaction proceeds to the left. In other words, the rate of the reverse reaction is greater than the rate of the forward reaction. Thus: [CH 3 OH] decreases, [CO] increases, [H 2 ] increases Q C decreases until Q C = K C When Q C = K C, equilibrium is reached. The forward and reverse rates are equal. York University CHEM 1001 3.0 Chemical Equilibrium - 26
More on Using the Reaction Quotient What if conditions change? CO(g) + 2H 2 (g) CH 3 OH(g) At equilibrium, Q C = K C. What if we then add more CO? Adding CO makes Q C smaller. Then Q C < K C. The reaction proceeds to the right. As the reaction proceeds [CH 3 OH], [CO], [H 2 ], Q C. Eventually Q C = K C and we have a new equilibrium. All concentrations are different from our first equilibrium. York University CHEM 1001 3.0 Chemical Equilibrium - 27
Le Châtelier s Principle When an equilibrium system is subjected to a change, it responds by attaining a new equilibrium that partially offsets the impact of the change. Input (Change) increase inside pressure increase outside pressure increase in temperature increase in concentration Response increase volume reduce volume absorb heat shift away from that species (as directed by Q) York University CHEM 1001 3.0 Chemical Equilibrium - 28
Le Châtelier s Principle - Example Closed container with liquid water and water vapor in equilibrium: H 2 O(l) H 2 O(g) Increase external pressure: Volume decreases, some H 2 O(g) changes to H 2 O(l) (shift to left). Add He(g) to increase internal pressure: Volume increases, some H 2 O(l) changes to H 2 O(g) (shift to right). Increase temperature: Some H 2 O(l) changes to H 2 O(g) (shift to right absorbs heat). York University CHEM 1001 3.0 Chemical Equilibrium - 29
Adding/Removing Reactants or Products The equilibrium will shift to partially offset the change. CO(g) + 2H 2 (g) CH 3 OH(g) Change: Response: Change: Response: Change: Response: adding CH 3 OH equilibrium shifts to left adding CO equilibrium shifts to right removing CH 3 OH equilibrium shifts to right York University CHEM 1001 3.0 Chemical Equilibrium - 30
Effect of Changing the Volume The equilibrium will shift to partially offset the change. CO(g) + 2H 2 (g) CH 3 OH(g) A shift to the right reduces the pressure (replaces 3 moles of gas with 1 mole of gas). A shift to the left increases the pressure. Change: Response: Change: Response: decreased volume (increased pressure) equilibrium shifts to right increased volume (decreased pressure) equilibrium shifts to left York University CHEM 1001 3.0 Chemical Equilibrium - 31
Effect of Inert Gas CO(g) + 2H 2 (g) CH 3 OH(g) (1) Add He(g) while keeping the volume fixed. No effect on the reactants or products. No effect on the equilibrium. (2) Add He(g) while keeping the total pressure fixed. Volume increases. Partial pressures of reactants and products decrease. Equilibrium shifts to left. York University CHEM 1001 3.0 Chemical Equilibrium - 32
Exothermic and Endothermic Reactions Breaking chemical bonds requires energy. Forming bonds releases energy. Chemical reactions may absorb or release energy. This energy is often transferred as heat. Exothermic reactions release energy as heat: CH 4 + 2O 2 CO 2 + 2H 2 O H 2 SO 4 (l) + 2H 2 O(l) 2H 3 O + (aq) + SO 4 2- (aq) Endothermic reactions absorb energy as heat: H 2 O(l) H 2 O(g) NaOH(s) Na + (aq) + OH - (aq) York University CHEM 1001 3.0 Chemical Equilibrium - 33
Effect of Change in Temperature The equilibrium will shift to partially offset the change. For endothermic reactions, the equilibrium constant increases when the temperature goes up. For exothermic reactions, the equilibrium constant decreases when the temperature goes up. Examples: T (K) K P CaCO 3 (s) CaO(s) + CO 2 (g) 298 1.0 10-23 This reaction is endothermic. 1200 1.0 CH 4 + 2O 2 CO 2 + 2H 2 O The equilibrium constant will be smaller at higher T. York University CHEM 1001 3.0 Chemical Equilibrium - 34
Effect of a Catalyst A catalyst is a substance that increases the rate of a chemical reaction without being consumed by the reaction. A catalyst: Does not change the position of equilibrium. Can decrease the time it takes to reach equilibrium. Will not make an unfavorable reaction occur. Can help a favorable reaction to proceed faster. York University CHEM 1001 3.0 Chemical Equilibrium - 35
The Direction of Change - Summary Equilibrium is not achieved for just one set of concentrations. It is achieved for any concentrations that satisfy the equilibrium constant. The reaction quotient, Q, gives the direction of change from a specific set of initial conditions. Le Châtelier s Principle: An equilibrium system responds to a change in conditions by partially offsetting the change. The effect of changes in amount, pressure, or volume can also be determined by using Q. The effect of temperature on K depends on whether the reaction is endothermic or exothermic. York University CHEM 1001 3.0 Chemical Equilibrium - 36
Equilibrium Calculations Many types of reactions rapidly reach equilibrium: acid/base reactions salts entering solution formation of complex ions We need to learn how to calculate: equilibrium constants from composition data composition from equilibrium constants Some of these calculations require the use of stoichiometry and mass balance. York University CHEM 1001 3.0 Chemical Equilibrium - 37
Determining K C from Equilibrium Data Example: N 2 O 4 (g) 2NO 2 (g) At equilibrium at 25 C, a 3.00 L vessel contains 7.64 g of N 2 O 4 and 1.56 g of NO 2. Find K C. Solution: Calculate the concentrations of reactants and products: [N 2 O 4 ] = 0.0277 M [NO 2 ] = 0.0113 M Evaluate the equilibrium constant: K C = [NO 2 ] 2 / [N 2 O 4 ] = 4.61 10-3 York University CHEM 1001 3.0 Chemical Equilibrium - 38
K P from Initial and Final Amounts of a Substance Example: 2SO 3 (g) 2SO 2 (g) + O 2 (g) Initially, 0.0200 mol of SO 3 are put into an evacuated 1.52 L vessel. At equilibrium at 900 K, there are 0.0142 mol SO 3. Find K P. Solution: (1) Calculate equilibrium moles of all reactants and products. (2) Calculate partial pressures of all species. (3) Evaluate the equilibrium constant: K P = P 2 SO2P O2 /P 2 SO3 York University CHEM 1001 3.0 Chemical Equilibrium - 39
K P from Initial and Final Amounts - continued (1) The ICE Table Reaction 2SO 3 (g) 2SO 2 (g) + O 2 (g) Initial 0.0200 mol 0.0000 mol 0.0000 mol Change -0.0058 mol 0.0058 mol 0.0029 mol Equilibrium 0.0142 mol 0.0058 mol 0.0029 mol (2) Partial pressures: 0.690 atm 0.282 atm 0.141 atm This used the ideal gas law: P = nrt/v = (48.6 atm mol -1 ) n (3) Evaluate K P = P 2 SO2P O2 /P 2 SO3 K P = 0.0235 = 0.024 (2 significant figures) York University CHEM 1001 3.0 Chemical Equilibrium - 40
Variations in the ICE Table The ICE table is based on moles. Other quantities (concentrations, partial pressures) may be used if they are strictly proportional to moles. Examples: Concentration = may be used if V is fixed. Partial pressure may be used if T and V are fixed: York University CHEM 1001 3.0 Chemical Equilibrium - 41
Equilibrium Partial Pressures from K P Example: NH 4 HS(s) NH 3 (g) + H 2 S(g) At 25 C, K P = 0.108 Excess solid NH 4 HS is placed in an evacuated flask. Find the partial pressures of NH 3 and H 2 S at equilibrium. Solution: K P = P NH3 P H2S Reaction NH 4 HS(s) NH 3 (g) + H 2 S(g) Initial - 0.0000 atm 0.0000 atm Change - x atm x atm Equilibrium - x atm x atm x 2 = K P = 0.108 P NH3 = 0.329 atm and P H2S = 0.329 atm York University CHEM 1001 3.0 Chemical Equilibrium - 42
Equilibrium from Initial Conditions and K C Example: N 2 O 4 (g) 2NO 2 (g) At 25 C, K C = 4.61 10-3 0.0240 mole of N 2 O 4 is placed in a 0.372 L flask. Find the moles of N 2 O 4 at equilibrium. Solution: K C = [NO 2 ] 2 / [N 2 O 4 ] Reaction N 2 O 4 (g) 2NO 2 (g) Initial 0.0240 mol 0.0000 mol Change -x mol +2x mol Equilibrium (0.0240-x) mol +2x mol K C = (2x/V) 2 / ((0.0240-x)/V) York University CHEM 1001 3.0 Chemical Equilibrium - 43
Solving for the Unknown Concentration From the previous slide: VK C (0.0240-x) = 4x 2 Method 1: If only a little N 2 O 4 dissociates, then x 0.0240 0.0240-x 0.0240 4x 2 0.0240VK C So x 0.00321 and n N2O4 = 0.0240-x = 0.0208 mol Check: x/0.0240 = 0.13, so assumption may be OK. Method 2: Use the result from Method 1 to improve the approximation: 0.0240-x 0.0240-0.0032 = 0.0208 So 4x 2 0.0208VK C x 0.00299 and n N2O4 = 0.0210 mol Can repeat until a constant value is reached. York University CHEM 1001 3.0 Chemical Equilibrium - 44
Solving for Unknown Concentration - continued Method 3: Use quadratic equation. If 0 = ax 2 + bx + c the 0 = 4x 2 + VK C x - 0.0240VK C a = 4 b = VK C c = -0.0240VK C x = 0.00300 and n N2O4 = 0.0210 mol Check result: K C = [NO 2 ] 2 / [N 2 O 4 ] [N 2 O 4 ] = n N2O4 /V = 0.0564 M, [NO 2 ] = 2x/V = 0.016 M K C = 4.61 10-3 which is what it should be York University CHEM 1001 3.0 Chemical Equilibrium - 45
Initial Concentrations Given for Multiple Species Example: V 3+ (aq) + Cr 2+ (aq) V 2+ (aq) + Cr 3+ (aq) [V 3+ ] = [Cr 2+ ] = 0.0100 M [V 2+ ] = [Cr 3+ ] = 0.150 M K C = 720 Find the equilibrium concentrations. Solution: Decide in which direction the reaction will proceed: Since Q C < K C, the reaction will proceed to the right. York University CHEM 1001 3.0 Chemical Equilibrium - 46
Initial Concen. for Multiple Species - continued Reaction V 3+ (aq) + Cr 2+ (aq) V 2+ (aq) + Cr 3+ (aq) Initial 0.0100 M 0.0100 M 0.150 M 0.150 M Change -x M -x M x M x M Equil. (0.0100-x) (0.0100-x) (0.150+x) (0.150+x) x = 4.25 10 --3 [V 3+ ] = [Cr 2+ ] = 0.0057 M [V 2+ ] = [Cr 3+ ] = 0.154 M York University CHEM 1001 3.0 Chemical Equilibrium - 47
Equilibrium Calculations - Summary To calculate equilibrium constants from composition data: Calculate equilibrium amounts all species. Evaluate the equilibrium constant. To calculate equilibrium composition from equilibrium constants and initial composition: Construct the ICE table to get the all the equilibrium amounts in terms of a single unknown, x. Substitute into the expression for K, then solve for x. Evaluate all the equilibrium amounts. Amounts can be moles, concentrations, or partial pressures. York University CHEM 1001 3.0 Chemical Equilibrium - 48