S5 Chemistry Notes PDF

UNIT 2: Energy Changes, Thermochemistry & Rates of Reaction 1. Different Forms of Energy Heat, work and energy all have units of Joules, J. The Law of Conservation of Mass-Energy: states that energy can change form. ○ eg. The chemical potential energ...

UNIT 2: Energy Changes, Thermochemistry & Rates of Reaction 1. Different Forms of Energy Heat, work and energy all have units of Joules, J. The Law of Conservation of Mass-Energy: states that energy can change form. ○ eg. The chemical potential energy in food is converted to mechanical energy to do work, such as lifting an object. 1. Heat Energy 𝑞 = 𝑚𝐶∆𝑇, where 𝑞 = heat (j) → to determine the rate of transfer (power, P), divide by time. 𝑚 = mass (g) 𝐽 𝐶 = specific heat capacity, different for each material ( 𝑔·𝑐 ) ∆𝑇 = change in temperature (°C or °F) 2. Mechanical Energy required to lift an object 𝑊 = 𝐹𝑑, where 𝐹 = 𝑚𝑎, where 𝑊 = work (J): 𝑚 = mass (kg) 𝐹 = force (N) 𝑎 = acceleration (usually 9.8m/s2) 𝑔 𝑑 = density ( 𝑚𝐿 ) 3. Mechanical Energy required to expand the volume of a cylinder 𝑊 = 𝐹𝑑 = 𝑃𝐴𝑑 = 𝑚𝑎𝑑 =− 𝑃∆𝑉 =− 𝑛𝑅𝑇 5 1 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑒 (𝑎𝑡𝑚) = 1. 013 × 10 𝑃𝑎 𝐹 = 𝑚𝑎 −2 𝐹 1 𝑃𝑎 = 1 𝑁𝑚 𝑃= 𝐴 3 1000 𝐿 = 1 𝑚 2. Intro to Thermochemistry + Enthalpy Thermodynamics: the science of the relationships between heat and other forms of energy. ○ 1st Law of Thermodynamics: Energy cannot be created or destroyed. ○ 2nd Law of Thermodynamics: For a spontaneous process, the entropy of the universe increases. ○ 3rd Law of Thermodynamics: A perfect crystal at zero Kelvin has zero entropy Thermochemistry: one area of thermodynamics. Energy: the potential or capacity to move matter (Units of energy are Joules) Intensive vs Extensive Properties Extensive properties (lowercase): vary with the amount of the substance ○ mass, weight, and volume. Intensive properties (uppercase): do not depend (independent) on the amount of the substance; ○ color, melting point, boiling point, electrical conductivity, physical state at a given temperature. Types of Energy 1. Kinetic Energy: Energy associated with the motion of 2. Potential Energy Energy stored in an object due to its an object. position. (for gravitational potential energy) 𝐸𝑘 = 2 𝑚𝑣 1 2 𝐸𝑝 = 𝑚𝑔ℎ 2 2 𝑘𝑔·𝑚 units: J (SI base units: 𝑘𝑔·𝑚 ) units: J (SI base units: 2 ) 𝑠 2 𝑠 3. Internal Energy, 𝑈: Refers to the kinetic and potential energy of the individual molecules, electrons, nuclei in the sample. For a beaker of water sitting on the benchtop, 𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐸𝑘 + 𝐸𝑝 + 𝑈 , and ∆𝑈 = 𝑄 + 𝑊 (positive W means work was done on the system, Q=△H at constant pressure), and 𝑊 =− 𝑃∆𝑉 =− 𝑛𝑅𝑇 (if moles are different in products and reactants, n=moles of product - moles of reactants) Since the sample is not moving, and has no gravitational potential energy relative to the benchtop, we have 𝐸𝑡𝑜𝑡𝑎𝑙 = 𝑈 The total energy of a substance equals its internal energy. Law of Conservation of Energy Energy may be converted from one form to another, but the total quantity of energy remains constant. This is the First Law of Thermodynamics. The (thermodynamic) system: the reactants and products The surroundings: everything else that is nearby. ○ eg. the beaker, the benchtop, you, the air in the room, etc [ℎ𝑒𝑎𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑠𝑦𝑠𝑡𝑒𝑚] =− [ℎ𝑒𝑎𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠] ○ Note the negative sign, indicating exothermic - losing energy Heat, q: the energy that flows into, or out of, a system because of a difference in temperature between the thermodynamic system and the surroundings (extensive property) Temperature: a measure of the average kinetic energy of the particles in a sample (intensive property) Heat of Reaction: the value of q (heat) required to return a system to a given temperature (usually the temp of the room) at the completion of the reaction. Exothermic process: a chemical reaction or physical change in which heat is evolved. That is, the system loses heat; -q ○ System gets colder, surroundings get hotter Endothermic process: a chemical reaction of physical change in which heat is absorbed. That is, the system gains heat; +q  ○ System gets hotter, surroundings get colder Enthalpy Diagrams Here is an enthalpy diagram There is a peak that shows the energy required to overcome the activation barrier (endothermic) Then a dip to show the release of energy as the reaction progresses (exothermic) If the products are lower than the reactants, the reaction is exothermic (-q). If the products are higher than the reactants, the reaction is endothermic. (+q) NOTE: if the temp increases, the reaction was exothermic, -q In high school, we carry out chemical reactions in vessels that are open to the atmosphere → constant pressure. But what if a gas is produced in a reaction? We know that the expanding gas does work against the atmosphere. For most reactions, this amount of work is pretty small, so we’ll not sweat it. Enthalpy (H) an extensive property can be used to obtain heat absorbed or given off in a chemical reaction is a state function: ○ a property that depends only on its present state, determined by T and P, for example. A state function is independent of the history of the system. Enthalpy Change of a Reaction ∆Hr = Hproducts - Hreactants or ∆H = qp ∆H with reactants = endothermic, with products = exothermic If the rxn flips, the sign of ∆H will flip If the reaction doubles, ∆H will double The enthalpy change in a reaction equals the heat of reaction at constant pressure. Since H is a state function, it is independent of the history of the system. 3. Heats of Fusion & Vapourization Things that apply for both heats: 𝑒𝑛𝑒𝑟𝑔𝑦 (𝑞) 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑏𝑦 𝑤𝑎𝑡𝑒𝑟 / 𝑡𝑖𝑚𝑒 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦: 𝑒𝑛𝑒𝑟𝑔𝑦 (𝑤𝑎𝑡𝑡𝑠) 𝑖𝑛𝑝𝑢𝑡𝑡𝑒𝑑 𝑏𝑦 𝑘𝑒𝑡𝑡𝑙𝑒 Temperature vs Time graph Heat of Vaporization The energy required to convert a substance from liquid to gas at its boiling point 1. 𝑞 = 𝑚∆𝐻𝑣𝑎𝑝, where q is the energy required for phase change (j) m is the mass of substance (g) ΔHvap is the heat of vaporization (j/g) Heat of Fusion The energy required to convert a substance from solid to liquid at its melting point Formulas 1. 𝑞 = 𝑚∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛, where 2. 𝑞 = 𝑛∆𝐻𝑚𝑜𝑙𝑎𝑟 𝑓𝑢𝑠𝑖𝑜𝑛, where q is the energy required for phase change (j) q is the energy required for phase change (j) m is the mass of substance (g) n is the number of moles ∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛the heat of fusion (j/g) 𝐻𝑚𝑜𝑙𝑎𝑟 𝑓𝑢𝑠𝑖𝑜𝑛 is the molar heat of fusion (j/mol) For the straight part of the graph 3. ∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛 = ∆𝐻𝑚𝑜𝑙𝑎𝑟 𝑓𝑢𝑠𝑖𝑜𝑛 ÷ 𝑚𝑚, where Calculation Steps 1. Heat solid to melting point ΔHfusion is the heat of fusion (j/g) 2. Phase change at melting point ΔHfusion (molar) is the molar heat of fusion (j/mol) 3. Heat liquid from melting point to final temperature mm is the molar mass of substance (g) Important Concepts Occurs at constant temperature (melting point)  Endothermic process for melting (ΔH positive) Exothermic process for freezing (ΔH negative) Represents breaking of intermolecular forces in solids 4. Calorimetry Thermal Equilibrium Thermal equilibrium occurs when two or more systems in thermal contact reach a state where their temperatures become equal, resulting in no net heat transfer between them. Heat naturally flows from a hotter object to a cooler one until they reach the same temperature. This process continues until thermal equilibrium is achieved. Therefore, 𝑇𝑓𝑖𝑛𝑎𝑙 can be the same for the system and the surroundings in some cases. Two types of calorimeters 1. Constant pressure (Coffee Cup) Calorimeters are open to the atmosphere (𝑞𝑃) We use a coffee-cup calorimeter to measure enthalpy changes for reactions that occur in aqueous solution. We measure ∆T of the soln—part of the surroundings, hence ∆𝐻 = 𝑞𝑃 =− [ 𝑚𝐶∆𝑇], where ∆𝐻 is enthalpy (kj/mol) → divide by moles Assumptions when using a coffee cup calorimeter Negligible heat loss. 𝐽 We assume dilute aq. solutions to have density (1.0 g/mL) and specific heat capacity of 4.0 𝑔·𝑐 NOTE: ∆𝑇 = 𝑇𝑓𝑖𝑛𝑎𝑙 − 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 , and 𝑇𝑓𝑖𝑛𝑎𝑙 for both system and surroundings should be the same 2. Constant volume (Bomb) Calorimeters Not open to the atmosphere, constant volume (𝑞𝑣) Often used for reactions involving gasses Heat of reaction absorbed by the water in the calorimeter and by the calorimeter itself, ∆𝐻 = 𝑞𝑣 =− [ 𝐶∆𝑇], where ∆𝐻 is enthalpy (kj/mol) → divide by moles NOTE: if the heat constant of the calorimeter isn't given, calculate the q of water and q of the reaction, and add. 5. Determining the Heat Capacity of a Constant P Calorimeter Thought Lab #1 Thought Lab #2 The constant pressure calorimeter “takes a certain cut” of the total heat. The first law of thermodynamics states that For an exothermic reaction, measured ∆T is less than actual ∆T. the change in internal energy, U, of a system measured heat = actual heat – heat “stolen”, or equals heat, q, plus work, w. qmeasured = qactual – heat “stolen”, or ○ ∆𝑈 = 𝑞 + 𝑤, or ∆𝑈 = 𝑞 − 𝑃∆𝑉 ∆𝐻𝑟 =− [𝑚𝐶∆𝑇 + (𝐶𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟∆𝑇)], and (𝑃∆𝑉 is negative because gas produced (ℎ𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 𝑏𝑦 ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟)−(ℎ𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟) in the reaction (system) is doing the ℎ𝑐𝑐 (𝑎𝑘𝑎 𝐶𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟) = ∆𝑇𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 work on the surroundings) 𝐽 We empirically calculated 𝑃∆𝑉 (using Units of ℎ𝑐𝑐 are °𝐶 𝑜𝑟 °𝐾 𝑃∆𝑉 = 𝑛𝑅𝑇), determining that it is extremely small and therefore not that important. 6. Hess's Law Hess’s Law of Heat Summation: For a chemical equation that can be written as the sum of two or more steps, the ∆H for the overall equation equals the sum of ∆H for the individual steps. His experiments on various hydrates of sulfuric acid showed that the heat released when they formed was always the same, whether the reactions proceeded directly or through intermediates. This is explained by enthalpy diagrams. There are three ways to apply Hess’s Law to calculate ∆H for a given chemical reaction: 1. ∆Hr using bond dissociation energy (BDE) values. Example: Calculate, using BDE values, the molar enthalpy change for the complete combustion of propane, C3H8. 1) Write out the balanced chemical equation C3H8 + 5 O2 → 3 CO2 + 4 H2O 2) Draw the lewis structures to determine how many of each bond there is. 3) Fill this table out: Bonds broken E input (kJ) Bonds formed E output (kJ) Put each bonds from the Multiply the quantity of each Put each bonds from the Multiply the quantity of each reactants in a separate column. bond by its bond energy. products in a separate column bond by its bond energy = Total broken = Total formed ◦ 4) ∆𝐻 = 𝑇𝑜𝑡𝑎𝑙𝑏𝑟𝑜𝑘𝑒𝑛 − 𝑇𝑜𝑡𝑎𝑙𝑓𝑜𝑟𝑚𝑒𝑑 OR ∆𝐻𝑟 = Σ𝑛(𝑏𝑜𝑛𝑑𝑠 𝑏𝑟𝑜𝑘𝑒𝑛) − Σ𝑚(𝑏𝑜𝑛𝑑𝑠 𝑓𝑜𝑟𝑚𝑒𝑑) NOTE: breaking bonds is endothermic, forming bonds is exothermic 2. ∆Hr by algebraic addition of thermo-chemical equations Combine given—and relevant—equations algebraically to get the “target” equation. Each equation can be doubled, tripled, halved, etc. →The ΔH values must then also be doubled, tripled, halved, etc. If an equation is reversed the sign of ΔH° must be “flipped”. ○ The ° just means it was conducted at SATP, it doesn’t change the values. 1) Look at the target equation and its molecules. Find them in the given equations. 2) Manipulate each given equation to match the coefficient of the desired identical molecule from the target equation. Write out what you did and the resulting equation. Remember to do the same to ΔH values. 3) Add all the equations, crossing out things that appear in the products of one equation and the reactants of another, but they must be the same state. 4) The resulting equation should be equal to the target one. Now just add the manipulated ΔH values. 3. ∆Hr using standard heat of formation (∆H°f) data The enthalpy change for the formation of 1.00 mol of that substance in its standard state from its elements in their reference form (usually the most stable allotrope) in their standard state. For a standard state, ∆H°f = 0 For a non standard state, ∆H°f ≠ 0 ◦ ◦ ◦ ∆𝐻𝑟 = Σ𝑛∆𝐻𝑓(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − Σ𝑚∆𝐻𝑓(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) 7. Entropy and Gibbs Free Energy Pt 1 (video) 𝐽 Entropy (symbol: S; units: 𝑚𝑜𝑙·𝐾 ) The energy of the universe has the capacity to flow as a result of its inhomogeneous distribution. Entropy is a measure of the capacity energy may have to transform or flow from one place to another → refers to energy dispersion. Spontaneous endothermic reactions: reactions that once begun proceed without a constant input of energy. They are driven by increasing entropy → by a drive towards dispersion of energy. Think of entropy as dispersion of energy among available microscopic energy levels (states) at a particular macroscopic state. Microstate: A single configurations of the particles among the states (positions and kinetic e) Greater number of configurations of the particles among states (microscopic energy levels) = more microstates the system has = greater entropy NOTE: entropy refers to Energy that is NOT available for useful work Number of microstates will increase with an increase in volume/temperature/number of molecules because these increase the possible positions and kinetic energies of the molecules. ○ Example: Gas expanding to a larger area. ○ Low entropy = energy is concentrated ○ High entropy = energy is dispersed Qualitative Determination of ∆S°r ∆𝑆 > 0 for the following processes: 1. When more molecules are produced: E is more widely distributed ○ How to tell: more moles (greater sum of coefficients) on products than recatants 2. In exothermic reactions: ○ When the phase change solid to liquid occurs: more microstates “contained” in liquid than solid phase ○ When the phase change liquid to gas occurs: more microstates “contained” in gas than liquid phase. 8. Entropy and Gibbs Free Energy Pt 2 (video) Quantitative Determination of ∆S°r Hess’s Law applies to entropy: ◦ ◦ ◦ ∆𝑆𝑟 = Σ𝑛∆𝑆 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − Σ𝑚∆𝑆 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) Gibb’s Free Energy Equation Free Energy includes changes related to chemical bonding and to the dispersal of energy among states. ◦ ◦ ◦ ∆𝐺 = ∆𝐻 − 𝑇∆𝑆 Total energy spread out in Energy from the reaction Energy in the system via states, in the products system+surroundings from system to surroundings as compared to reactants. Free energy of a rxn; measure of spont. of rxn: Enthalpy T: absolute T, °K ◦ ∆𝐺 < 0 → spont. kj/mol 𝐽 S: change in entropy ( in appendix), 𝑚𝑜𝑙·𝐾 ◦ ∆𝐺 > 0 → non-spont. ◦ ∆𝐺 = 0 → rxn at equilibrium kj/mol NOTE: Spontaneous = thermodynamically favorable, will continue on its own. Hess’s Law also applies to Gibb’s free energy: ◦ ◦ ◦ ∆𝐺 = Σ𝑛∆𝐺 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − Σ𝑚∆𝐺 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) NOTE: G is also a measure of the max amount of work a reaction can do. Spontaneity of a Reaction ∆𝐻 ∆𝑆 ∆𝐺 Explanation Spontaneous? 0 (+) 0 (+) → endo 0 (+) (+) − (+)(−) = (+) Non-spont @ all T (+) + (+) = (+) 0 (+) >0 (+) (+) − (+)(+) = (+) 𝑜𝑟 (−) ◦ Spont if 𝑇∆𝑆 > ∆𝐻 9. Rates of Chemical Reactions We study rates because some reactions (like precipitation or double displacement) are fast due to the presence of ions with strong bonds or a medium through which to react. Some reactions are slow like iron rusting or cement hardening. A knowledge of the factors that affect the rate of a chemical reaction can tell us about how the reaction occurs - the steps through which the reaction proceeds - called the reaction mechanism Factors that affect Reaction rates 1. Nature of reactants (ie. electron configuration, states, resonance, etc) 2. Concentration of reactants. 3. Presence and concentration of a catalyst. 4. Temperature: For every 10°C↑ = 2X increase in recession rate 5. Surface area of solid reactants or heterogeneous catalyst. Rate of reactions 𝑚𝑜𝑙 Units are 𝐿·𝑠 For the reaction A → B: The rate of reaction is not constant over time - its fasted at the beginning because that's when we have the most A Secant: m= average rate of concentration change Tangent: m = Instantaneous rate of concentration change How could we monitor the rate of reaction? 1. Change in mass due to state changes 2. pH change 3. Pressure change (least idea) How could we measure the rate of reaction? 1. Absorbance of light 2. Conductivity Rate Law Equation For a chemical equation A (aq or g) + B (aq or g) → products 𝑥 𝑦 The rate law equation is 𝑟𝑎𝑡𝑒 = 𝑘[𝐴] [𝐵] , where 𝑘 = the rate constant, 1−𝑛 depends 𝑛−1 on temperature - T must be specified. 𝑚𝑜𝑙 ·𝐿 ○ Units are 𝑠 , where n =reaction order X and Y are empirically determined - NOT the molar coefficients from a balanced equation ○ Their values are most often 0, 1, 2. ○ If x=0, the reaction is zero order with respect to A ○ If y=1, the reaction is first order with respect to B. ○ OVERALL reaction order is the sum of the exponents ○ If x or y is 0, remove that whole value from the equation since anything to the exponent 0 is just 1. This means the rate of the reaction is independent of [A] and linearly dependent on [B], for example. So a rate vs [A] graph would be a straight horizontal line, and the rate vs [B] graph would be linear In general, ○ Zero order = no relationship, just a horizontal line ○ First order = linear relationship ○ Second order = exponential Experimental Determination of Reaction Order We use the method of initial rates, and determine the slope of the tangent of a rate vs concentration graph at the start of the reaction Example 1: Determine the rate law equation for the reaction 2N2O5(g) → 4NO2(g) + O2(g) experiment [N2O5] mol/L 𝑚𝑜𝑙 Initial rate of disappearance ( 𝐿·𝑠 ) Here we see that as [N2O5] doubles, the rate of reaction 1 1.0E-2 4.8E-6 doubles. This is a linear relationship, so the order is 1. 1 ∴ 𝑟𝑎𝑡𝑒 = 𝑘[𝑁2𝑂5] 2 2.0E-2 9.8E-6 Example 2: Determine the rate law equation for the reaction H2O2(aq) + 3I-(aq) + 2H+(aq) → I3-(aq) + 2H2O(l) - experiment [H2O2] mol/L [I ] mol/L [H-] mol/L 𝑚𝑜𝑙 Initial rate of ( 𝐿·𝑠 ) Since there are multiple reactants, when comparing 1 0.010 0.010 0.00050 1.15E-6 two concentration values for one reactant from two different experiments to their respective rates, all 2 0.020 0.010 0.00050 2.30E-6 other values must be equal: For [N2O2]: exp 1 & 2 → 1st order 3 0.010 0.020 0.00050 2.30E-6 For [I-]: exp 1 & 3 → 1st order For [H-]: exp 1 & 4 → no order 4 0.010 0.010 0.00100 1.15E-6 1 − 1 ∴ 𝑟𝑎𝑡𝑒 = 𝑘[𝐻2𝑂2] [𝐻 ]