CHEM 108 1st EditionExam # 2 Study Guide Chapters: 15 - 16 Chapter 15: Chemical equilibriumChemical equilibrium: Dynamic process in which the concentrations of reactants and produces remain constant over time.- Also the rate of its forward reaction is equal to the rate its backward reaction.- At equilibrium Reaction: 2 NO(g) ⇌ N2O4 (g): o Rate f = kf [NO2]2o Rate r = kr [N2O4]o kf [NO2]2 = kr [N2O4]- (k1/k2)=([A]/[B]) = Constant KEquilibrium Constant Expression: Ratio of equilibrium concentrations or partial pressures of products to reactants, each term raised to a power equal to the coefficient of that substance in the balanced chemical equation.- K = ([C]c[D]d)/([A]a[B]b) = (Products/reactants)o where [A], [B], etc., are equilibrium concentrations, could also be partial pressure- K>>1 equilibrium lies to the right: the concentration of the products is greater than the concentration of the reactants - K<<1 equilibrium lies to the left: the concentration of the reactants is greater than the concentration of the products- K=0 no reaction- Kf = (1/ Kr)- *The Molar concentrations of pure liquids and solids are not in the equilibrium constant expression because those concentrations stay constant. Reaction Quotient Q:- The K constant of reactions is only when the reaction is in equilibrium. When it is not in equilibrium the value becomes the reaction quotient- Q = ([C]c[D]d)/([A]a[B]b)o Q < K, reaction goes left to right, favors the products (forward reaction). To get to equilibrium must add more reactants o Q > K, reaction goes right to left, favors the reactants (reverse reaction). To get to equilibrium must add more products.o Q = K, reaction is at equilibriumo With this you can predict where the reaction is going to go.Equilibrium in the Gas Phase:These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- Kp = Kc (RT)Δn- Kp = Kc /(RT)Le Châtelier’s Principle - “A system at equilibrium responds to a stress in such a way that it relieves that stress.” - Factors that will change the relative rates of forward/reverse reactions, or change the value of Q compared to K, will cause a shift in the position of equilibrium.- If you add or take away products or reactants, the reaction would shift until equilibrium was reachedo Ex: H2O(g) + CO(g) ⇌ H2 (g) + CO2 (g) Remove CO2 (g): o Rate of reverse reaction decreases; reaction proceeds in forward direction to establish new equilibrium.o Q = ([H2][CO2])/([H2O][CO])o Removing CO2, Q < K; reaction shifts right (toward products).Effects of Temperature and pressure on equilibrium:- The value of K for an endothermic reaction increases with increasing temperature, and the valueof K for an exothermic reaction decreases with increasing temperature. - Equilibrium shifts in response to an increase or decrease in volume caused by a decrease or increase in pressure toward the side with fewer moles of gases.Calculating concentrations/ partial pressures of reactants and products:- Use the I.C.E table to develop algebraic terms for each reactants and products concentration/partial pressure at equilibrium. Let x be the change in concentration/partial pressure of each component of the reaction. Express the changes in terms of x and substitute these terms into the expression for K and solve for x. - Quadratic equation: Chapter 16 Acid–Base and Solubility EquilibriumArrhenius Acids and Bases- Arrhenius acid: an Arrhenius acid will increase proton concentration in water - Arrhenius base: an Arrhenius base will increase hydroxide concentration in waterBrønsted-Lowry Model:- Brønsted-Lowry acid = H+ ion donor. - Brønsted-Lowry base = H+ ion acceptor- NH3 (aq) + H2O(l)  NH4+ (aq) + OH (aq) H2O is acid, NH3 is base- Water can act as both a Brønsted-Lowry acid and baseo H2O + H2O  H3O+ + OH-Lewis Acids and Bases:- Lewis acid: electron pair acceptor - Lewis base: electron pair donor - H+ and OH do not necessarily have to be involved - Electron pairs of one reactant (the Lewis base) attracted to positively charged regions of another reactant (the Lewis acid) o positively charged transition metal ions act as Lewis acidsAcid Ionization Constant, Ka- Acid strength is measured by the size of the equilibrium constant when it reacts with H2O o HAcid + H2O  Acid− + H3O- The equilibrium constant for this reaction is called the acid ionization constant, Kao Ka= ([Acid−][ H3O])/( HAcid[H2O])o strong acids completely ionized in water, Strong acids: Ka >>1o weak acids only partially ionized in water, Weak Acids: Ka < 1Conjugate acid-base pairs:- Differ from each other only by the presence or absence of a protono The strongest acid have the weakest conjugate base o The weakest acid has the strongest conjugate base.o But in between there are weak acids that are paired with weak bases The pH Scale- Quantitative measure of acidity: pH  log [H+ ], where [H+ ] is in M o Strong acids have a lower pH since strong acids fully dissociate in water so it increases the [H+] concentration, and decreases the pHo lower pH, stronger acid (weaker base) o higher pH, weaker acid (stronger base) o Usually, pH defined for aqueous solutions- pOH = −log[OH-]- Kw = 1 × 10−14 = [H+ ][OH− ]o Log form of Kw : −log(Kw )= −log([H+] [OH− ]) - Concentration of [H+] = [OH− ] Solution is neutral- 14.00 = pH + pOHpK: A way of expressing the strength of an acid or base - pKa = −log(Ka ), Ka = 10−pKao The stronger the acid, the smaller the pKa o larger Ka = smaller pKa » because it is the –logCalculating pH of:o strong acid; [H+] = [Acid]initial o Weak acid; [H+] = calculated by solving equilibrium equation using I.C.E table. Once we have the equilibrium concentrations we can find the pH because we have [H+] concentration.o Obtain [OH−] at equilibrium using I.C.E table, then use Kw relationship to find the concentration of [H+] for the pH.Polyprotic acid:- Polyprotic acid: can donate more than one proton - There is a different value of Ka for each ionizable level o Ka1 > Ka2 (> Ka3 …) - More difficult to remove H+ ion (positive charge) from negatively charged anion. - pH of Polyprotic acid solutions: Typically only first ionizable H atom affects pH (Ka1).pH of salt solutions:- F− (aq) + H2O(l) ⇌ HF(aq) + OH− (aq)- Determine Ka or Kbo Kb= ([HF][ OH− ])/[ F− ] using I.C.E tableso pOH=-log[OH− ]o pH= pKw- pOH ( pKw = 14)- By Solving