The graph represents changes in concentrations with time for the reaction H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) when equimolar quantities of H 2 O(g) and CO(g) are mixed. H 2 O(g) + CO(g) H 2 (g) + CO 2 (g)
(a) H 2 O and CO are mixed in equal numbers and begin to react (b) to form CO 2 and H 2. After time has passed, equilibrium is reached (c) and the numbers of reactant and product molecules then remain constant over time (d).
H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) or X Y At equilibrium, X forms Y at the same rate as Y forms X. Therefore, the concentration of [X] and [Y] will remain constant. Also, when equilibrium is reached, the rates of the forward reaction will equal the rates of the reverse reaction.
The Equilibrium Constant, K eq (also called the mass action expression) Two methods for describing the equilibrium of a reaction. 1.K c - describes the equilibrium of a reaction where the concentrations of the materials is known. 2.K p - describes the equilibrium of a gaseous reaction using partial pressures instead of concentrations.
K c c = concentration aA + bB cC + cD Coefficients in the chemical equation become exponents in the equilibrium constant expression. Include only substances in the gas or aqueous phase. Solid’s and liquid’s concentrations do not change during a chemical reaction.
Write the equilibrium expression for the reaction below. Ni (s) + 4CO (g) Ni(CO) 4 (g)
Write the equilibrium expressions for the following reactions: a.H 2 (g) + I 2 (g) 2HI (g) b.2NH 3 (g) + 3 CuO(s) 3H 2 O (g) + N 2 (g) + 3Cu (s) c.NH 4 Cl (s) NH 3 (g) + HCl (g)
K p p= pressure Equilibrium is described in terms of the partial pressures of the reactants and products. 2CO 2 (g) 2CO (g) + O 2 (g)
Write the equilibrium constant for the following mixture of gases at equilibrium. 2NO (g) + O 2 (g) 2NO 2 (g)
The Relationship Between K c and K p Where n = (total moles of gaseous products)-(total moles of gaseous reactants) R= ideal gas law constant (.08206 L atm / mol K T in Kelvin
Example: A 3:1 starting mixture of hydrogen, H 2, and nitrogen, N 2, comes to equilibrium at 500 o C. The mixture at equilibrium is 35.06% NH 3, 96.143% N 2 and 0.3506% H 2 by volume. The total pressure in the reaction vessel was 50.0 atm. What is the value of K p and K c for this reaction? K p = K c =
Example: 2NO (g) + Cl 2 (g) 2NOCl (g) In the above reaction, the total pressure of the mixture of gases at equilibrium is 1.55 atm. The percentages of each gas in the mixture areas follows: NOCl = 77.4%, NO = 3.20%, and Cl 2 = 19.4%. Calculate K p and K c for the reaction.
Interpretation of the Equilibrium Constant For K>1, we can state that the amount of product is greater than the amount of reactant. The equilibrium is favored to the right. Likewise, for K<1, the amount of reactants is much greater than the products. The equilibrium favored to the left.
The Reaction Quotient- Q There are times when a chemist may be given information about a chemical reaction that may or may not be at equilibrium. A Q calculation can be used to determine the direction the reaction is heading to reach equilibrium. If Q > K, then the reaction has too many products and is heading towards the reactant side. If Q < K, then the reaction has too many reactants and is heading towards the product side. If Q = K, then the reaction is at equilibrium.
Example: N 2 (g) + 3H 2 (g) 2NH 3 (g) In the reaction shown above, the value of K c at 500 0 C is 6.0 X 10 -2. At some point during the reaction, the concentrations of each material were measured. At this point, the concentrations of each substance were [N 2 ] = 1.0 X 10 -5 M, [H 2 ] = 1.5 X 10 -3 M, and [NH 3 ] = 1.5 X 10 -3 M. Calculate the value of Q, and determine the direction that the reaction was most likely to proceed when the measurements were taken.
Solving Problems When Not All Equilibrium Concentrations are Known H 2 (g) + I 2 (g) 2HI (g) When 2.00 mole each of hydrogen and iodine are mixed in an evacuated vessel, 3.50 mole of HI are produced. What is the value of the equilibrium constant, K c ? It is important to realize that not all of the reactants are converted to products in reversible reactions. Some reactants will remain at equilibrium.
Use the mole ratios to determine the change concentrations of the reactants.
H 2 (mol)I 2 (mol)2HI (mol) Start2.0 0 Change (∆) +3.50 Final3.50 Since we are in 1.0 L, 0.25 mol = 0.25 M
Use the final concentrations to calculate the value of K c
Example: Nitric oxide gas, NO, and oxygen gas, O 2, react to form the poisonout gas nitrogen dioxide, NO 2, in the reaction shown below: 2NO (g) + O 2 (g) 2NO 2 (g) 10.0 moles of NO and 6.00 moles of O 2 are placed into an evacuated 1.00 L vessel, where they begin to react. At equilibrium, there are 8.80 moles of NO 2 present. Calculate the value of K c, assuming that the temperature remains constant throughout the reaction. K c =
Determining K When Only Initial Concentrations Are Known H 2 (g) + I 2 (g) 2HI 0.500 mol of hydrogen and 0.500 mol of iodine are added to a 1.00 liter reaction vessel. The mixture is heated to 498 0 C and allowed to reach equilibrium according to the reaction shown above. At this temperature, K c = 49.7. What is the composition of the reaction mixture at equilibrium in this system?
H 2 (g) + I 2 (g) 2HI H 2 (mol)I 2 (mol)2HI (mol) Start0.50 0 Change (∆) Final
Example: 0.500 moles of NO gas are placed into a 1.00 L reaction vessel. The gas is heated to an extremely high temperature, where it decomposes according to the reaction shown below: 2NO (g) N 2 (g) + O 2 (g) At equilibrium, K c = 2.4 X 10 3. Determine the equilibrium composition of the mixture. 5.1 X 10 -3 M NO, 0.25 M N 2, 0.25 M O 2
Le Chatelier’s Principle When a system at equilibrium is disturbed, it will react in a way that minimizes the disturbance. Factors that can influence the position of an equilibrium. –1. Concentration (aq or g) –2. Temperature –3. Pressure (gases) –4. Addition of a catalyst
Example: Describe how the reaction below adjusts its equilibrium in each of the scenarios. CO (g) + 2H 2 (g) CH 3 OH (g) H 0 = -21.7 Kcal a.Some of the methanol vapor is condensed and removed from the reaction vessel. b.The pressure is increased by decreasing the volume of the reaction vessel. c.The temperature is increased.