PHA612 LEC TRANSES 2.pdf

PHARMACEUTICAL INORGANIC CHEMISTRY (WITH QUALITATIVE ANALYSIS) PHA 612 (LEC) B.S. Pharmacy | 1E-PH | Term 1 2024-2025 Ms. Ysabel Ann C. De Ocampo / Mr. Adrien Kyle M. Jacinto seems ○ Sharing a p...

PHARMACEUTICAL INORGANIC CHEMISTRY (WITH QUALITATIVE ANALYSIS) PHA 612 (LEC) B.S. Pharmacy | 1E-PH | Term 1 2024-2025 Ms. Ysabel Ann C. De Ocampo / Mr. Adrien Kyle M. Jacinto seems ○ Sharing a pair of electrons between 2 atoms ○ Strongest IMFA IMFA (Intramolecular Forces of Attraction) ○ Interaction within a molecule Homogenous Catalysis Reactants and catalysts are dispersed in a single phase, usu. liquid ○ Acid catalysis – Lead Chamber Process (H2SO4) ○ Base catalysis ○ Enzyme catalysis – Substrate (reactants), enzyme, and products are Chemical Equilibrium aqueous solutions Reaction that occurs in both Enzymes directions. ○ Biological catalysts (usually Is attained when: protein in nature) ○ Forward and backward/ ○ Selective: Acts on certain reverse reactions are of same substrate rate ○ Reactant and product concentration remain constant ○ No net change of reactant Homogenous Catalysis advantages over and product molecules Heterogenous Catalysis ex. 𝐻2𝑂 (𝑙) ↔ 𝐻2𝑂 (𝑔) or Can be carried out under atmospheric 𝑁2𝑂4 (𝑔) ↔ 2𝑁𝑂4 (𝑔) conditions Law of Mass Action (Effect of Function selectively for a particular Concentration on Equilibrium) reaction type Law of Mass Action Costs less than precious metals (Pt ○ States that the rate of and Au) chemical reaction is proportional to the mass/ Chapter 2: Chemical Equilibrium (Part concentration of reactants 1) ○ The equilibrium constant is constant for a given reaction Chemical Reaction at a constant temperature Ionic Bond Equilibrium Constant (Kc, Keq) ○ Removal of valence electrons, Quantity that relates the adding electrons to a partly concentrations of reactants & products filled valence shell) at equilibrium ○ NM + M It is the ratio of product concentration over the reactant concentration, each raised to the power of their coefficients in a balanced chemical equation. Covalent Bond Characteristics: 7 PHARMACEUTICAL INORGANIC CHEMISTRY (WITH QUALITATIVE ANALYSIS) PHA 612 (LEC) B.S. Pharmacy | 1E-PH | Term 1 2024-2025 Ms. Ysabel Ann C. De Ocampo / Mr. Adrien Kyle M. Jacinto ○ Dependent on ∆n = Products moles (g) - Reactant temperature moles (g) ○ Independent of initial Homogeneous Equilibrium concentrations of Reactants & products exist in the reactant and products same phase General equation: aA + bB ↔ cC + dD Ex. N2O4 (g) ↔ 2NO2 (g) Kc & Kp? Writing Equilibrium Constant Expressions Ex. CH3COOH (aq) + H2O (l) ↔ CH3COO- (aq) M or mol/L (M or atm in case of + H3O+ (aq) Kc? gases) ○ Concentration unit of all reacting species not included in expression Pure Solids and Liquids (including solvents) ○ Concentrations are fixed by Heterogeneous Equilibrium their density & molar mass Reactants & products exist in the (constants) & do not vary with different phases the amount. Not included in expression Ex. CaCO3 (s) ↔ CaO (s) + CO2 (g) Equilibrium Constants is Dimensionless ○ In quoting Kc, specify the balanced equation and temperature Multiple Equilibria ○ Kc for the overall reaction is given by the product of Kc of individual reactions. Equilibrium Partial Pressure (Kp) For gases Expression is written using equilibrium partial pressure of reactants & products. In most cases, Kc ≠ Kp. Since pressure and molarity are related by I. Computations: Kc & Kp ideal gas law, the following equation 1. The equilibrium concentrations for the relates Kp & Kc: reaction between carbon monoxide and molecular chlorine to form COCl2 Kp = Kc(RT)∆n (g) at 74oC are [CO] = 0.012 M, [Cl2] = Where: 0.054 M, and [COCl2] = 0.14 M. R = 0.08205 L. atm/mol. K Calculate the equilibrium constants Kc T = in Kelvin and Kp. 8 PHARMACEUTICAL INORGANIC CHEMISTRY (WITH QUALITATIVE ANALYSIS) PHA 612 (LEC) B.S. Pharmacy | 1E-PH | Term 1 2024-2025 Ms. Ysabel Ann C. De Ocampo / Mr. Adrien Kyle M. Jacinto 2. The value of K also depends on how the equilibrium equation is balanced (i.e., the stoichiometric coefficients used). 2. The equilibrium constant Kc for the reaction: 2NO2 (g) ↔ 2NO (g) + O2 (g) is 158 at 1000K. What is the equilibrium Chemical Kinetics & Chemical Equilibrium pressure of O2 if the PNO2 = 0.400 atm and PNO = 0.270 atm? Chemical Equilibrium Kf = Kb ○ Rate of forward reaction (Kf) = Rate of backward/reverse reaction (Kb or Kr) Multiple Equilibria Equilibrium Constant (Kc) From a chemical kinetics viewpoint, it If a reaction can be expressed as the is the ratio of the rate constants of sum of ≥2 reactions, the Kc for the forward and reverse reactions. overall reaction is given by the product This explains why Kc is constant but of Kc of individual reactions. value changes with temperature. Product molecules in one equilibrium system are involved in a second equilibrium process: II. Computations: Multiple Equilibria 1. The ff. equilibrium constants have been determined for carbonic acid Chapter 2: Chemical Equilibrium (Part (H2CO3) at 25oC: 2) K’c = 4.2 x 10-7 K’’c = 4.8 x 10-11 Uses of Equilibrium constant (Kc) What is the equilibrium constant (Kc) for the overall reaction? Calculate equilibrium concentrations Predict the direction of reaction (to the left, to the right, or no shift) Important Rules About Writing Equilibrium Reaction quotient (Qc) Constants Calculated by substituting the initial 1. When the reversible reaction equation concentrations (in M or mol/L) of is written in the opposite direction, the reactants and products into the Kc equilibrium constant becomes the expression reciprocal of the original equilibrium Very similar with Kc constant Equilibrium shift Kc & Qc (to reach new Favors equilibrium) Kc = Qc No shift (at - equilibrium) 9 PHARMACEUTICAL INORGANIC CHEMISTRY (WITH QUALITATIVE ANALYSIS) PHA 612 (LEC) B.S. Pharmacy | 1E-PH | Term 1 2024-2025 Ms. Ysabel Ann C. De Ocampo / Mr. Adrien Kyle M. Jacinto Steps in Calculating Equilibrium Kc > Qc To the right Product Concentrations: formation 1. Set up ICE table following the Kc < Qc To the left Reactant reaction’s stoichiometry: (reverse/backward formation I = Initial conc: Initial reactant reaction) concs are usually given; initial [Product]'s are assumed to be 0 unless otherwise specified C = Change in conc: Assign change as variable x; use the stoichiometry of the reaction to assign changes for all species E = Equilibrium conc: E = I + III. Computations: Reaction Quotient C 1. At the start of a reaction, there are 2. Set up the Kc expression: substitute 0.249 mol N2, 3.21 x 10-2 mol H2, and equilibrium concentrations from the 6.42 x 10-4 mol NH3 in a 3.50-L ICE table into Kc reaction vessel at 375°C. If the 3. Solve for x then plug the solution for x equilibrium constant (Kc) for the back into the equilibrium reaction: N2 (g) + 3H2 (g) ↔ 2NH3 (g) concentration expressions. There is 1.2 at this temperature, decide are 2 methods for solving x: whether the system is at equilibrium. If Perfect square root system to it is not, predict which way the net solve for x reaction will proceed. ○ Take the square root of both Given: sides when the math Initial concentrations in expression for Kc is a mole → Convert to M (mol/ L) perfect square (i.e., starting Kc at 375ºC concs of reactants are Solution: equal) Initial concentrations of reacting Quadratic expression to solve species in M: for x ○ If the Kc expression is not a perfect square ○ Two computations are needed, one with a negative sign and one with a positive, but only one final answer can be used ○ ax2 + bx + c = 0 ○ Kc (1.2) > Qc (0.611) ○ System is not at equilibrium; IV. Computations: Equilibrium net reaction will proceed from Concentrations left to right (forward reaction) 4. A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L 10 PHARMACEUTICAL INORGANIC CHEMISTRY (WITH QUALITATIVE ANALYSIS) PHA 612 (LEC) B.S. Pharmacy | 1E-PH | Term 1 2024-2025 Ms. Ysabel Ann C. De Ocampo / Mr. Adrien Kyle M. Jacinto stainless-steel flask at 430°C. The N2(g) + 3H2(g) ↔ 2NH3(g) equilibrium constant Kc for the is Kc = 2.37 x 10-3. In a certain reaction: H2 (g) + I2 (g) ↔ 2HI (g) is experiment, the equilibrium 54.3 at this temperature. Calculate the concentrations are [N2] = 0.683 M, concentrations of H2, I2, and HI at [H2] = 8.80 M, and [NH3] = 1.05 M. equilibrium. Suppose some NH3 is added to the STEP 1: Reaction’s Stoichiometry is: mixture so that its concentration is 1 mol H2 + 1 mol I2 ↔ 2 mol HI increased to 3.65 M. Let x be the depletion in concentration (mol/L) of H2, I2 and 2x for HI must be A. Use Le Chatelier’s principle to at equilibrium. predict the shift in direction of the net reaction to reach a new equilibrium. B. Confirm your prediction by calculating the reaction quotient Qc and comparing its value with Kc. STEP 2: Get the Kc then solve for x Given: N2(g) + 3H2(g) ↔ 2NH3(g) Kc = 2.37 x 10-3 (At 720°C) [N2] = 0.683 M [H2] = 8.80 M [NH3] = 1.05 M [NH3] = Conc. increased to 3.65 M Multiply (0.500 - x) with 7.37 and solve for x Solution: STEP 3: Concentrations at equilibrium Kc (2.37 x 10-3) < Qc (2.86 x 10-2) Equilibrium shifts to the left until CHECKING equilibrium is reached (Kc = Qc) By calculating Kc using the equilibrium N2(g) + 3H2(g) ← 2NH3(g) concentrations (Kc is a constant for a particular reaction at a given temperature). 2 [𝐻𝐼] 𝐾𝑐 = [𝐻2][𝐼2] 2 [0.786] 𝐾𝑐 = [0.107][0.107] 𝐾𝑐 = 53. 96 Given Kc = 54.3 Computed Kc = 53.96 ○ Acceptable ○ ± 1 should be the approximate difference V. Computations: Changes in Concentration 5. At 720°C, the equilibrium constant for the reaction: 11