UNIT 2 Understanding Chemical Constants

CHEM 1412 UNIT 2 Typically, chemical reactions involve placing the reactants on the left side of formula and products on right side. Reactants → Products . Reversible reactions, proceed in both forward (left to right) and reverse (right to left) directions when the system (reaction) is at equilibrium: the rates of the forward and reverse reactions are equal. [Also at equilibrium, the concentrations of the reactant and product species remain constant over time] -Although a chemical reaction may react quickly/fast that doesn't necessarily mean that there will be a complete conversion of reactants to products: A reaction is only complete when stoichiometric amounts of reactants are mixed and, after reaction, there are no measurable amounts of reactants left. Chemical equilibria are dynamic; a reaction at equilibrium has not “stopped,” but is proceeding in the forward and reverse directions at the same rate. [The concentration for recants and products do not have to be same, only the rate of reactions] N2O4(g)⇌2NO2(g) Note the difference in graph between concentration vs rates. Remember that for elementary reactions, rate laws for the forward and reverse reactions may be derived directly from the reaction stoichiometry 1. As the reaction begins (t = 0), the concentration of the reactant(s) is finite and that of the product is zero, so the forward reaction proceeds at a finite rate while the reverse reaction rate is zero. 2. As time passes, the reactant(s) is consumed and its concentration falls (wont decrease to 0 but will eventually become constant) , while the product(s) is produced and its concentration increases. The decreasing concentration of the reactant slows the forward reaction rate, and the increasing product concentration speeds the reverse reaction rate. 3. This process continues until the forward and reverse reaction rates become equal (the concentration of the products become constant when the concentration of reactants become constant), at which time the reaction has reached equilibrium, as characterized by constant concentrations of its reactants and products. Physical changes, such as phase transitions, are also reversible and may establish equilibria: Br2(l)⇌Br2(g) : When liquid bromine is added to an empty container and the container is sealed, the forward (vaporization) will commence and continue at a roughly constant rate as long as the exposed surface area of the liquid and its temperature remain constant. As increasing amounts of gaseous bromine are produced, the rate of the reverse process (condensation) will increase until it equals the rate of vaporization and equilibrium is established. A+B ⇌C+D 1. ratef =kf [A][B] and rateb=kb[C][D] (f=forward, b=backward) 2. ratef =rateb Kc is referred to as equilibrium constant while the entire equation plus equilibrium constant is referred to as equilibrium constant expression or law of mass action. If the reaction is a multistep then you must multiply each concentration by the power of their coefficient. When substituting values into equilibrium constant expression, the concentrations must be equilibrium concentrations. The equilibrium constant is constant as long as temperature is constant (Because forward and backward rate constants are dependent on temperature so is Kc) A+B is exothermic because ΔH<0 C+D is endothermic because ΔH>0 [(Ea)f - (Ea)b] f=forward, b=backward A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. Homogeneous equilibria take place in solutions and therefore aqueous and gas phases. A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples: We do not include pure solids or pure liquids in the Kc expression because the concentration of a pure solid or pure liquid is constant: 1. The concentration of pure solid and liquid is already incorporated into Kc expression. The relative concentrations of reactants and products in equilibrium systems vary depending on equilibrium constant reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product (more products compared to reactants) Terminology: equilibrium lies to the right, reaction is complete A small K indicates the reaction achieves equilibrium after very little reactant has been converted (more reactant than products). “equilibrium lies to the left, reaction is incomplete” If the equilibrium constant is neither too large or small, then there are comparable amounts of products and reactants at equilibrium. We do not use units for Kc ! The “c” in the equilibrium constant means the concentration are in moles per liter. 1 mol = 22.4 L @ STP For certain gas phase reactions (reactants/products can either all be gasses, or some be gasses) it can be easier to find equilibrium constants when expressed in terms of partial pressures of gases, Kp. p=Partial pressure in atm [These reactions take place in a vessel or tank, and although we could still use concentration, finding the molarity, volume, liter of each gas over time can be time consuming] Kp is constant as long as temperature is constant Just like Kc, Kp has the same expression but with pressure instead of concentration. When substituting values into equilibrium constant expression, the pressures must be equilibrium pressures. Also the pressures must always be in atm. 1 atm=760 torr. 1 atm=760 mmHg. 1 atm=101.325 kPA. 1 atm=14.7 psi Ideal Gas Law: PV=nRT (dividing the volume on both sides yields a proportional relationship between pressure and molarity. If molarity increases then pressure increases) P=MRT. The relationship between Kc and KP is: (The equation requires factoring out MRT) Δn is the difference in the molar amounts of product and reactant gases. (Not solid or liquid but gasses). Temperature must be in K. R=0.00821. Kc = KP when n=0 The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient (Qc). Similar to the Kc and KP expression the reaction quotient follows the same formula; however any molar concentration can be substituted not just equilibrium concentration. At a given temperature, the reaction quotient for a system at equilibrium is constant. The constant value of Q exhibited by a system at equilibrium is called the equilibrium constant, K: K≡Q at equilibrium The numerical value of Q varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. Consider the oxidation of sulfur dioxide: 2SO2(g)+O2(g)⇌2SO3(g) At t=0 sec , the reaction begins with a mixture of reactants only, Q is initially equal to zero As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Qc), product concentration increases (as does the numerator of Qc), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Qc. If the reaction begins with only product present, the value of Qc is initially undefined (immeasurably large, or infinite): In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Qc decrease with time, the reactant concentrations and the denominator of Qc increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium. If Qc > Kc There is too much product. Net reaction to the left until Qc = Kc Qc < Kc There is too much reactant. Net reaction to the right until Qc = Kc Qc = Kc. The system is at equilibrium. If the system is not at equilibrium, it will try to reach equilibrium. Many equilibria systems involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. ~ Because the law of mass action allows for a straightforward derivation of Kc expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. 1. Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation. A⇌B Kc=[B]/[A] B⇌A Kc’=[A]/[B] Kc’=1/Kc Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor: A⇌B Kc =[B]/[A] xA ⇌ xB Kc’ =[B]x/[A]x If two or more chemical reactions are added together to give another chemical reaction, the equilibrium constant for the chemical reaction is the product of the equilibrium constants of the chemical equations added. The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies: In some instances we may need to combine two different manipulation methods. Study how the equilibrium expression of each reaction, ultimately yields Kc The initial concentration of a reactant/product affects the equilibrium concentrations for a chemical system You can't go directly from the initial concentration to Kc. Equilibrium computations are essential in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect. The acronym ICE (Initial Concentration, Change in Concentration, Equilibrium Concentration) is a mathematical approach used to solve Equilibria Calculations. We must have all the data in molarity (M) If we are not explicitly given the initial concentration of a reactant/product, we can assume its 0. Many equilibrium calculations require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometric factors of the reaction. A reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry 2NH3(g)⇌N2(g)+3H2(g) If the nitrogen concentration increases by an amount x: Δ[N2]=+ x The corresponding changes in the other species concentrations are In the off chance you have a solid/liquid reactant or product do not fill any data out for that specific section You can only simplify with a square root if both numerator and denominator are squared. Given initial concentrations and an equilibrium constant. A four-step approach is typically useful: 1. 2. 3. 4. Identify the direction in which the reaction will proceed to reach equilibrium (using Qc equation) or if there is only on recta Develop an ICE table. Calculate the concentration changes and, subsequently, the equilibrium concentrations. Confirm the calculated equilibrium concentrations. Because only the product is present initially Qc = 0 and the reaction will proceed to the left. Also 0.01726 is a non-sense answer b/c if we substitute it into the equilibrium concentration formula of 2 NO2 we will get a negative answer. At 668 K, 0.350 mol CO, 0.350 mol C12, and 0.485 mol COC12 are placed in a 10.0-L vessel. What are the equilibrium concentrations of each gas? For the equation shown, Kc = 1.23 x 103 CO(g)+Cl2(g)⇌ COCl2 (g) Remember that a system is at equilibrium when the forward and reverse reactions taking place at equal. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. Le Châtelier’s principle: if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium. Changes in concentration of a reactant or product species is a stress that can shift an equilibrium. (Note the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action) Consider the equilibrium reaction: H2(g)+I2(g)⇌2HI(g) Kc=50.0 at 400°C The rate laws for the forward and reverse reactions are Forward: H2(g)+I2(g)→2HI(g) ratef=kf[H2]m[I2]n Reverse: 2HI(g)→H2(g)+I2(g) rater=kr[HI]x If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction: ratef>rater . The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction: ratef Kc and the equilibrium will shift left Adding or removing some of a pure liquid or pure solid has no effect on the equilibrium (as long as some of the liquid or solid is left) because the concentration of a pure liquid or pure solid does not change. For gas-phase equilibria, because the partial pressure P of an ideal gas is proportional to its molar concentration M, changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. The pressures (concentrations) of species in a gas-phase equilibrium can be changed by increasing or decreasing the volume. Increasing the pressure by decreasing the volume shifts the equilibrium toward the side that has the smaller # moles of gas. (KEY WORD: GAS) For instance, If the volume occupied by an equilibrium mixture is decreased by a factor of 3, the partial pressures of all three species will be increased by a factor of 3 Given the reaction: 2NO2(g) ⇌ 2 NO(g) + O2(g) Since the change in volume/partial pressure results in a reaction quotient greater than the equilibrium constant, the equilibrium will shift left. If the # moles of gas on each side is the same, then changing the pressure by changing the volume has no effect. If the volume occupied by an equilibrium mixture is decreased by a factor of 3, the partial pressures of all three species will be increased by a factor of 3 Given the reaction: H2(g) + I2(g) ⇌ 2HI(g) Decreasing the pressure by increasing the volume shifts the equilibrium toward the side that has the larger # moles of gas. Adding an inert gas at a constant volume can change the pressure. However, an inert gas does not react with either reactants or products. Adding an inert gas at a constant volume has no effect on the equilibrium. Scientists may add an inert gas to a chemical equation to increase the pressure of a system without causing a shift in the equilibrium. Effect of a Change in Temperature-When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant. (It also shifts the reaction depending on whether the reaction is exothermic or endothermic) The equilibrium constant is seen to be a mathematical function of the rate constants for the forward and reverse reactions. Since the rate constants vary with temperature as described by the Arrhenius equation, is stands to reason that the equilibrium constant will likewise vary with temperature (assuming the rate constants are affected to different extents by the temperature change). Predicting the shift an equilibrium will experience in response to a change in temperature is most conveniently accomplished by considering the enthalpy change of the reaction. Given an exothermic reaction where heat is given off 2H2 + O2 ⇌ 2H2O ΔH=-572 kj 2H2 + O2 ⇌ 2H2O+572 kj 2H2 + O2 ⇌ 2H2O+ heat Raising the temperature of the system is akin to increasing the amount of heat, and so the equilibrium/reaction will shift to the left. If the temperature is decreased, the reaction will shift to the right. (Although heat and temperature are not the same thing, but we can think of a decrease in temperature as decrease in heat.) Endothermic Reactions: Heat is a reactant (heat is required) 3O2 ⇌ 2O3 ΔH=286 kj 286kj+3O2 ⇌ 2O3 heat +3O2 ⇌ 2O3 If the temperature is increased, the reaction will shift to the right. If the temperature is decreased, the reaction will shift to the left A catalyst is a substance that enables a reaction to proceed via a different mechanism with an accelerated rate. The catalyzed reaction mechanism involves a lower energy transition state than the uncatalyzed reaction, resulting in a lower activation energy, Ea, and a correspondingly greater rate constant. To discern the effect of catalysis on an equilibrium system, the lowered transition state energy of the catalyzed reaction results in lowered activation energies for both the forward and the reverse reactions. Consequently, both forward and reverse reactions are accelerated, and equilibrium is achieved more quickly but without a change in the equilibrium constant. Summary: Adding a catalyst to an equilibrium system has no effect. A catalyst speeds up both the forward and reverse reactions. The equilibrium constant is not affected.