CM011 Academic Review Session PDF

CM011 Finals Review Session (Course Outcome 1) Engr. Daniel Eldrei D. Loresca, M.Sc. November 8, 2024 Coverage Formula Stoichiometry Reaction Stoichiometry Calorimetry Enthalpy of Formation and Reaction Hess Law CM011 Review Session 2 Formula Stoichiometry Review: Molar Mass = sum of the atomic masses of a compound Atomic mass of an element can be found in the periodic table CM011 Review Session 3 Formula Stoichiometry Examples: Determine the molar mass of: a. CO2 1 carbon atom and 2 oxygen atoms Molar Mass = 1(12.011 g/mol) + 2(15.999 g/mol) Molar Mass = 44.009 g/mol b. H2SO4 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms Molar Mass = 2(1.0079) + 1(32.065) + 4(15.999) Molar Mass = 98.0768 g/mol CM011 Review Session 4 Formula Stoichiometry Review: Gram-Mole-Atom/Molecule Calculations To express the mass of a compound in terms of number of moles and atoms/molecules, a conversion factor must be used. where NA (Avogadro’s number) = 6.022 x 1023 atoms/mol CM011 Review Session 5 Formula Stoichiometry Example: How many moles of CO2 are in a 150.0 g sample? Mass → Moles MM of CO2 = 44.009 g/mol 1 𝑚𝑜𝑙 𝐶𝑂2 Conversion Factor: 44.009 𝑔 1 𝑚𝑜𝑙 𝐶𝑂2 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑂2 𝑛𝐶𝑂2 = 150.0 𝑔 𝐶𝑂2 × 44.009 𝑔 𝒏𝑪𝑶𝟐 = 𝟑. 𝟒𝟎𝟖 𝒎𝒐𝒍𝒆𝒔 𝑪𝑶𝟐 CM011 Review Session 6 Formula Stoichiometry Example: How many molecules of CO2 are in a 150.0 g sample? Mass → Moles→ Molecules MM of CO2 = 44.009 g/mol 1 𝑚𝑜𝑙 𝐶𝑂2 6.022×1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝑂2 Conversion Factor: , 44.009 𝑔 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝐶𝑂2 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑂2 𝑛𝐶𝑂2 = 150.0 𝑔 𝐶𝑂2 × 44.009 𝑔 𝑛𝐶𝑂2 = 3.408 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 6.022 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝑂2 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝐶𝑂2 = 3.408 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 × 𝑚𝑜𝑙 = 𝟐. 𝟎𝟓𝟑 × 𝟏𝟎𝟐𝟒 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑪𝑶𝟐 CM011 Review Session 7 Formula Stoichiometry Review: Percent Composition To get the percent composition by mass (or mass percent) of an element in a compound, divide the mass of the element present to the total mass of the compound. 𝒏 × 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑋 %𝑋 = × 100% 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 n refers to the number of moles of the element present in 1 mole of the compound. e.g. There are 2 moles of O present in 1 mole of CO2. CM011 Review Session 8 Formula Stoichiometry Example: What is the mass percent of each element in sulfuric acid, H2SO4? List down first each element and get individual molar masses present in the compound For each element, divide the obtained individual molar mass to the molar mass of the compound. MM of H in H2SO4 = 2(1.0079 g/mol) = 2.0158 g/mol MM of S in H2SO4 = 1(32.065 g/mol) = 32.065 g/mol MM of O in H2SO4 = 4(15.999 g/mol) = 63.996 g/mol MM of H2SO4 = 98.0768 g/mol CM011 Review Session 9 Formula Stoichiometry Example (Continuation): What is the mass percent of each element in sulfuric acid, H2SO4? 2.0158 𝑔/𝑚𝑜𝑙 %𝐻 = × 100% 98.0768 𝑔/𝑚𝑜𝑙 % 𝑯 = 𝟐. 𝟎𝟓𝟓% 32.065 𝑔/𝑚𝑜𝑙 %𝑆 = × 100% 98.0768 𝑔/𝑚𝑜𝑙 % 𝑺 = 𝟑𝟐. 𝟔𝟗𝟒% 63.996 𝑔/𝑚𝑜𝑙 %𝑂 = × 100% 98.0768 𝑔/𝑚𝑜𝑙 % 𝑶 = 𝟔𝟓. 𝟐𝟓𝟏% Checking, 2.055+32.694+65.251 = 100% CM011 Review Session 10 Reaction Stoichiometry Review: When dealing with stoichiometric calculations involving chemical reactions, a general relationship can be used: For the reaction: xA→yB CM011 Review Session 11 Reaction Stoichiometry Example: A typical reaction is that between lithium and water: 2𝐿𝑖 𝑠 + 2𝐻2 𝑂 𝑙 → 2𝐿𝑖𝑂𝐻 𝑎𝑞 + 𝐻2 (𝑔) How many grams of Li are needed to produce 9.89 g of H2? Approach: If not given, write the chemical equation Always check if the chemical equation is already balanced Using the coefficients of the balanced equation, make a mole ratio between the desired product and reactant which will serve as your conversion factor Proceed to moles to mass calculations CM011 Review Session 12 Reaction Stoichiometry Given: 2𝐿𝑖 𝑠 + 2𝐻2 𝑂 𝑙 → 2𝐿𝑖𝑂𝐻 𝑎𝑞 + 𝐻2 (𝑔) Mass of H2 produced = 9.89 g Unknown: Mass of Li Solution: Chemical equation is already balanced! From balanced equation, we know that for every 2 moles of Li consumed in the reaction, 1 mole of H2 is produced Starting with the mass of H2, we can get the mass of Li with the following steps: mass H2 → moles H2 → moles Li → mass Li CM011 Review Session 13 Reaction Stoichiometry Given: 2𝐿𝑖 𝑠 + 2𝐻2 𝑂 𝑙 → 2𝐿𝑖𝑂𝐻 𝑎𝑞 + 𝐻2 (𝑔) Mass of H2 produced = 9.89 g Unknown: Mass of Li Solution: 1 𝑚𝑜𝑙 𝐻2 2 𝑚𝑜𝑙 𝐿𝑖 6.941 𝑔 𝐿𝑖 𝑚𝑎𝑠𝑠 𝐿𝑖 = 9.89 𝑔 𝐻2 × × × 2.0158 𝑔 1 𝑚𝑜𝑙 𝐻2 𝑚𝑜𝑙 𝒎𝒂𝒔𝒔 𝑳𝒊 = 𝟔𝟖. 𝟏 𝒈 𝑳𝒊 CM011 Review Session 14 Reaction Stoichiometry Review: Limiting and Excess Reactants Limiting Reactant – reactant that is used up first which limits the amount of product that can be made. Excess Reactant – reactant that is not completely used up. To determine the which of the reactants are limiting and excess: Convert the mass of the reactants into the equivalent moles of product produced The reactant with the smaller moles of product produced would be the limiting reactant. CM011 Review Session 15 Reaction Stoichiometry Review: Percent Yield Theoretical Yield – amount of product predicted by stoichiometry. Actual Yield – quantity of desired product actually obtained. 𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 %𝑌𝑖𝑒𝑙𝑑 = × 100% 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 CM011 Review Session 16 Reaction Stoichiometry Example: 2.50 g of copper heated with 2.12 g of sulfur made 2.53 g of copper (I) sulfide following the reaction: 16 𝐶𝑢 𝑠 + 𝑆8 𝑠 → 8𝐶𝑢2 𝑆 (𝑠) a. What are the limiting and excess reactants? b. What was the percent yield for this reaction? Given: mCu = 2.50 g mS8 = 2.12 g mCu2S = 2.53 g (actual yield!) CM011 Review Session 17 Reaction Stoichiometry Given: 16 𝐶𝑢 𝑠 + 𝑆8 𝑠 → 8𝐶𝑢2 𝑆 (𝑠) mCu = 2.50 g, mS8 = 2.12 g, mCu2S = 2.53 g Approach: Determine the limiting and excess reactants by getting the moles Cu2S produced if each reactant was consumed completely Using the limiting reactant, calculate the mass of Cu2S produced. This will be the Theoretical Yield. Calculate the %Yield. CM011 Review Session 18 Reaction Stoichiometry Given: 16 𝐶𝑢 𝑠 + 𝑆8 𝑠 → 8𝐶𝑢2 𝑆 (𝑠) mCu = 2.50 g, mS8 = 2.12 g, mCu2S = 2.53 g Solution: Moles Cu2S if Cu is limiting: 1 𝑚𝑜𝑙 𝐶𝑢 8 𝑚𝑜𝑙 𝐶𝑢2 𝑆 𝑛𝐶𝑢2𝑆 = 2.50 𝑔 𝐶𝑢 × × 63.55 𝑔 16 𝑚𝑜𝑙 𝐶𝑢 𝑛𝐶𝑢2𝑆 = 0.01967 𝑚𝑜𝑙 Moles Cu2S if S8 is limiting: 1 𝑚𝑜𝑙 𝑆8 8 𝑚𝑜𝑙 𝐶𝑢2 𝑆 𝑛𝐶𝑢2𝑆 = 2.12 𝑔 𝑆8 × × 8 32.07 𝑔 1 𝑚𝑜𝑙 𝑆8 𝑛𝐶𝑢2𝑆 = 0.0661 𝑚𝑜𝑙 Since moles Cu2S produced from Cu is smaller, Cu is the limiting reactant and S8 is the excess reactant. CM011 Review Session 19 Reaction Stoichiometry Given: 16 𝐶𝑢 𝑠 + 𝑆8 𝑠 → 8𝐶𝑢2 𝑆 (𝑠) mCu = 2.50 g, mS8 = 2.12 g, mCu2S = 2.53 g Solution: Since Cu is limiting, we can solve for mass Cu2S 159.2 𝑔 𝑚𝑎𝑠𝑠 𝐶𝑢2 𝑆 = 0.0197 𝑚𝑜𝑙 𝐶𝑢2 𝑆 × 1 𝑚𝑜𝑙 𝑚𝑎𝑠𝑠 𝐶𝑢2 𝑆 = 3.131 𝑔 𝐶𝑢2 𝑆 Theoretical Yield is 3.131 g Solving for the %Yield 𝐴𝑐𝑡𝑢𝑎𝑙 𝑌𝑖𝑒𝑙𝑑 %𝑌𝑖𝑒𝑙𝑑 = × 100% 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑 2.53 𝑔 %𝑌𝑖𝑒𝑙𝑑 = × 100% 3.131 𝑔 %𝒀𝒊𝒆𝒍𝒅 = 𝟖𝟎. 𝟖𝟎% CM011 Review Session 20 Calorimetry Review: Heat Capacity (C) - amount of heat required to raise the temperature of a substance by 1°C. Unit is J/°C 𝑞 = 𝐶∆𝑇 Specific heat capacity (c) - amount of heat required to raise the temperature of 1 g of a substance by 1°C. Unit is J/g-°C 𝑞 = 𝑚𝑐∆𝑇 Commonly known specific heat is H2O(l) = 4.184 J/g-°C CM011 Review Session 21 Calorimetry Review: Bomb Calorimeter (Constant Volume Calorimeter) rigid steel container. filled with O2(g) and a small sample to be burnt. Conservation of Energy: qreaction + qbomb + qwater = 0 Rearranging, −qreaction = qbomb + qwater qbomb = CcalΔT CM011 Review Session 22 Calorimetry Example: A bomb calorimeter has a heat capacity of 815 J/°C and contains 156 g of water. How much energy is evolved or absorbed when the temperature of the calorimeter goes from 20.24°C to 28.15°C? Given: Cbomb = 815 J/°C mwater = 156 g Ti = 20.24°C Tf = 28.15°C Unknown: qrxn CM011 Review Session 23 Calorimetry Solution: Conservation of Energy: qreaction + qbomb + qwater = 0 qreaction = -(qbomb + qwater) Get qbomb and qwater 𝐽 𝑞𝑏𝑜𝑚𝑏 = 𝐶𝑏𝑜𝑚𝑏 ∆𝑇 = 815 28.15℃ − 20.24℃ ℃ 𝑞𝑏𝑜𝑚𝑏 = 6446.65 𝐽 𝐽 𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑚𝑤𝑎𝑡𝑒𝑟 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇 = 156 𝑔 4.184 28.15℃ − 20.24℃ 𝑔℃ 𝑞𝑤𝑎𝑡𝑒𝑟 = 5162.89 𝐽 𝑞𝑟𝑥𝑛 = − 6446.65 + 5162.89 𝐽 𝒒𝒓𝒙𝒏 = −𝟏𝟏𝟔𝟎𝟗. 𝟓𝟒 𝑱 q is negative so heat was evolved CM011 Review Session 24 Enthalpy of Formation and Reaction Review: Standard Formation Enthalpy Denoted by ΔHf° ΔHf° for any element in its standard state is zero. The standard state of an element is its most stable form when P = 1 bar. Some standard states (at 25 °C): Carbon = graphite (not diamond). Nitrogen = N2(g) Bromine = Br2 (l) (not gas) CM011 Review Session 25 Enthalpy of Formation and Reaction Review: Standard Reaction Enthalpy Denoted by ΔHr° Can be calculated using the Standard Formation Enthalpies of compounds ∆𝐻𝑟° = ∑ 𝑛𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 ∆𝐻𝑓° −∑ 𝑛𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 ∆𝐻𝑓° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 CM011 Review Session 26 Enthalpy of Formation and Reaction Example: The standard enthalpies of formation for several substances are given below: ∆𝐻𝑓° = 50.6 𝑘𝐽/𝑚𝑜𝑙 𝑁2 𝐻4 𝑙 ° ∆𝐻𝑓 = 95.4 𝑘𝐽/𝑚𝑜𝑙 𝑁2 𝐻4 𝑔 ∆𝐻𝑓° = −46.2 𝑘𝐽/𝑚𝑜𝑙 𝑁𝐻3 𝑔 Determine the ΔH° for the reaction: 𝑁2 𝐻4(𝑔) + 𝐻2 𝑔 → 2𝑁𝐻3 𝑔 CM011 Review Session 27 Enthalpy of Formation and Reaction Solution: ∆𝐻𝑟° = ∑ 𝑛𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 ∆𝐻𝑓° −∑ 𝑛𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 ∆𝐻𝑓° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 ∆𝐻𝑟° = 2 ∆𝐻𝑓° − 1 ∆𝐻𝑓° + 1 ∆𝐻𝑓° 𝑁𝐻3 𝑔 𝐻2 𝑔 𝑁2 𝐻4 𝑔 At standard state, ∆𝐻𝑓° =0 𝐻2 𝑔 𝑘𝐽 𝑘𝐽 ∆𝐻𝑟° = 2 −46.2 − 1 0 + 1 95.4 𝑚𝑜𝑙 𝑚𝑜𝑙 ° 𝒌𝑱 ∆𝑯𝒓 = −𝟏𝟖𝟕. 𝟖 𝒎𝒐𝒍 CM011 Review Session 28 Hess Law Review: “If the equation for a reaction is the sum of the equations for two or more other reactions, then ΔrH° for the 1st reaction must be the sum of the ΔrH° values of the other reactions.” ΔH° for a reaction is the same whether it takes place in a single step or several steps! CM011 Review Session 29 Hess Law Review: Properties to consider: Multiply a reaction, multiply ΔHr°. Reverse a reaction, change the sign of ΔHr° 2 CO(g) + O2(g) → 2 CO2 (g) ΔrH° = −566.0 kJ/mol Then 2 CO2(g) → 2 CO(g) + O2(g) ΔrH° = -1(-566.0 kJ/mol) = +566.0 kJ/mol 4 CO2(g) → 4 CO(g) + 2 O2(g) ΔrH° = -2(-566.0 kJ/mol) = +1132.0 kJ/mol CM011 Review Session 30 Hess Law Example: Determine the ΔH° for the reaction 𝐶𝐻4 𝑔 + 𝑁𝐻3 𝑔 → 𝐻𝐶𝑁 𝑔 + 3𝐻2 𝑔 Using 𝑁2 𝑔 + 3𝐻2 𝑔 → 2𝑁𝐻3 𝑔 ∆𝐻° = −91.8 𝑘𝐽 𝐶 𝑠 + 2𝐻2 𝑔 → 𝐶𝐻4 𝑔 ∆𝐻° = −74.9 𝑘𝐽 𝐻2 𝑔 + 2𝐶 𝑠 + 𝑁2 𝑔 → 2𝐻𝐶𝑁 𝑔 ∆𝐻° = +270.3 𝑘𝐽 CM011 Review Session 31 Hess Law Given: 𝐶𝐻4 𝑔 + 𝑁𝐻3 𝑔 → 𝐻𝐶𝑁 𝑔 + 3𝐻2 𝑔 𝑁2 𝑔 + 3𝐻2 𝑔 → 2𝑁𝐻3 𝑔 ∆𝐻° = −91.8 𝑘𝐽 𝐶 𝑠 + 2𝐻2 𝑔 → 𝐶𝐻4 𝑔 ∆𝐻° = −74.9 𝑘𝐽 𝐻2 𝑔 + 2𝐶 𝑠 + 𝑁2 𝑔 → 2𝐻𝐶𝑁 𝑔 ∆𝐻° = +270.3 𝑘𝐽 Approach: Label each reaction as A, B, and C Adjust the 3 reactions by multiplying and/or reversing such that the sum of the reactions will be the desired reaction Apply the changes done in their respective ΔHr° values and get the sum CM011 Review Session 32 Hess Law Given: 𝐶𝐻4 𝑔 + 𝑁𝐻3 𝑔 → 𝐻𝐶𝑁 𝑔 + 3𝐻2 𝑔 A 𝑁2 𝑔 + 3𝐻2 𝑔 → 2𝑁𝐻3 𝑔 ∆𝐻° = −91.8 𝑘𝐽 B𝐶 𝑠 + 2𝐻2 𝑔 → 𝐶𝐻4 𝑔 ∆𝐻° = −74.9 𝑘𝐽 C 𝐻2 𝑔 + 2𝐶 𝑠 + 𝑁2 𝑔 → 2𝐻𝐶𝑁 𝑔 ∆𝐻° = +270.3 𝑘𝐽 Solution: 1 3 1 -½ x A 𝑁𝐻3 𝑔 → 𝑁 + 𝐻2 𝑔 ∆𝐻° = − −91.8 𝑘𝐽 2 2 𝑔 2 2 -1 x B 𝐶𝐻4 𝑔 →𝐶 𝑠 + 2𝐻2 𝑔 ∆𝐻° = −1 −74.9 𝑘𝐽 1 1 1 ½xC 𝐻 +𝐶 + 𝑁2 → 𝐻𝐶𝑁 ∆𝐻° = (+270.3 𝑘𝐽) 2 2 𝑔 𝑠 2 𝑔 𝑔 2 𝐶𝐻4 𝑔 + 𝑁𝐻3 𝑔 → 𝐻𝐶𝑁 𝑔 + 3𝐻2 𝑔 ∆𝑯° = +𝟐𝟓𝟓. 𝟗𝟓 𝒌𝑱 CM011 Review Session 33 Chemistry: The Molecular Science Moore, Stanitski and Jurs Chemical Kinetics: Rates of Reactions © 2008 Brooks/Cole 1 Chemical Kinetics “The study of speeds of reactions and the nanoscale pathways or rearrangements by which atoms and molecules are transformed to products” Chemical kinetics is also called reaction kinetics or kinetics. © 2008 Brooks/Cole 2 Reaction Rate Factors affecting the speed of a reaction: Properties of reactants and products ▪ especially their structure and bonding. Concentrations of reactants (and products). Temperature Catalysts ▪ and if present, their concentration. Reactions are either: Homogeneous - reactants & products in one phase. Heterogeneous - species in multiple phases. © 2008 Brooks/Cole 3 Reaction Rates and Stoichiometry For any general reaction: aA + bB cC + dD The overall rate of reaction is: 1 Δ[A] 1 Δ[B] 1 Δ[C] 1 Δ[D] Rate = − = − = + = + d Δt a Δt b Δt c Δt Reactants decrease with time. Products increase with time. Negative sign. Positive sign © 2008 Brooks/Cole 4 Determining Rate Laws from Initial Rates Data for the reaction of methyl acetate with base: CH3COOCH3 + OH- CH3COO- + CH3OH Initial concentration (M) Expt. [CH3COOCH3] [OH-] Initial rate (M/s) 1 0.040 0.040 2.2 x 10-4 2 0.040 0.080 4.5 x 10-4 3 0.080 0.080 9.0 x 10-4 Rate law: rate = k [CH3COOCH3]m [OH-]n © 2008 Brooks/Cole 5 Determining Rate Laws from Initial Rates Initial concentration (M) Expt. [CH3COOCH3] [OH-] Initial rate (M/s) 1 0.040 0.040 2.2 x 10-4 2 0.040 0.080 4.5 x 10-4 3 0.080 0.080 9.0 x 10-4 Dividing the first two data sets: 4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n 2.2 x 10-4 M/s = k (0.040 M)m(0.040 M)n Thus: 2.05 = (1)m (2.00)n 2.05 = (2.00)n 1 raised to any and n=1 power = 1 It is 1st order with respect to OH-. © 2008 Brooks/Cole 6 Determining Rate Laws from Initial Rates Use experiments 2 & 3 to find m: 9.0 x 10-4 M/s = k (0.080 M)m(0.080 M)n 4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n So: 2.00 = (2.00)m (1)n 2.00 = (2.00)m and m=1 Also 1st order with respect to CH3COOCH3. © 2008 Brooks/Cole 7 Determining Rate Laws from Initial Rates The rate law is: rate = k [CH3COOCH3][OH-] Overall order for the reaction is: m+n = 1+1 = 2 The reaction is: 2nd order overall. 1st order in OH- 1st order in CH3COOCH3 © 2008 Brooks/Cole 8 Determining Rate Laws from Initial Rates If a rate law is known, k can be determined: rate k= [CH3COOCH3 ][OH-] 2.2 x 10-4 M/s Using run 1: k= (0.040 M)(0.040 M) k = 0.1375 M-1 s-1 = 0.1375 L mol-1 s-1 Could repeat for each run, take an average… But a graphical method is better. © 2008 Brooks/Cole 9 The Integrated Rate Law Calculus is used to integrate a rate law. Consider a 1st-order reaction: A products Δ[A] rate = – = k [A] Δt d [A] =– = k [A] (as a differential equation) dt Integrates to: ln [A]t = −k t + ln [A]0 y = mx + b (straight line) If a reaction is 1st-order, a plot of ln [A] vs. t will be linear. © 2008 Brooks/Cole 10 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Concentration-Time Order Rate Law Equation Half-Life [A]0 0 rate = k [A] = [A]0 - kt t½ = 2k 1 rate = k [A] ln[A] = ln[A]0 - kt t½ = ln 2 k 1 1 1 2 rate = k [A]2 = + kt t½ = [A] [A]0 k[A]0 11 Example 14.6 Iodine atoms combine to form molecular iodine in the gas phase This reaction follows second-order kinetics and has the high rate constant 7.0 × 109/M · s at 23°C. (a) If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 min. (b) Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M. Example 14.6 Strategy (a) The relationship between the concentrations of a reactant at different times is given by the integrated rate law. Because this is a second-order reaction, we use Equation (14.6). (b) We are asked to calculate the half-life. The half-life for a second-order reaction is given by Equation (14.7). Solution (a) To calculate the concentration of a species at a later time of a second−order reaction, we need the initial concentration and the rate constant. Applying Equation (14.6) Example 14.6 where [A]t is the concentration at t = 2.0 min. Solving the equation, we get This is such a low concentration that it is virtually undetectable. The very large rate constant for the reaction means that nearly all the I atoms combine after only 2.0 min of reaction time. (b) We need Equation (14.7) for this part. For [I]0 = 0.60 M Example 14.6 For [I]0 = 0.42 M Check These results confirm that the half-life of a second- order reaction, unlike that of a first-order reaction, is not a constant but depends on the initial concentration of the reactant(s). Does it make sense that a larger initial concentration should have a shorter half-life? Calculating [ ] or t from a Rate Law In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (a) Calculate [reactant] ,1600.s after initiation. 1st order: k = ln 2/ t½ = 0.6931/(400. s) = 1.733x10-3 s-1 and ln [A]t = -kt + ln [A]0 so ln[A]t = -(0.001733 s-1)(1600 s) +ln(0.500) ln[A]t = -2.773 + -0.693 = -3.466 [A]t = e-3.466 = 0.0312 mol/L © 2008 Brooks/Cole 16 Calculating [ ] or t from a Rate Law In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (b) Calculate t for [reactant] to drop to 1/16th of its initial value. [reactant]0 1 [reactant] t1/2 0 2 1 1 [reactant] [reactant]0 0 t1/2 2 4 1 1 4 [reactant]0 8 [reactant]0 t1/2 1 1 [reactant]0 [reactant]0 t1/2 8 16 4 t1/2 = 4 (400 s) = 1600 s Note: part (a) could be solved in a similar way. 1600 s = 4 t1/2 so 0.500 → 0.250 → 0.125 → 0.0625 → 0.0313 M. © 2008 Brooks/Cole 17 Calculating [ ] or t from a Rate Law In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (c) Calculate t for [reactant] to drop to 0.0500 mol/L ? From part (a): k = 1.733 x 10-3 s-1 ln [A]t = -kt + ln [A]0 then ln (0.0500) = -(0.001733 s-1) t + ln(0.500) -2.996 = -(0.001733 s-1) t – 0.693 t = -2.303 -0.001733 s-1 t = 1.33 x 103 s © 2008 Brooks/Cole 18 Determining Activation Energy Take the natural logarithms of both sides: -Ea / RT ln k = ln A e ln ab = ln a + ln b -Ea / RT ln k = ln A + ln e ln e = 1 Ea ln k = ln A + − RT ln e Ea 1 ln k = − R + ln A T A plot of ln k vs. 1/T is linear (slope = −Ea/R). © 2008 Brooks/Cole 19 Example 14.8 The rate constant of a first-order reaction is 3.46 × 10−2 s−1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? Example 14.8 Strategy A modified form of the Arrhenius equation relates two rate constants at two different temperatures [see Equation (14.13)]. Make sure the units of R and Ea are consistent. Solution The data are Substituting in Equation (14.13), Example 14.8 We convert Ea to units of J/mol to match the units of R. Solving the equation gives Check The rate constant is expected to be greater at a higher temperature. Therefore, the answer is reasonable. Chemistry: The Molecular Science Moore, Stanitski and Jurs Chapter 14: Chemical Equilibrium © 2008 Brooks/Cole 1 Equilibrium is Independent of Direction of Approach N2(g) + 3 H2(g) 2 NH3(g) or 2 mol NH3 Start with either 1 mol N2 + 3 mol H2 and get the same equilibrium mixture. © 2008 Brooks/Cole 2 The Equilibrium Constant For the 2-butene isomerization: H3C CH3 H3C H C=C C=C H H H CH3 At equilibrium: rate forward = rate in reverse An elementary reaction, so: kforward[cis] = kreverse[trans] © 2008 Brooks/Cole 3 The Equilibrium Constant We had: kforward[cis] = kreverse[trans] kforward [trans] or = [cis] kreverse This ratio is named the equilibrium constant, Kc: kforward [trans] Kc = = = 1.65 (at 500 K) kreverse [cis] “c” for concentration based © 2008 Brooks/Cole 4 The Equilibrium Constant For a general reaction: aA+bB cC+dD Products raised to stoichiometric powers… kforward [C]c [D]d …divided by reactants Kc = = kreverse [A]a [B]b raised to their stoichiometric powers © 2008 Brooks/Cole 5 The Equilibrium Constant Examples [NO]2 N2(g) + O2(g) 2 NO(g) Kc = [N ] [O ] 2 2 [SO2] 1 8 S8(s) + O2(g) SO2(g) Kc = [O ] 2 S8 is ignored. All pure solids are omitted from Kc. © 2008 Brooks/Cole 6 The Equilibrium Constant What is the equilibrium constant for: SiH4(g) + 2 O2(g) SiO2(s) + 2 H2O(g) ? Omit SiO2 (a solid) [H2O]2 Include H2O (a gas) [SiH4][O2]2 Write down Kc for: 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l) [Na2SO4] Omit H2O (aq. solution) [H2SO4][NaOH]2 © 2008 Brooks/Cole 7 Kc for Related Reactions [NH3]2 N2(g) + 3 H2(g) 2 NH3(g) Kc = [N2][H2]3 Change the stoichiometry… …change Kc ½ N2(g) + 3 2 H2(g) NH3(g) [NH3] Kc = ½ 3 [N2] [H2] 2 reaction divided by 2 This new Kc is the square root of the original Kc. Multiply an equation by a factor… … Raise Kc to the power of that factor. © 2008 Brooks/Cole 8 Kc for Related Reactions N2(g) + 3 H2(g) 2 NH3(g) [NH3]2 Kc = [N2][H2]3 Reverse a reaction… … Invert Kc: 2 NH3(g) N2(g) + 3 H2(g) [N2][H2]3 Kc = [NH3]2 © 2008 Brooks/Cole 9 Kc for a Reaction that Combines Reactions If a reaction can be written as a series of steps: [NO]2 step 1 N2(g) + O2(g) 2NO(g) Kc = [N2][O2] [NO2]2 step 2 2 NO(g) + O2(g) 2 NO2(g) Kc = [NO]2[O2] [NO2]2 overall N2(g) + 2 O2(g) 2 NO2(g) Kc = [N2][O2]2 The overall Kc is the product of the steps: 2 2 [NO ] 2 Kc(step1) x Kc(step2) = [NO] [NO2 ] 2 = [N2][O2] [NO]2[O2] [N2][O2]2 © 2008 Brooks/Cole 10 Equilibrium Constants in Terms of Pressure In a constant-V reaction, partial pressures change as concentrations change. At constant T, P is proportional to molar conc.: Ideal gas: PV = nRT and for gas A: PAV = nART PA= nART = [A]RT V P is easily measured, so another K is used… © 2008 Brooks/Cole 11 Equilibrium Constants in Terms of Pressure For a gas-phase reaction: aA+bB cC+dD “p” for pressure PCc PDd Kp = based PAa PBb In general, Kp ≠ Kc. Your text shows: Kp = Kc(RT)Δngas Δngas = (c + d) – (a + b) =(moles of gaseous products) − (moles of gaseous reactants) © 2008 Brooks/Cole 12 Equilibrium Constants in Terms of Pressure Calculate Kp from Kc for the reaction: N2(g) + 3 H2(g) 2 NH3(g) Kc = 5.8 x 105 at 25°C. Kp = Kc(RT)Δngas R must have L/mol units. T = 25 + 273 = 298 K ([ ] has mol/L units). Δngas = 2 – (3 + 1) = −2 -2 Kp = 5.8 x 105 0.0821 L atm mol-1 K-1(298 K) = 9.7 x 102 mol2 L-2 atm-2 mol and L units are assumed to cancel (units omitted from Kc). Kp = 9.7 x 102 atm-2 © 2008 Brooks/Cole 13 Determining Equilibrium Constants If all the equilibrium concentrations are known it’s easy to calculate Kc for a reaction…. For: aA+bB cC+dD kforward [C]c [D]d Kc = = kreverse [A]a [B]b Example Consider: 2 A (aq) B (aq) At equilibrium [A] = 2.0 M & [B] = 4.0 M. What is Kc? Kc = [B]/[A]2 = 4.0/(2.0)2 = 1.0 © 2008 Brooks/Cole 14 Determining Equilibrium Constants … In other cases stoichiometry is used to find some concentrations. Example H2(g) and I2(g) were added to a heated container. [H2]initial = 0.0100 mol/L and [I2]initial = 0.00800 mol/L. At equilibrium [I2] = 0.00560 mol/L. Determine Kc. H2(g) + I2(g) 2 HI(g) © 2008 Brooks/Cole 15 Determining Equilibrium Constants Write a balanced equation; fill in [ ]initial: H2(g) + I2(g) 2 HI(g) [ ]initial 0.01000 0.00800 0 Concentrations change – use stoichiometry: [ ]change -x -x 2x Stoichiometry: 1H2 ≡ 2HI If 1I2 ≡ 2HI Let HI made HI made = 2x, H2 lost = x = 2x I2 lost = x © 2008 Brooks/Cole 16 Determining Equilibrium Constants Add [ ]initial and [ ]change to get [ ]equilib H2(g) + I2(g) 2 HI(g) [ ]equilib 0.01000 - x 0.00800 - x 2x Since [I2] = 0.00560 M at equilibrium 0.00560 = 0.00800 - x -0.00240 M = - x x = 0.00240 M © 2008 Brooks/Cole 17 Determining Equilibrium Constants Calculate [ ]equilib and then Kc [H2]equilib = 0.0100 - (0.00240) = 0.0076 M [I2]equilib = 0.00800 - (0.00240) = 0.00560 M [HI]equilib = 2(0.00240) = 0.00480 M [HI]2 (0.00480)2 Kc = = [H2][I2] (0.00760)(0.00560) Kc = 0.541 © 2008 Brooks/Cole 18 Meaning of the Equilibrium Constant When: Kc >> 1 Reaction is strongly product favored. very little reactant remains often written as a forward reaction only. assume reaction goes to completion. Kc 20: N / Z gradually increases. 209Bi (Z = 83) is the heaviest stable nucleus. Even-Z isotopes are more common than odd. Even-N isotopes are more common than odd. 200 “even-even”; 120 “odd-even”; 4 “odd-odd” Unstable isotopes decay so that the daughter will enter the “peninsula of stability”. © 2008 Brooks/Cole 11 12 Predicting Nuclear Decay Elements with Z > 83 Most decay by alpha emission. Elements with Z 112 do not have permanent names. Temporary names: 112 ununbium (Uub) 113 ununtrium (Uut) 114 ununquadium (Uuq) etc. © 2008 Brooks/Cole 31 Nuclear Fission Hahn and Strassman (1938) fired n0 at 135U. Ba was produced! Nuclear fission had occurred. 235 1 236 141 92 1 92 U + 0n 92 U 56 Ba + 36 Kr + 3 0 n 3 n0 produced Very exothermic © 2008 Brooks/Cole 32 Nuclear Fission Chain reactions are possible: Small amounts of 235U do not capture all of the n0. (stays under control). Nuclear bombs exceed the critical mass; the chain reaction grows explosively. © 2008 Brooks/Cole 33 Energy from Fission Efission(235U) = 2 x 1013 J/mol. 1 kg of 235U ≈ 33 kilotons of TNT. Natural U is 99.3% 238U (not fissile). Reactor fuel rods are enriched to 3% 235U. Weapons-grade is > 90% 235U. © 2008 Brooks/Cole 34 Energy from Fission Nuclear power-plants produce “clean” energy. No atmospheric pollution. No CO2 emission. But… yield highly radioactive waste Tens of thousands of tons in storage Long half-lives (239Pu, t1/2 = 24,400 yr) Can be vitrified (encased in “glass”) Vwaste = 2 m3/reactor/yr. Yucca Mountain, NV (salt dome). 104 nuclear plants in the U.S. None built since 1979 (Three Mile Island). © 2008 Brooks/Cole 35 Nuclear Fusion Light atoms can be joined: 1 4 0 4 1H 2 He + 2 +1e Nuclear fusion. Very exothermic (ΔE = -2.5 x 109 kJ/mol ). The energy source for stars. Laboratory fusion is an attractive power source: Hydrogen (the fuel) can be extracted from oceans. Waste products are short-lived, low-mass isotopes. © 2008 Brooks/Cole 36 Nuclear Fusion Unfortunately, fusion is hard to produce on earth: H-atoms must be converted into a plasma – a soup of bare nuclei and e-. T > 108 K required. The plasma is hard to contain –magnetic “bottles” are used. Commercial fusion reactors are not very likely to occur in the near future. © 2008 Brooks/Cole 37 Nuclear Radiation: Effects & Units rad radiation absorbed dose 1 rad = 0.010 J absorbed/kg of material gray (Gy) SI unit. 1 Gy = 1 J absorbed/kg of material 1 Gy = 100 rad Roentgen (R) dosage of X-ray and -radiation. R = 9.33 µJ deposited/g of tissue © 2008 Brooks/Cole 38 Nuclear Radiation: Effects & Units , , and  have different biological effects, so… rem roentgen equivalent in man. dose in rem = (quality factor) x (dose in rads) seivert (Sv) SI version. 1 Sv = 100 rem Quality factors:  = 10 - 20,  = 1,  = 1 © 2008 Brooks/Cole 39 Applications of Radioactivity Food Irradiation -rays kill bacteria, molds, spores… Food spoils much less rapidly. It does not make food radioactive. Tracers Chemicals made with radioactive atoms. Introduced into plants, animals… Concentrate where used (rapid growth regions). Uptake can be monitored with a Geiger counter. © 2008 Brooks/Cole 40 Applications of Radioactivity Medical Imaging -emitters are often used (e.g. 99mTc) ▪ Gamma rays can exit the body ▪ Less damaging than α or β. Tracers are used by organs, bones… PET (positron emission tomography) A β+ emitter is injected e + +10e → 2 0 -1 The -rays emit in opposite directions. Detectors show the origin of the -rays © 2008 Brooks/Cole 41 Applications of Radioactivity Chemotherapy = use of radiation to treat cancer. Rapidly growing cells are more susceptible to radiation than mature cells. Cancerous cells divide and grow more rapidly than normal cells Malignant cells are more likely to be killed than normal cells. © 2008 Brooks/Cole 42 Chemistry: The Molecular Science Moore, Stanitski and Jurs Polymers © 2008 Brooks/Cole 1 Synthetic Organic Polymers A polymer is a large molecule formed by linking a series of monomers (smaller units). mono = one poly = many mer = part Polymers are also called macromolecules. Classified by the way they respond to heat: thermoplastics soften and flow, then harden on cooling. thermosetting plastics initially soften, but then form a rigid structure that will not remelt. © 2008 Brooks/Cole 2 Synthetic Organic Polymers Addition Polymers Monomer units add directly to each other. Ethylene (ethene) forms polyethylene Initiation A peroxide (R-O-O-R) breaks into two RO and attacks a monomer: H H H H | | | | RO + C –C RO – C – C | | | | a free H H H H radical © 2008 Brooks/Cole 3 Addition Polymers Propagation Polymer growth. H H H H H H H H | | | | | | | | RO – C – C + C –C → RO – C – C – C – C | | | | | | | | H H H H H H H H H H | | n CH2=CH2 → → → → → –C–C– | | H H n © 2008 Brooks/Cole 4 Addition Polymers Termination Two radicals combine. H H H H H H H | | | | | | | RO – C – C – C C – C – C – OR RO – C – C – OR | | | | | | | H H nH H Hm H H n+m 1000 – 50,000 units © 2008 Brooks/Cole 5 Addition Polymers Polyethylene is the world’s most widely used polymer. High density polyethylene (HDPE) Linear chains. High density, high molar mass. Hard and semirigid. Used in plastic milk jugs © 2008 Brooks/Cole 6 Addition Polymers Low density polyethylene (LDPE) Branched chains. Cannot pack as closely. Soft and pliable (sandwich bags) Cross-linked polyethylene (CLPE) Short -CH2- groups connect adjacent chains. Very tough (soft-drink caps) © 2008 Brooks/Cole 7 © 2008 Brooks/Cole 8 © 2008 Brooks/Cole 9 Addition Polymers Substituted ethylene can be used: Monomer Polymer Name F F CF2=CF2 | | –C–C– tetrafluoroethylene | | polytetrafluoroethylene F F n (PTFE, Teflon) H H H | | CH2=C– –C–C– polystyrene | styrene H n © 2008 Brooks/Cole 10 Addition Polymers H H Many other substituted ethylenes are important C=C H R R Common Name Polymer Name Uses H ethylene polyethylene Squeeze bottles, films, toys, (Polythene) electrical insulation CH3 propylene polypropylene Bottles, films, indoor-outdoor (Vectra, Herculon) carpet Cl vinyl chloride poly(vinyl chloride) Floor tile, raincoats, pipe (PVC) CN acrylonitrile polyacrylonitrile Rugs, fabrics (Orlon, Acrilan) C6H6 styrene polystyrene Food and drink coolers, (Styrofoam, Styron) building insulation OCOCH3 poly(vinyl acetate) Latex paint, adhesives, vinyl acetate (PVA) textile coatings COOCH3 poly(methyl methacrylate) Transparent objects, latex methyl methacrylate (PMMA, Plexiglass) paint, contact lenses © 2008 Brooks/Cole 11 Natural and Synthetic Rubbers Latex oozes from rubber trees (water/rubber particle emulsion). It becomes gum rubber when the water is removed - a sticky, elastic and water repellant mass. Charles Goodyear (1839) discovered vulcanization: Gum rubber + sulfur + heat → vulcanized rubber No longer sticky… (still elastic and water-repellant) © 2008 Brooks/Cole 12 Natural and Synthetic Rubbers Rubber is poly-cis-isoprene H CH3 C=C H H C=C isoprene (2-methyl-1,3,-butadiene) H H Poly-trans-isoprene (gutta-percha) comes from another tree. Hard and brittle – used in golf-ball covers. © 2008 Brooks/Cole 13 Natural and Synthetic Rubbers Rubber is an elastomer – no permanent change when stretched. Sulfur cross-links help maintain a shape. Today, cis-isoprene is synthesized and polymerized when natural rubber is in short supply. © 2008 Brooks/Cole 14 Copolymers Copolymers are made from mixed monomers: Styrene-butadiene rubber (SBR) is used to make tires: H H H CH2=C – + C=C H H C=C butadiene (B) styrene (S) H H S+B →→→→→ –S–B– n ABS (acrylonitrile-butadiene-styrene) is used to make car bumpers and computer cases. © 2008 Brooks/Cole 15 Condensation Polymers Condensation reactions release a small molecule (often water). e.g. acid + alcohol ester + water Polyesters form between molecules with two acid groups and two alcohol groups: HOOC – – COOH + HO – CH2 – CH2 – OH ethylene glycol terephthalic acid HOOC – – COO – CH2 – CH2 – OH + H2O © 2008 Brooks/Cole 16 Condensation Polymers HOOC – – COO – CH2 – CH2 – OH This product still has two reactive groups, and can polymerize. –O–C– – C – O – CH2 – CH2 – || || O O n © 2008 Brooks/Cole 17 Condensation Polymers Carboxylic acid + amine → amide + water © 2008 Brooks/Cole 18 Condensation Polymers Dicarboxylic acids react with diamines to produce polyamide. Nylon-66 adipic acid hexamethylenediamine HO-C-(CH2)4-C-OH + H2N-(CH2)6-NH2 || || O O 6 C’s in the acid, 6 C’s in the amine… HO-C-(CH2)4-C-NH-(CH2)6-NH2 + H2O || || O O Two reactive groups available… … polymerization occurs © 2008 Brooks/Cole 19 Hydrogen Bonding H-bonds between chains align the polymer into fibers. Nylon-66 © 2008 Brooks/Cole 20 Recycling Plastics © 2008 Brooks/Cole 21 Biopolymers: Proteins and Polysaccharides Polymers are common in nature. Glucose (monomer) → cellulose & starch (polymers) Amino acids (monomers) → proteins (polymers) Amino acid Amino acids undergo condensation reactions (lose H2O). The amide link = peptide link in protein chemistry. Small ( < 50 ) amino acid polymers are polypeptides © 2008 Brooks/Cole 22 Common amino acids *essential amino acid (must be part of diet). ‡growing children also require arginine © 2008 Brooks/Cole 23 Biopolymers: Proteins and Polysaccharides There are two ways to link a pair of amino acids: peptide link H O H O H O H O | || | || | || | || H2N–C–C–OH + H–N–C–C–OH H2N–C–C–N–C–C–OH + H2O | | | | | | H H CH3 H H CH3 glycine alanine glycylalanine H O H O H O H O | || | || | || | || H2N–C–C–OH + H–N–C–C–OH H2N–C–C–N–C–C–OH + H2O | | | | | | CH3 H H CH3 H H alanine glycine alanylglycine © 2008 Brooks/Cole 24 Biopolymers: Proteins and Polysaccharides n different amino acids link in n! different ways. e.g. 3! = 3 x 2 x 1 = 6 polypeptides Phe-Ala-Ser Ser-Ala-Phe Ala-Ser-Phe Phe-Ser-Ala Ser-Phe-Ala Ala-Phe-Ser 5! = 120 polypeptides Proteins include multiple copies of amino acids. Huge number of possible combinations. © 2008 Brooks/Cole 25 Primary, Secondary & Tertiary Structure of a Protein The sequence of amino acids along as chain is the primary structure of the protein. The secondary structure describes regular patterns caused by hydrogen bonding. Types: alpha (α)-helix beta (β)-pleated sheet Tertiary structure is the overall shape of the protein. It is maintained by side-chain interactions and disulfide bonds (-S-S-). © 2008 Brooks/Cole 26 Secondary Structure of a Protein β-sheet α-helix H-bonds between long H-bonds between N-H and segments hold the beta a nearby C=O hold the sheet together helix coils in shape © 2008 Brooks/Cole 27 Tertiary Structure of a Protein A globular protein (chymotrypsin): α-helix (blue) β-sheet (green) random coils (copper) © 2008 Brooks/Cole 28 Monosaccharides to Polysaccharides Saccharide: scientific name for sugar © 2008 Brooks/Cole 29 Polysaccharides: Starches and Glycogen Plant starch is stored in protein-covered granules. Two starches release on heating: amylose (~25%) ▪ ~200 glucose monomers in a straight-chain polymer. amylopectin (~75%). ▪ ~1000 glucose monomers in a branched-chain polymer. © 2008 Brooks/Cole 30 Cellulose, a Polysaccharide Cellulose is similar to amylose. Straight chain polymer of ~280 glucose units Glycosidic links between units alternate in direction. Every other glucose unit is turned over. In amylose Glycosidic links are in the same direction. Humans can digest starch, but not cellulose… © 2008 Brooks/Cole 31 Chemistry: The Molecular Science Moore, Stanitski and Jurs Chapter 19: Electrochemistry and its Applications © 2008 Brooks/Cole 1 Redox Reactions Electrochemistry The study and use of e- flow in chemical reactions: Redox reactions generate (and use) e- Those e- can be harnessed (batteries). Corrosion is an electrochemical reaction. Applied e- flow Can drive reactant-favored reactions toward products. Rechargeable batteries, electrolysis, and electroplating… © 2008 Brooks/Cole 2 Redox Reactions Oxidation Number Refresher Pure element = 0. Monatomic ion = charge of ion. (ox. numbers in a species) = overall charge. Element ox. no. exceptions? F −1 None Cl, Br, I −1 Interhalogens H +1 Metal hydrides = -1 O −2 Metal peroxides = -1 Halogen oxides © 2008 Brooks/Cole 3 Redox Reactions Oxidation & reduction (Redox) always occur together. Reduction = gain of e- = decrease in ox. no. Oxidation = loss of e- = increase in ox. no. +2 e- 2HCl(aq) + Mg(s) H2(g) + MgCl2 (aq) +1 -1 0 0 +2 -1 -2 e- H+ is reduced, Mg is oxidized. © 2008 Brooks/Cole 4 Redox Reactions Give oxidation numbers for each atom. Identify the oxidizing and reducing agents: 6 Fe2+ + Cr2O72- + 14 H3O+ 6 Fe3+ + 2 Cr3+ + 21 H2O Species Ox. number Explanation Fe2+ +2 charge on ion Cr2O72- O = -2; Cr = +6 O is usually -2; 2(Cr) + 7(-2) = -2 H 3 O+ O = -2; H = +1 O is usually -2; H is usually +1 Fe3+ +3 charge on ion Cr3+ +3 charge on ion H 2O O = -2; H = +1 O is usually -2, H is usually +1 Fe2+ → Fe3+ oxidation Fe2+ = reducing agent Cr(+6) → Cr3+ reduction Cr2O72- = oxidizing agent © 2008 Brooks/Cole 5 Galvanic Cells anode cathode oxidation reduction spontaneous redox reaction 6 Using Half-Reactions to Understand Redox Half-reactions may include different numbers of e- Al(s) Al3+(aq) + 3 e- Zn2+(aq) + 2 e- Zn(s) e- must balance in the full reaction. 2[ Al(s) Al3+(aq) + 3 e- ] 3[ Zn2+(aq) + 2 e- Zn(s) ] 2 Al(s) + 3 Zn2+(aq) 2 Al3+(aq) + 3 Zn(s) © 2008 Brooks/Cole 7 Balancing Redox Equations Redox in acidic or basic solutions are harder (H2O, H3O+ or OH- omitted…). Balance: H3AsO4 + I2 → HAsO2 + IO3- (aqueous acidic solution) © 2008 Brooks/Cole 8 Balancing Redox Equations in Acidic Solution H3AsO4 + I2 HAsO2 + IO3- (acidic solution) (i) Oxidized? Reduced? I2 (I = 0) → IO3- (I = +5) oxidation H3AsO4 (As = +5) → HAsO2 (As = +3) reduction (ii) Unbalanced half-reactions: H3AsO4 → HAsO2 I2 → IO3- (iii) Balance (except H and O). H3AsO4 → HAsO2 I2 → 2 IO3- © 2008 Brooks/Cole 9 Balancing Redox Equations in Acidic Solution (iv) Balance O (add H2O as needed) H3AsO4 → HAsO2 + 2 H2O I2 + 6 H2O → 2 IO3- (v) Balance H (add H+ as needed) H3AsO4 + 2 H+ → HAsO2 +2 H2O I2 + 6 H2O → 2 IO3- + 12 H+ (vi) Balance charges (add e- ) H3AsO4 + 2 H+ + 2 e- → HAsO2 +2 H2O I2 + 6 H2O → 2 IO3- + 12 H+ + 10 e- zero charge {2(-1) + 12(+1)} 10(-1) 0 = 10 + -10 © 2008 Brooks/Cole 10 Balancing Redox Equations in Acidic Solution (vii) Equalize e- and add 5 [ H3AsO4 + 2 H+ + 2 e- → HAsO2 +2 H2O ] 1[ I2 + 6 H2O → 2 IO3- + 12 H+ + 10 e- ] 5 H3AsO4 + 10 H+ + 10 e- + I2 + 6 H2O 2H+ → 5 HAsO2 + 10 H2O + 2 IO3- + 12 H+ + 10 e- 4H2O 5 H3AsO4 + I2 → 5 HAsO2 + 4 H2O + 2 IO3- + 2 H+ (viii) Make H3O+ (H2O + H+). Add H2O if needed 5 H3AsO4 + I2 → 5 HAsO2 + 2 H2O + 2 IO3- + 2 H3O+ © 2008 Brooks/Cole 11 Balancing Redox Equations in Basic Solution Balance the following (basic conditions): N2 + S2- S + N2H4 (i) Oxidized? Reduced? N2 (N = 0) → N2H4 (N = -2) reduction S2- (S = -2) → S (S = 0) oxidation (ii) Unbalanced half-reactions: N2 → N2H4 S2- → S (iii) Balance (except H and O). N2 → N2H4 S2- → S © 2008 Brooks/Cole 12 Balancing Redox Equations in Basic Solution (iv) Balance O (add H2O as needed) N2 → N2H4 S2- → S (v) Balance H (add H+ as needed) N2 + 4 H+ → N2H4 S2- → S (vi) Balance charges (add e- ) N2 + 4 H+ + 4 e- → N2H4 S2- → S + 2 e- © 2008 Brooks/Cole 13 Balancing Redox Equations in Basic Solution (vi) Equalize e- and add 1 [ N2 + 4 H+ + 4 e- → N2H4 ] 2 [ S2- → S + 2 e- ] N2 + 4 H+ + 4 e- + 2 S2- → N2H4 + 2 S + 4 e- N2 + 4 H+ + 2 S2- → N2H4 + 2 S (vii) Make H2O (H+ + OH-). Add OH- N2 + 4 H+ + 4 OH- + 2 S2- → N2H4 + 2 S + 4 OH- N2 + 4 H2O + 2 S2- → N2H4 + 2 S + 4 OH- © 2008 Brooks/Cole 14 Electrochemical Cells Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) e- e- Linked oxidation and reduction reactions. e- move across an external conductor. Also called a voltaic cell or a battery. (A battery is strictly a series of linked voltaic cells) © 2008 Brooks/Cole 15 Electrochemical Cells Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) e- e- Electrodes (anode & cathode) Allow e- to pass in and out of solution. A salt bridge (or porous barrier) Salt is required... bridge Anode Cathode (oxidation) (reduction) © 2008 Brooks/Cole 16 Electrochemical Cells Salt bridge Contains a salt solution (e.g. K2SO4 ). Ions pass into the cells (restricts bulk flow). Stops charge buildup. porous plug Cu Zn SO42- K+ K2SO4 Zn2+ Cu2+ 2- SO4 released 2 K+ released as Zn → Zn2+ as Cu2+ → Cu © 2008 Brooks/Cole 17 Electrochemical Cells © 2008 Brooks/Cole 18 Electrochemical Cells Zinc is removed: Zn Zn2+ + 2 e- Oxidation at the anode (both vowels). Zn supplies e-. Anode has “-” charge. Copper is deposited: Cu2+ + 2 e- Cu Reduction at the cathode (both consonants). Cu2+ accepts e-. Cathode has “+” charge © 2008 Brooks/Cole 19 Electrochemical Cells A compact notation: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) anode cell cathode cell Current flows from anode to cathode. | = phase boundary || = salt bridge Details (e.g. concentration) listed after each species. © 2008 Brooks/Cole 20 Electrochemical Cells and Voltage Electrical work = charge x ΔEp = (number of e-) ΔEp SI Units Charge: 1 coulomb (C) = 1 ampere x second = 1 As Potential: 1 volt (V) = 1 J C-1 Voltage depends on cell chemistry ≠ size. Charge depends on nreactants ≈ size. © 2008 Brooks/Cole 21 Electrochemical Cells and Voltage Cell voltage varies if conditions vary. A standard voltage ( E° ) occurs if: All [solute] = 1 M. – or saturated if the solubility < 1 M. All gases have P = 1 bar. All solids are pure. © 2008 Brooks/Cole 22 Electrochemical Cells and Voltage E°cell is positive = product favored reaction (E°cell < 0 is reactant favored). Absolute voltages cannot be measured. They are measured relative to a standard electrode © 2008 Brooks/Cole 23 Electrochemical Cells and Voltage Standard hydrogen electrode (SHE) Pt | H2(1bar), 1M H3O+ || E° = 0 V (oxidation & reduction). 2 H3O+(aq, 1M) + 2 e- H2(g, 1 bar) + 2 H2O (l) © 2008 Brooks/Cole 24 Using Standard Cell Potentials Reduction Half Reaction E° (V) F2(g) + 2 e- → 2 F-(aq) +2.87 H2O2(aq) + 2 H3O + 2 e- → 4 H2O(l) +1.77 MnO4-(aq)+8 H3O+ + 5 e- → Mn2+(aq) + 12 H2O(l) +1.51 Tabulated as Cl2(g) + 2 e- → 2 Cl-(aq) +1.358 reductions Br2(g) + 2 e- → 2 Br-(aq) +1.066 Ag+(aq) + e- → Ag(s) +0.799 Cu2+(aq) + 2 e- → Cu(s) +0.337 2 H3O+(aq) + 2 e- → H2(g) + 2 H2O(l) 0.00 Ni2+(aq) + 2 e- → Ni(s) -0.25 Fe2+(aq) + 2 e- → Fe(s) -0.44 Zn2+(aq) + 2 e- → Zn(s) -0.763 Al3+(aq) + 3 e- → Al(s) -1.66 Li+(aq) + e- → Li(s) -3.045 © 2008 Brooks/Cole 25 Electrochemical Cells and Voltage Cu2+(aq) + 2 e- Cu(s) reduction Zn(s) Zn2+(aq) + 2 e- oxidation The overall voltage: E°cell = E°Zn2+, reduction + E°Cu, oxidation If an equation is reversed, E° → -1 x E° E°oxidation = - E°reduction. E° tables are reduction values, so: E°cell = E°cathode - E°anode E°cell = E°reduction + E°oxidation © 2008 Brooks/Cole 26 Electrochemical Cells and Voltage Reaction Process E°red (Table) E° Zn(s) Zn2+(aq) + 2 e- oxid. (anode) -0.76 V +0.76 V Cu2+(aq) + 2 e- Cu(s) red. (cathode) +0.34 V +0.34 V E°cell = E°cathode - E°anode E°cell = 0.34 – (-0.76) V = 1.10 V Or E°cell = E°oxid + E°red = 0.34 + 0.76 V = 1.10 V © 2008 Brooks/Cole 27 Electrochemical Cells and Voltage What is E° for a Ni(s) | Ni2+|| Ag+ | Ag(s) cell? Reduction Half Reaction E° (V) Ag+(aq) + e- → Ag(s) +0.799 Cu2+(aq) + 2 e- → Cu(s) +0.337 2 H3O+(aq) + 2 e- → H2(g) + 2 H2O(l) 0.00 Ni2+(aq) + 2 e- → Ni(s) -0.25 Anode written on the left; cathode the right. E° = E°cathode - E°anode = 0.799 – (-0.25) V = 1.05 V © 2008 Brooks/Cole 28 Electrochemical Cells and Voltage If Ni(s)| Ni2+(aq, 1M) is connected to SHE, the Ni electrode loses mass over time. E°cell = 0.25 V. Is Ni oxidized or reduced? What is E° for the Ni half cell? Ni loses mass: Ni(s) → Ni2+(aq) + 2 e- Ni is oxidized (anode). Since: E°cell = E°cathode - E°anode 0.25 V = E°SHE - E°anode = 0 - E°anode E°anode = -0.25 V (Note: this is the tabulated reduction value) © 2008 Brooks/Cole 29 Using Standard Cell Potentials 1. Tabulated half-cell E° are reductions. 2. Reactions can be reversed. 3. Larger E° = easier reduction. 4. Smaller E° = easier oxidation (reverse reaction). 5. A “left” species, will oxidize any “right” below it. 6. E° depends on [reactant] and [product], but not on nreactant or nproduct (i.e not stoichiometric coefficient). © 2008 Brooks/Cole 30 Using Standard Cell Potentials Will Zn(s) react with a 1 M iron(III) solution? If so what is E° for the reaction? Reduction Half Reaction E° (V) F2(g) +2 e- 2 F-(aq) +2.87 Fe3+ + e- Fe2+ +0.771 2 H3O+ + 2 e- H2(g) + 2 H2O(l) 0.00 Zn2+ + 2 e- Zn(s) -0.763 Yes! Fe3+ (“left”) is higher than Zn (“right”). © 2008 Brooks/Cole 31 Using Standard Cell Potentials Will Zn react with a 1 M iron(III) solution? If so what is E° for the reaction. 2 Fe3+ + 2 e- → Fe2+ E° = +0.771 V Zn → Zn2+ + 2 e- E° = -1(-0.763 V) 2 Fe3+ + Zn → 2 Fe2+ + Zn2+ E°cell = +1.534 V Note 2 x (Fe3+ reaction) to balance e-. E° (Fe3+) is not doubled. © 2008 Brooks/Cole 32 Using Standard Cell Potentials a) Will Al(s) react with a 1 M tin(IV) solution? b) Will 1 M Fe2+ react with Sn(s)? Sn4+ + 2 e- → Sn2+ (s) +0.15 V Sn2+ + 2 e- → Sn (s) -0.14 V Fe2+ + 2 e- → Fe (s) -0.44 V Al3+ + 3 e- → Al (s) -1.66 V (a) Yes – “left” (Sn4+) above “right” (Al) (b) No – “left” (Fe2+) below “right” (Sn) © 2008 Brooks/Cole 33 Using Standard Cell Potentials What’s the voltage of: Al(s)|Al3+,1M || Sn2+,1M|Sn(s) ? Sn2+ + 2 e- → Sn E° = -0.14 V Al → Al3+ + 3 e- E° = -1(-1.66 V) Balance e-: 3(Sn2+ + 2 e- → Sn) E° = -0.14 V 2(Al → Al3+ + 3 e- ) E° = +1.66 V 3 Sn2+ + 2 Al → 3 Sn + 2 Al3+ E°cell = +1.52 V © 2008 Brooks/Cole 34 E° and Gibbs Free Energy Product-favored reactions: ΔG° < 0 Spontaneous cell reactions: E°cell > 0. ΔG° = –n F E°cell with n = moles of e- transferred, F = Faraday constant = charge/(mol of e-). = (e- charge) x (Avogadro’s number) = (1.60218 x 10-19 C)(6.02214 x 1023 mol-1) F = 96,485 C/mol = 96,500 C/mol (3 sig. fig.) © 2008 Brooks/Cole 35 E° and Gibbs Free Energy Cu2+ + Zn(s) → Cu(s) + Zn2+ E°cell = 1.10 V Spontaneous. ΔG° = − nFE°cell = −2 mol (96500 C/mol)(1.10 V) = −2.12 x105 J (1 J = 1 C V) = −212 kJ © 2008 Brooks/Cole 36 E° and Gibbs Free Energy Since: ΔG° = − RT ln K° = −nFE°cell E°cell =R T R ln K° = (2.303) T log K° nF nF At 298 K: E°cell =0.0257 V ln K° or, n 0.0592 V n E°cell = log K° © 2008 Brooks/Cole 37 E° and Gibbs Free Energy Determine K° for: Cu2+ + Zn(s) → Cu(s) + Zn2+ E°cell = 1.10 V E°cell =0.0592 V log K° n 1.10 V = 0.0592 V log K° 2 log K° = 37.16 K° = 1037.16 = 1.5 x 1037 © 2008 Brooks/Cole 38 Effect of Concentration on Cell Potential E° values apply if [solute] = 1 M (or saturated). Other conditions: RT Ecell = E°cell − ln Q (Nernst equation) nF At 298 K: Ecell = E°cell − 0.0592 V log Q n Ecell = E°cell − 0.0257 V ln Q n © 2008 Brooks/Cole 39 Effect of Concentration on Cell Potential What is the voltage for Cu2+ + Zn(s) → Cu(s) + Zn2+ if [Cu2+] = 0.1 M and [Zn2+] = 5.0 M. E°cell = 1.10 V. 0.0592 [Zn2+] E = E° − log 2 [Cu2+] 0.0592 5.0 E = 1.10 − log 2 0.1 = 1.05 V © 2008 Brooks/Cole 40 Concentration Cells Concentration dependence leads to: Zn | Zn2+ (dilute) || Zn2+ (conc.) | Zn Ecell ≠ 0 V Example Zn | Zn2+ (0.01M) || Zn2+ (1M) | Zn anode = oxidation cathode = reduction Zn(s) → Zn2+ (0.01M) + 2 e- Zn2+ (1M) + 2 e- → Zn(s) Zn2+(1M) → Zn2+(0.01M) (net reaction) © 2008 Brooks/Cole 41 Concentration Cells 0.0592 [Zn2+]dilute E = E° − log 2 [Zn2+]conc 0.0592 0.010 E = 0.0 − log 2 1 0.0592 E = 0.0 − log 10-2 2 E = 0.0592 V © 2008 Brooks/Cole 42 Common Batteries Primary battery One time use. Not easily rechargeable Secondary battery Rechargeable battery. © 2008 Brooks/Cole 43 Primary Batteries Leclanché Dry Cell © 2008 Brooks/Cole 44 Primary Batteries Mercury battery Zn(s) + 2 OH- → ZnO(aq) + H2O(l) + 2 e- HgO(s) + H2O(l) + 2 e- → Hg(l) + 2 OH-(aq) Zn(s) + HgO(s) → Hg(l) + ZnO(aq) Ecell = 1.35 V © 2008 Brooks/Cole 45 Secondary Batteries Lead-Acid Battery (high capacity, high current). Pb(s) + HSO4-(aq) + H2O(l) → PbSO4(s) + H3O+(aq) + 2 e- PbO2(s) + 3 H3O+(aq) + HSO4-(aq) + 2 e- → PbSO4(s) + 5 H2O Pb + PbO2(s) + 2 H3O+ + 2 HSO4-(aq) → 2 PbSO4(s) + 4 H2O Net E° = +2.041 V Insoluble PbSO4 stays on the electrodes The reaction is reversed by recharging. © 2008 Brooks/Cole 46 Lead-Acid Storage Battery 6 cells in series (12 V). © 2008 Brooks/Cole 47 Fuel Cells Convert bond energy into electricity. Proton-Exchange V e- Membrane (PEM) fuel cell. e- Pt catalyst O2 in H2 → 2 +2 H+ e- H2 in ½ O2 + 2 H+ + 2 e- → H2O gases flow H+ through channels H+ Graphite electrodes. H2 out H2O out Pt catalyst coated on both sides of the membrane. Anode Cathode H+ exchange membrane © 2008 Brooks/Cole 48 Electrolysis Electrolytic cell: Applied voltage forces a reaction to occur. e.g. electrolysis of molten NaCl: 2 Na+ + 2 e- → 2Na (l) E° = -2.714 V 2 Cl- → Cl2 (g) + 2 e- E° = -1.358 V 2 Na+ + 2 Cl- → Cl2 (g) + 2 Na(l) E° = -4.072 V Na(l) and Cl2(g) produced if > 4.1 V is applied. However, melting NaCl takes lots of energy… © 2008 Brooks/Cole 49 Electrolysis Aqueous solution? Other reactions can occur. Consider KI(aq): K+(aq) + e- → K(s) E° = -2.925 V 2 H- 2O(l) + 2 e- → H2(g)- + 2 OH-(aq) E° = -0.828 2 I (aq) → I2(aq) + 2 e E° = -0.535 V V 6 H2O(l) → O2(g) + 4 H3O+(aq) + 4 e- E° = -1.229 V (Written Most as oxidations; positive E occurs…E°Overall: → -1 x E°) 2 H2O + 2 I- → H2 + I2 + 2 OH- Ecell = -1.363 V © 2008 Brooks/Cole 50 Electrolysis Brown = I2 ; Purple = OH- (phenolphthalein) © 2008 Brooks/Cole 51 Reduction Half Reaction E° (V) Electrolysis F2(g) +2 e- → 2 F-(aq) +2.87 Aqueous Pb4+(aq) +2 e- → Pb2+(aq) +1.8 H2O2(aq) + 2 H3O+2 e- → 4 H2O(l) +1.77 oxidation will Cl2(g) +2 e- → 2 Cl-(aq) +1.358 not occur. O2(g) + 4 H3O+(aq) + 4 e- → 6 H2O(l) +1.229 Br2(g) +2 e- → 2 Br-(aq) +1.066 Ag+(aq) + e- → Ag(s) +0.799 2 H3O+(aq) + 2 e- → H2(g) + 2 H2O(l) 0.00 Ni2+(aq) + 2 e- → Ni(s) -0.25 Fe2+(aq) + 2 e- → Fe(s) -0.44 Zn2+(aq) + 2 e- → Zn(s) -0.763 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) -0.8277 Aqueous Al3+(aq) + 3 e- → Al(s) -1.66 reduction will Na+(aq) + e- → Na(s) -2.714 not occur. K+(aq) + e- → K(s) -2.925 © 2008 Brooks/Cole 52 Electrolysis Summary Metal ions are reduced if E°red > −0.8 V Aqueous Na+, K+, Mg2+, Al3+ … cannot be reduced. Anions can be oxidized if E°red < +1.2 V Aqueous F- … cannot be oxidized. In practice Erequired > E calculated. Overvoltage is needed. (Cl-(aq) can be oxidized to Cl2(g)) © 2008 Brooks/Cole 53 Counting Electrons Cu2+(aq) + 2 e- Cu(s) 2 mol e- ≡ 1 mol Cu. Also charge = current x time 1 coulomb = 1 ampere x 1 second 1C=1As © 2008 Brooks/Cole 54 Counting Electrons Determine the mass of Cu plated onto an electrode from a Cu2+ solution by the application of a 10. A current for 10. minutes. Cu2+(aq) + 2 e- Cu(s) Charge = 10. A x 10. min x (60 s/min) = 6.0 x 103 C 1 mol e- 1 mol Cu 63.55 g 6.0 x103 C = 2.0 g 96500 C 2 mol e- 1 mol Cu © 2008 Brooks/Cole 55 Counting Electrons What is the cost to produce 14. g of Al (mass of a soda can) by the reduction of Al3+. Assume V = 3.0 V and 1 kWh of electricity costs 10 cents. Al3+ + 3 e- Al 14. g 1 mol Al 3 mol e- 96500 C = 1.5 x 105 C 26.98 g 1 mol Al 1 mol e- 1J 1 kWh Energy used = (1.5 x105 C)(3.0V) 1 C x 1V 3.60 x 106 J = 0.13 kWh (or 1.3 cents) © 2008 Brooks/Cole 56 Corrosion Corrosion is the term usually applied to the deterioration of metals by an electrochemical process. 57 Cathodic Protection of an Iron Storage Tank 58 Corrosion: Product-Favored Reactions Anode: M(s) Mn+ + n e- Cathode: often involve water and/or O2 O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq) 2 H2O(l) + 2 e- → 2 OH-(aq) + H2(g) Anode and cathode must be electrically connected. (The metal itself acts as the conductor). © 2008 Brooks/Cole 59 Corrosion: Product-Favored Reactions Anode: 2[Fe(s) Fe2+ + 2 e-] Cathode: O2(g) + 2 H2O(l) + 4 e- 4 OH-(aq) 2 Fe(s) + O2(g) + 2 H2O(l) 2 Fe(OH)2(s) Converted to rust by oxygen: 4 Fe(OH)2(s) + O2(g) 2 Fe2O3·H2O(s) + 2 H2O(l) © 2008 Brooks/Cole 60 Corrosion: Product-Favored Reactions Blue = Fe2+ indicator ; Purple = OH- indicator © 2008 Brooks/Cole 61 Corrosion Protection Anodic inhibition – paint or coat the surface. Cathodic protection – A more reactive metal is attached to the metal. The reactive metal corrodes – acts as a sacrificial anode. A block of Mg is attached to gasoline storage tanks. Mg is oxidized (Mg → Mg2+ + 2 e- E° = 2.87 V) not the iron tank (Fe → Fe2+ + 2 e- E° = 0.44V). © 2008 Brooks/Cole 62 Corrosion Prevention Iron is galvanized (coated with Zn): Zn(OH)2 (insoluble film) is produced on the surface. Zn → Zn2++ 2 e- E° = 0.763 V Fe → Fe2+ + 2 e- E° = 0.44 V © 2008 Brooks/Cole 63

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