Chemical Equilibrium PDF

Field of Pharmacy Sciences Bachelor of Pharmacy-PharmD (Clinical Pharmacy) Program Pharmaceutical Inorganic & Physical Lecture 8: Chemical Equilibrium Chemistry(PMC101) Dr. Amir Shaaban Farag Date :...

Field of Pharmacy Sciences Bachelor of Pharmacy-PharmD (Clinical Pharmacy) Program Pharmaceutical Inorganic & Physical Lecture 8: Chemical Equilibrium Chemistry(PMC101) Dr. Amir Shaaban Farag Date : Dec./04/2024  EQUILIBRIUM  In many chemical reactions both a forward and reverse reaction occur simultaneously. When the rate of forward and reverse reactions are equalized, the system is at equilibrium.  Not all reactions can reach equilibrium Consider a burning log: The products of the combustion (the smoke and ashes) can never reunite to form the reactants (the log and the oxygen)  By contrast some common processes are equilibrium systems, such as,  Formation of a saturated solution.  Vaporization of a liquid in a sealed container. Eg. SOLUBILITY EQUILIBRIUM RATE OF DISSOLVING = RATE OF PRCIPITATION RATE OF FORWARD PROCESS = RATE OF REVERSE PROCESS The amount of substance dissolved remains constant The system appears macroscopically static but it’s constantly  RECOGNIZING EQUILIBRIUM When equilibrium systems do exist:  They may be recognized by their apparent static nature (it looks as if nothing is happening).  All equilibrium systems must be closed, that is, nothing let in and nothing let out including energy.  Equilibrium appear unchanging when observed on a large scale (macroscopically) they are in constant change on the molecular level (microscopically). Reactants are continually forming products and products are continually forming reactants at equal rates. Law of mass action  correlates the rate of a chemical reaction and the concentration of the reactants, stating that, at a constant temperature and pressure, the rate of a chemical reaction is directly proportional to the product of the molar concentration of the reactants each raised to a power (equal to the corresponding stoichiometric coefficient in the balanced chemical equation). Rate equation for a reaction is: Rate = k x [reactants]n  When the rates of opposing processes are equal, equilibrium is established. Given the reaction: a A + b B  c C + d D If the rate of A reacting with B = the rate of C reacting with D, the system is at equilibrium CHEMICAL EQUILIBRIUM At equilibrium: Rate forward = Rate reverse kf [reactants] = kr[products] By rearranging the equation: kf / kr = [products] / [reactants] Dividing a constant by a constant gives another constant, so: Ke = [PRODUCTS] / [REACTANTS] at equilibrium for the given reaction, Ke = ([C]c x [D]d) / ([A]a x [B]b) equilibrium expression Notice that in this equation, the coefficients of the balanced equation serve as  EQUILIBRIUM EXPRESSION  For a system at equilibrium, the value of the equilibrium constant (Ke) remains constant unless the temperature is changed.  The concentrations of products and reactants may change as a result of “equilibrium shifts” but the ratio of products to reactants as the equilibrium expression remains unchanged.  Let’s find the equilibrium expression for the following reaction: 2 H2(g) + O2(g)  2 H2O(g) (double arrow  means equilibrium) Ke = [H2O]2 / ([H2]2 x [O2])  How about H2(g) + 1/2 O2(g)  H2O(g) ? The reaction is the same, but it is balanced with a different set of coefficients, therefore the equilibrium expression is different: Ke = [H2O] / ([H2] x [O2]1/2) The coefficients used in balancing the reaction alters the equilibrium expression and also the value of the equilibrium constant.  For eg., using the equilibrium equation: H2(g) + I2(g)  2 HI(g) Let assume the equilibrium concentrations [H2] = 0.094 M, [I2] = 0.094 and [HI] = 0.012 M. Equilibrium expression is: Ke= [HI]2 / ([H2] x [I2]) Ke = (0.012)2 / ((0.094) x (0.094)) = 0.016 If the equation was balanced as: ½ H2(g) + ½ I2(g)  HI(g) Equilibrium expression would be: Ke = [HI] / ([H2]1/2 x [I2]1/2 ) Ke = (0.012)/ ((0.094)1/2 x (0.094)1/2 ) = 0.127 HOW ARE Ke VALUES RELATED?  In this example as coefficients of balanced equation were halved, Ke value changed. But how? Notice that the square root (half power) of 0.016 = 0.127  If for some reason the equation was balanced as: 2 H2(g) + 2 I2(g)  4 HI(g), The constant would change to the square of the original value or 0.0162 and Ke would equal 0.064  When the balancing coefficients of an equation are changed by a factor, Ke value changes by that factor as a power of the original Ke How else may the original balanced equation be altered? Often equations are reversed (interchanging products with reactants). For eg.: H2(g) + I2(g)  2 HI(g) may be rewritten as: 2 HI(g)  H2(g) + I2(g)  The equilibrium expression for the first equation is: Ke = [HI]2 / ([H2] x [I2])  for the second it is: the reciprocal of the first. Ke = ([H2] x [I2]) / [HI]2 Therefore, when an equation is reversed, Ke becomes the reciprocal of the original Ke value. Ke (reverse) = 1 / K (forward)  HETEROGENEOUS EQUILIBRIUM SYSTEMS Heterogeneous and homogeneous are terms applied to the physical state (phases) of the components in equilibrium systems. A system consisting of all gases is homogeneous while including, for eg. both solids and gases is heterogeneous. H2(g) + I2(g)  2 HI(g) (homogeneous – all gases) H2(g) + ½ O2(g)  H2O(l) (heterogeneous – gas & liquid) Solids and liquids have concentrations which can vary only slightly due to temperature changes, so we can suppose the concentrations of solids and liquids are fixed. Only gases and dissolved substances can change concentration. Since the equilibrium expression involves only components which change concentration, solids and liquids are not included in its format. When solids or liquids are encountered in an equilibrium equation, a “1” is substituted in the equilibrium expression for that component. 2 H2(g) + O2(g)  2 H2O(l) Ke = 1/ ([H2]2 x [O2]) CaCO3(s)  CaO(s) + CO2(g) Ke = 1 x [CO2] / 1 or simply Ke = [CO2]  MAGNITUDE OF Ke AND EQUILIBRIUM CONCENTRATIONS  Equilibrium means that the rate of the forward and the reverse reaction are equal.  The concentrations of the products and the reactants need NOT be equal and rarely are equal at equilibrium.  When the equilibrium constant is small, the concentrations of the reactants are large as compared to the products at equilibrium. (The equilibrium favors the reactants). 𝑷𝒓𝒐𝒅𝒖𝒄𝒕𝒔 𝑲𝒆 = = a small value, K> 1, formation of products is favored 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔  When the equilibrium constant is about one, the concentrations of the reactants and the products are about equal at equilibrium. 𝑷𝒓𝒐𝒅𝒖𝒄𝒕𝒔 𝑲𝒆 = = about one (1) 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔  MEASURING Ke VALUES  Concentration may be measured for solutions in moles per liter and for gases in terms of pressures (atms), so two unit systems are possible for Ke;  When concentration for calculating Ke are in terms of molar units, the constant is referred to as Kc , while when concentration is measured in pressure units (atms), the constant is referred to as Kp. Relating Kc and Kp The equilibrium constant in concentration terms (Kc) can be converted to an equivalent value in (Kp). Kp = Kc(R x T)n R = 0.0821 L atm / moles K, T = Kelvin temp. n= moles of gaseous products – moles of gaseous reactants in the balanced equation. Problem: How to CONVERT KP AND KC VALUES N2(g) + 3 H2(g)  2 NH3(g) , Kc = 0.105 at 472 0C. Find Kp at this temperature. Kp = Kc(R x T)n R = 0.0821 L atm/ mole K, T= 472 + 273 = 745 K, n= 2 – 4 = - 2 Kp = 0.105 x (0.0821 x 745)-2 = 2.81 x 10-5 NB. Special Equilibrium Expressions Kw (dissociation of water) Ksp (solubility product of salts in saturated solutions) Ka (acid dissociation) Kb (base hydrolysis) LeChatelier’s Principle and Chemical Equilibrium describes how equilibrium systems change when equilibrium conditions are changed. The changes imposed on an equilibrium are called stresses. The response to these changes (stresses) are called shifts. Stresses consist of changes in concentrations, pressures and temperatures. Shifts may be, “shift towards products” (shift right) which means increased rate of forward reaction or “shift towards reactants” (shift left) which means increased rate of reverse reaction. LeChatelier’s Principle defines the relationship between applied stresses and equilibrium shifts.  1. The effect of change in concentration  In the reaction of A + B  C + D If at equilibrium state, concentration of one or more substances in the equilibrium mixture changes, the equilibrium system will no longer remain in equilibrium state. The system will undergo changes in concentrations of various substances so as to minimize the effect or restore equilibrium state.  So, if we increase amount of A or B at equilibrium, the equilibrium system will be disturbed. o According to Le-chatelier's principle, to restore equilibrium, the reaction will shift in the forward direction to cancel the effect of change in concentration. o To regain equilibrium, the reaction will shift in the forward direction where concentration of A and B will decrease and concentrations of C and D will increase.  A+B  C+D  A stress is applied, the concentration of reactant A is increased. Equilibrium shifts forward and concentrations of products (C and D) increases. Concentration of reactant B is reduced as it must be consumed to allow the shift forward. The concentration of A will be larger than the original, although some of it is consumed in the forward shift. The equilibrium constant value remains unchanged despite the shift.  A stress is applied, the concentration of product C is increased. Equilibrium shifts reverse and concentrations of reactants (A and B) increases. Concentration of reactant D is reduced as it must be consumed to allow the shift reverse. The concentration of C will be larger than the original, although some of it is consumed in the reverse shift. The equilibrium constant value remains unchanged despite the shift.  A stress is applied, the concentration of product C is decreased The reaction rate in the forward direction remains unchanged. The rate of the reverse reaction slows due to the reduced concentration of C and the net reaction shifts forward making more D and also increasing C while consuming reactants A and B. The value of the equilibrium constant remains unchanged.  2. The effect of change in temperature When changing temperature is the applied stress, the exothermic or endothermic nature of the reaction must be considered. a A + b B  c C + d D For endothermic reaction Kc increases with the increase in temperature. Increase in temperature (heat is added): favors the reaction in the forward direction. At equilibrium, conc. of A and B will decrease and conc. of C and D will increase. Decrease in temperature (heat is removed): favors the reaction in the backward direction. At equilibrium, conc. of C and D will decrease and conc. of A and B will increase. For exothermic reaction Kc decreases with the increase in temperature. Increase in temperature (heat is added): favors the reaction in the backward direction. At equilibrium, conc. of C and D will decrease and conc. of A and B will increase. Decrease in temperature (heat is removed): favors the reaction in the forward direction. At equilibrium, conc. of A and B will decrease and conc. of C and D will increase.  3. The effect of change in pressure When gaseous systems are compressed (inc. P, thus dec. Volume) the equilibrium shifts towards the side of the reaction containing the fewer number of gas phase molecules. When gaseous systems are expanded (dec. P, thus inc. V), the equilibrium shifts towards the side of the reaction containing the greater number of gas phase molecules For example: N2(g) + 3 H2(g)  2 NH3(g)  compressing this system would cause a shift to the right (products) which contains only two gas phase molecules (2 ammonia NH3).  Expanding this system would cause a shift to the left (reactants) which contains four gas phase molecules (1 nitrogen and 3 hydrogen) Le Chatelier's Principle and catalysts adding a catalyst makes absolutely no difference to the position of equilibrium, and Le Chatelier's Principle doesn't apply to them. This is because a catalyst speeds up the forward and back reaction to the same extent. For a dynamic equilibrium to be set up, the rates of the forward reaction and the back reaction have to become equal. This doesn't happen instantly. For a very slow reaction, it could take years. A catalyst speeds up the rate at which a reaction reaches dynamic equilibrium.  Solubility Products (Ksp) Formation of Precipitate Solubility Equilibria Many natural processes depend on the precipitation or dissolving of a slightly soluble ionic substance. When we consider the equilibria of slightly soluble, or nearly insoluble, ionic compounds, their equilibrium constants can be used to answer questions regarding solubility and precipitation. Solubility product constant, Ksp, is the way of expressing solubility in terms of an equilibrium constant Rules for Ksp are same as for all equilibrium constants – When an excess of a slightly soluble ionic compound is mixed with water, an equilibrium is established between the solid and the ions in the saturated solution. – For the salt calcium oxalate, CaC2O4, you have the following equilibrium. CaC2O4(s)  Ca2+(aq) + C2O42-(aq) – The equilibrium constant for the process is termed the solubility product constant (Ksp) Ksp = [Ca2+][C2O42-]  The common ion effect The presence of a common ion decreases the solubility of the salt XY (s)  X+ (aq) + Y- (aq) Increasing either ion suppresses the amount of dissolved salt (in terms of paired ions in solution) Increasing concentration of one ion decreases concentration of the other ion. Remember the Principle of LeChatelier!! If a stress (change of condition) is applied to a system at dynamic equilibrium, the system shifts in the direction that reduces the stress.  About Common ion effect Suppression of ionization of a weak electrolyte by the presence in the same solution of a strong electrolyte containing one of the same ions as the weak electrolyte Common ion effect is a special case of LeChatelier principle Addition of a common ion is equivalent to adding a stress to the system. The system responds to the stress by reducing the solubility of one of the ions and keeping the Ksp constant  Acids and Bases (review)… Brønsted-Lowry Definition: - Acids are proton donors - Bases are proton acceptors Lewis Definition - Acids are electron-pair acceptors. - Bases are electron-pair donors Arrhenius Definition - Acids react in water to release a proton - Bases react in water to release hydroxide ion Acid + Base  Salt + Water Acid + Base  Conjugate Base + Conjugate Acid  Some solvents are amphiprotic (describes a substance that can both accept and donate a proton or H+. An amphiprotic molecule has characteristics of both and acid and a base and can act as either). – Water can act as an acid and a base! – Methanol can act as an acid and a base! Autoprotolysis Some solvents can react with themselves to produce an acid and a base Water is a classical example Weak acids and weak bases undergo partial hydrolysis and dissociate partially. Strong acids and strong bases are strong electrolytes  Kw (Dissociation of Water) Water is amphiprotic it also undergoes autoprotolysis  hydronium ion is a very strong acid, and the hydroxide ion is a very strong base; As fast as they are formed, they react to produce water again.  The net effect is that an equilibrium is set up. What is the concentration of hydronium and hydroxide ions in neutral solution? What is the pH? What is the pOH?  In pure water, hydrogen ion (hydronium ion) conc. must be equal to hydroxide ion conc..  For every hydrogen ion formed, there is a hydroxide ion formed as well.  Kw value of 1.00 x 10-14 mol2 dm-6 at room temperature gives you a pKw value of 14.  This is where the pH scale we commonly use originates from…..  Like any other equilibrium constant, the value of Kw varies with temperature. Its value is usually taken to be 1.00 x 10-14 mol2 dm-6 at room temperature  The formation of hydrogen ions and hydroxide ions from water is an endothermic process. The forward reaction absorbs heat.  Defining pH pH is a measure of the concentration of hydrogen ions in a solution. A strong acid is one which is virtually 100% ionized in solution. A weak acid is one which doesn't ionise fully when it is dissolved in water.  The strength of an acid is related to the proportion of it reacting with water to produce ions. When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to produce a hydronium ion and a negative ion depending on what acid you are starting from.  These reactions are all reversible  The equilibrium constant can be written as:  if the acid is weak the concentration of water is virtually constant.  A new equilibrium constant is defined as Ka.  A strong base is 100% split up into metal ions and hydroxide ions in solution.  Each mole of a strong base dissolves to give a mole of hydroxide ions in solution.  A weak base is one which doesn't convert fully into OH- ions in solution.  The equilibrium constant (Kb) can be written as:  Eg. Weak Acid & Weak Base Equilibria  Ka is a “special” equilibrium constant for the hydrolysis or dissociation of a weak acid.  Kb is a “special” equilibrium constant for the hydrolysis or dissociation of a weak base. As Ka gets bigger, pKa gets smaller. The lower the value for pKa, the stronger the acid and vice versa. As K gets bigger, pK gets smaller.