Chem 10130 Applied Introductory and Physical Chemistry Past Paper PDF

CHEM 10130 Applied Introductory and Physical Chemistry AUTUMN 2024  CHEM 10130 INTRODUCTION TO GENERAL AND PHYSICAL CHEMISTRY Overview of topics 1) Sources of energy and nature of energy transfer 2) Nature of work and heat 3)...

CHEM 10130 Applied Introductory and Physical Chemistry AUTUMN 2024  CHEM 10130 INTRODUCTION TO GENERAL AND PHYSICAL CHEMISTRY Overview of topics 1) Sources of energy and nature of energy transfer 2) Nature of work and heat 3) Nature of enthalpy and entropy 4) Nature of spontaneous processes and relationship to Gibbs free energy 5) Nature of equilibrium reactions and the equilibrium constant 6) Nature of binding reactions 7) Nature of the relationship between equilibria and Gibbs free energy 8) Nature of rates of reaction and relationship to collision theory 9) Nature of activation energy and catalysis 10) Nature of enzymes, enzyme kinetics, and enzyme inhibition  What is energy? Energy is the capacity to supply heat (q) or do work (w). Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. In other words, during a chemical or physical change, energy can be neither created nor destroyed, although it can be changed in form. What law describes this observation? The First Law of Thermodynamics, also referred to as the The Law of Conservation of Energy. - Energy can neither be created nor destroyed - The total energy of the universe is constant - Expressed mathematically ΔU = 0 where ΔU = q + w (q is heat and w is work) How is heat (q) related to specific heat capacity (C)? q = C × m × ΔT § q = energy gained or lost as heat § m = mass of a substance § C = specific heat capacity § ΔT = change in temperature Δ𝑇 = Δ𝑇!"#$% − Δ𝑇"#"&"$% Energy and changes of state Melting à heat of fusion Energy transferred as heat to convert a substance from a solid at its melting point to a liquid is called heat of fusion Vaporization à heat of vaporization Energy transferred as heat to convert a substance from a liquid at its boiling point to a gas is called heat of vaporization Temperature is constant during a change of state! Temperature change as function of heat Water warms from −50 °C to 200 °C (1 Atm) Temp is constant during a change of state Enthalpy (H) § Enthalpy is the heat content of a chemical reaction carried out at constant pressure (like the pressure of the atmosphere). ΔH is the change in enthalpy. § At constant pressure DH = Hfinal – Hinitial = qp Subscript “p” here means “at constant pressure” § DH is a state function DH = Hproduct – Hreactant What is a state function? A state function is a property of a system that depends only on the state of the system and not upon how the system was brought to that state, or on the history of the system. Energy is an example of a state function. A state function’s value is fixed when temperature, pressure, composition, and physical form are specified. We distinguish between a state function and a path function. A path function is a property that depends on the preparation of the state. Work (w) and heat (q) are path functions. Hess’s Law and Energy diagrams C ( s ) + O 2 ( g ) ® CO 2 ( g ) + 393.5 kJ § Manipulate the equation § Add and subtract reactants and products § What you do to the reactants and products you do to the ΔH value § Cancel what appears on both sides of the equation § Look for what is on the appropriate side and what needs to be cancelled § Worry about numbers last ΔfH° is the enthalpy of formation DfH° is the heat required to form 1 mol of a compound from its elements in their standard state. ° identifies standard conditions (1 atm, 1 M, s, l, g) f stands for “formation” DfH° of an element = 0 (by definition) Given The Thermochemical Equation N 2 (g ) + 3 H 2 (g ) ® 2 NH 3 (g ) H ° = - 92.22 kJ Write a thermochemical equation for the following § Formation of 1 mol of NH3(g) § Decomposition of 4 mol of NH3(g) § Combination of 1 mol of H2(g) with a stoichiometric quantity of nitrogen Entropy (S) 1) In a spontaneous process, the energy of the final state is more dispersed. 2) The system moves to a higher state of disorder. 3) The Second Law of Thermodynamics states that a spontaneous process results in an increase in the entropy of the universe, DS > 0. Textbook Chemistry 2e link 4) The entropy of a perfect crystal approaches 0 as the temperature approaches 0 K. 5) Entropy increases with increasing temperature (𝑇). The entropy of a substance increases (ΔS > 0) as it transforms from a relatively ordered solid, to a less-ordered liquid, and then to a still less-ordered gas. Predicting Spontaneity under standard conditions  Quantifying Entropy Changes Problem: Some standard Molar Entropy Values at 298 K Consider the reaction of hydrogen and oxygen to form liquid water. What is the standard molar entropy for the reaction? 2H2 (g) + O2 (g) → 2H2O(l) J " J J# J Δr So = 2 × 69.6 − ' 2 × 130.7 + 205.3 ( = −326.9 K ) K K* K The enthalpy change is negative (net decrease in dispersion) due to the change in number of moles: 3 reduced to 2  Reaction Quotient For a general chemical reaction products aA + bB ⇌ cC + dD ' ( 𝐶 𝐷 𝑄' = $ ) 𝐴 𝐵 reactants The subscript “c” refers to concentrations Molar concentrations  Reaction Quotient If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures Partial pressures aA + bB ⇌ cC + dD ' , 𝑃+ 𝑃, 𝑄* = 𝑃- - 𝑃.. The subscript “P” refers to pressure Different equilibrium constants In general, for aA + bB ⇌ cC + dD an equilibrium constant may be expressed in terms of either the molar concentrations (Kc) or partial pressures (Kp) of the reactants and products, where Kp = Kc(RT)∆n where ∆n = (c+d)-(a+b) See derivation in Chemistry 2e Chapter 13.2 Conventionally, the equilibrium constant of water is termed Kw  Disturbing a Chemical Equilibrium: Le Châtelier’s principle The equilibrium between reactants and products may be disturbed in three ways: (1) by changing the temperature (2) by changing the concentration of a reactant (3) by changing the volume (for systems involving gases) If a change to a system in equilibrium imposes a “stress”, the equilibrium position will shift in a direction that tends to reduce that change. This statement is often referred to as Le Châtelier’s principle. Effect of the Addition or Removal of a Reactant or Product If the concentration of a reactant or product is changed from its equilibrium value at a given temperature, equilibrium will be reestablished eventually. The new equilibrium concentrations of reactants and products will be different, but the value of the equilibrium constant expression will still equal K  Example: Effect of a Change in Concentration Assume equilibrium has been established in a 1.00 L flask with [butane] = 0.0050 M and [isobutane] = 0.0125 M and Kc = 2.50. If 0.0150 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium? Reactants (Butane) Products (Isobutane) I 0.0050 0.0125 C 0.0050 + 0.0150 0.0125 ∆C -x +x E 0.0050 + 0.0150 - x 0.0125 + x What are [butane] and [isobutane] at new equilibrium?  What is pH? pH stands for “potentia hydrogenii” or “power hydrogen” p[H+] = -log10 [H+] p is a shorthand notation for “-log10" (the negative logarithm, base 10 )  2 H2O ⇌ H3O+ + OH- Kw = [H3O+]·[OH-] = [H3O+] [OH-] [H2O]2 *activity of pure solids and liquids is always 1 The equilibrium constant at T=25°C is Kw = 1.0 · 10−14 mol2/L2 Kw = [H3O+]·[OH-] = 1.0 · 10−14 = [H3O+]2 =[OH-]2= 1.0 · 10−14 [H3O+] = [OH-] = (1.0 · 10−14)1/2 = 1.0 · 10−7 mol/L  What is the pH of a 5·10−2 M NaOH solution at 25°C? Solution: [OH-] = 0.05 M Concentration known to one significant figure, i.e. 5 x 10-2 [H+]= Kw/[OH-] = 2.0 · 10−13 M middle of calculation so don’t change anything p[H+] = -log10 [H+] = 12.7 One decimal place reported because concentration has one significant figure From http://abacus.bates.edu/acad/depts/biobook/How-SigF.pdf  Henderson-Hasselbalch equation The Henderson-Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Ka, of acid and the ratio of the concentrations, of the acid (HA) and its conjugate base (A-) in an equilibrium. acid conjugate base acid conjugate base  Textbook Chemistry 2e link Gibbs Free Energy § Nature favors processes which have: −H and +S § These processes tend to occur passively or are spontaneous § The Gibbs free energy predicts if something is spontaneous Whether a process is spontaneous requires evaluation of enthalpy and entropy: ∆G = ∆ H – T ∆ S J. Willard Gibbs G = Gibbs Free energy of the system 1839−1903 H = system enthalpy S = the entropy of the system How can we calculate the Gibbs Free Energy? DG° is a state function DG° = SnDGf°(products) − SmDGf°(reactants) ∆G° = ∆ H° – T ∆ S° At constant temperature: DG° = Gibbs free energy change of the system DH° = enthalpy change DS° = the entropy change at Standard State conditions Similar to the standard enthalpy of formation, DG°is by definition zero for elemental substances in their standard states. Example: Given the following information, calculate ∆rG°for the reaction below at 25°C. Mg(s) + 1/2O2(g) → MgO(s) ∆rH° = −601.24 kJ/mol-rxn ∆rS° = −108.36 J/K T = 273 + 25 = 298 K ∆G° = ∆ H° – T ∆ S° = −601.24 kJ × (1000 J/kJ) – 298K(-108.36 J/K) = -568.95 kJ  Gibbs free energy of formation Standard molar free energies of formation of some substances at 298 K Note that ∆fG°for an element or for a substance in its elemental state is 0  Practice #1 Calculate D°G for the reaction 1C3H8(g) + 5O2(g) ⇌ 3CO2(g)+ 4H2O(g) Δr G° = ∑n ⋅ Δ f G°(products) −∑ m ⋅ Δ f G°(reactants) =[3mol(-394kJ/mol) + 4mol(-229kJ/mol)] – [1mol(-24 kJ/mol) + 5mol(0 kJ/mol)] = -2074 kJ From a look-up table: DGf° (C3H8 (g))= -24 kJ/mol DGf° (O2 (g))= 0 DGf° (CO2(g)) = -394 kJ/mol DGf° (H2O(g)) = -229 kJ How do we quantify DrG if we have non-standard state conditions? Example: CO(g) + 2H2 (g) → CH3OH (l) Calculate DG at 25°C for this reaction, in which carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol. DGrxn° = -2.9 x 104 J/mol rxn 0 0 𝑄= = = 2.2×1045 R = 8.3145 J K-1 mol-1 *"# *$% % 7.1 9.1% T=273 + 25 = 298 K ∆𝐺 = ∆𝐺 / + 𝑅𝑇𝑙𝑛 𝑄 01! 2 = −2.9× + 𝑅𝑇𝑙𝑛 2.2×1045 3/% 62 = −38 𝑟𝑥𝑛 3/%  Reaction Rate and Stoichiometry change in concentration Δ [product ] Δ [reactant ] Rate = = =− change in time Δ time Δ time In general, for the reaction: aA + bB ® cC + dD 1 D [A] 1 D [B] 1 D [C] 1 D [D] Rate of reaction = - =- = = a Dt b Dt c Dt d Dt Concentrations of reactants Concentrations of products decrease with time therefore increase with time therefore these rates are negative these rates are positive  Example: Peroxide decomposition 2H2O2 ® 2H2O + O2 [H2O2] decreases a reactant Two moles of H2O2 disappear for each mole of O2 that forms; the rate of disappearance of H2O2 is twice the rate of appearance of O2. [O2] increases a product 1 ∆ 𝐻5 𝑂5 ∆ 𝑂5 Reaction rate = − = 2 ∆𝑡 ∆𝑡 Recall for aA + bB ® cC + dD 1 D [A] 1 D [B] 1 D [C] 1 D [D] Rate of reaction = - =- = = a Dt b Dt c Dt d Dt Integrated Rate Laws For a zero-order process, where reactants “R” convert to products, the rate law can be written: reactants ® products -1 D[R] Rate(Ms ) = - = k[R] = k Dt k has units of Ms-1 For a zero-order process, the rate is the rate constant! Summary Order Rate Law Integrated Rate Law Half-life Straight line plot 0 Rate = k [𝑹] − [𝑹]𝟎 = −𝒌𝒕 𝑹𝒐 [𝑹] vs. 𝒕 𝟐𝒌 1 Rate = k[R] 𝒍𝒏[𝑹] − 𝒍𝒏[𝑹]𝟎 = −𝒌𝒕 𝟎. 𝟔𝟗𝟑/𝒌 𝒍𝒏[𝑹] vs. 𝒕 𝟏 𝟏 𝟏 𝟏 2 Rate = k[R]2 − = 𝒌𝒕 vs. 𝒕 [𝑹] 𝑹𝒐 𝒌𝑹 𝒐 [𝑹] For 1st order reaction, at what time is [R] exactly half of [R]o? The time required for the concentration of R to become exactly half of its initial concentration [R]o/2 is the half-life 𝒅𝑹 [𝑹]𝟎 𝒓𝒂𝒕𝒆𝑹 = − =𝒌𝑹 𝒍𝒏 = 𝒌𝒕 𝒅𝒕 𝑹 𝒅𝑹 [𝑹]𝟎 = −𝒌𝒅𝒕 𝒍𝒏 = 𝒌𝒕𝟏/𝟐 [𝑹] [𝑹]𝟎 /𝟐 𝑹 𝒕 𝒅𝑹 𝒍𝒏 𝟐 = 𝒌𝒕𝟏/𝟐. =. −𝒌 𝒅𝒕 𝑹𝟎 [𝑹] 𝟎 𝒕𝟏/𝟐 = 𝟎. 𝟔𝟗𝟑/𝒌 𝒍𝒏 𝑹 − 𝒍𝒏[𝑹]𝟎 = −𝒌𝒕 For first-order reactions, the half-life is independent of concentration. Activation energy Ea The potential energy difference between reactants and activated complex. The reaction of NO2 and CO (to give NO and CO2) has an activation energy barrier of 132 kJ/mol-rxn. The reverse reaction (NO + CO2 ® NO2 + CO) requires 358 kJ/mol-rxn. The net energy change for the reaction of NO2 and CO is -226 kJ/mol-rxn. Collision theory *Svante Arrhenius (1859 - 1927) Arrhenius discovered that most reaction-rate data obeyed an equation based on three factors: Nobel prize Chemistry 1903 (1) The number of collisions per unit time (2) The fraction of collisions that occur with the correct orientation (3) The fraction of the colliding molecules that have an energy greater than or equal to Ea From these observations Arrhenius developed the aptly named Arrhenius equation. Reactions are faster at higher temperatures... Arrhenius equation 4𝑬𝒂 𝒌= 𝑨𝒆 𝑹𝑻 k is the rate constant A is known the frequency or pre–exponential T is the temperature in K factor This factor is interpreted as the Ea is the activation energy fraction of molecules having the R is the ideal-gas constant (8.314 J/K×mol) minimum energy required for Both A and Ea are specific to a given reaction. reaction. It is always less than one. A relates to the frequency of collisions of the reactants and the orientation of a favorable collision probability What are the units of A? A carries the units of the rate constant! Catalysis mechanism changed Ea changed DHrxn unchanged Enzymes – Nature’s catalysts For an enzyme to catalyze a reaction, several key steps must occur: The reactant (the substrate) must bind to the enzyme. The chemical reaction must take place. The products(s) of the reaction must leave the enzyme so that more substrate can bind and the process can be repeated. Models for Substrate Binding to Enzymes lock-and-key model induced fit model UCD Student Feedback Promotional Resources for Lecturers UCD Student Feedback Surveys Have your say now! Scan QR code Check Brightspace Check email ucd.surveys.evasysplus.ie