Chemistry 2 Reviewer PDF

Chemistry 2 Reviewer Thermochemistry - concerned with the amount the system of heat absorbed or released during chemic...

Chemistry 2 Reviewer Thermochemistry - concerned with the amount the system of heat absorbed or released during chemical and physical changes Heat - is the energy transferred between the ENTHALPY FIRST LAW OF THERMODYNAMICS system and surroundings as a result of a temperature difference only; either absorbed or Enthalpy is a state function and is an extensive System - part of the universe that is being released property. (H) studied Exothermic Process (-) - heat is Formula: Surroundings - everything in the universe transferred from the system ΔH°rxn = H(products) – H(reactants) outside the system to the surroundings Endothermic Process (+) - heat is transferred HESS’S LAW Thermodynamics - science that studies energy from the surroundings to the system transfer and energy transformation State Function - a property whose value does Hess’s Law - states that the enthalpy change Three types of systems: not depend on the path taken to reach that for a reaction that occurs in a series of steps is - Open system: exchange specific value equal to the sum of the enthalpy changes of the mass and energy with the First Law of Thermodynamics - states that individual steps surroundings energy can be converted from one form to - Closed system: exchange of another, but cannot be created or destroyed When Applying Hess’s Law: energy but not mass Always specify the physical states of - Isolated system: does not reactants and products because they WORK AND HEAT exchange either mass or help determine the actual enthalpy energy Process Sign changes. Three laws of thermodynamics: When multiplying an equation by a - Energy cannot be created nor Heat absorbed by Positive q (heat) the system (endo) factor (n), multiply the ΔH value by destroyed. same factor. - For a spontaneous process, Heat released by Negative q (heat) Reversing an equation changes the the entropy of the universe the system (exo) sign but not the magnitude of ΔH. increases. Add the ΔH for each step after proper - A perfect crystal at 0 Kelvin Work is done on Positive w (work) manipulation. has no entropy the system Process is useful for calculating Work is done by Negative w (work) enthalpies that cannot be found directly. ℗jisuedo ENTROPY ΔH ΔS -TΔS ΔG = ΔH - - Molecules can only react if they collide TΔS with each other Entropy - a measure of the randomness of a - Only effective collisions can lead to a system; In general, greater disorder means – + – Always Spontaneo chemical reaction. negative us at all T - In order to have an effective collisions greater entropy. molecules must have correct + – + Always Nonsponta Formula: positive neous at orientation and sufficient energy ΔS° = ΣnΔS°(products) - ΣmΔS°(reactants) all T - Particles that do not have enough energy to react bounce apart Spontaneous Process - are processes that can – – + Negative Spon at unchanged when they collide. at low T low T proceed without any outside intervention; irrervesible; related to an increase in + + – Positive at Spon at Activated Complex/Transition Complex randomness low T high T - an unstable arrangement of atoms that forms for a moment at the peak of the Reversible - the system changes in such a way activation-energy barrier CHEMICAL KINETICS that the system and surroundings can be put - short-lived back in their original states by exactly reversing - Every reaction goes through its own Collision Theory, Factors Affecting Reaction Transition State the process Rates, Rate Law, Reaction Mechanism Irreversible - cannot be undone by exactly reversing the change to the system Reaction Coordinate Diagram Chemical Kinetics - It shows the energy of the reactants - Deals with speed or rate of reaction and products. GIBBS FREE ENERGY - Require varying lengths of time - The high point on the diagram is the ΔG = G(products) – G(reactants); - kinetics also sheds light on the transition state. ΔG = ΔH - TΔS reaction mechanism (exactly how the - The activation-energy barrier must be reaction occurs; pathway or series of crossed before reactants are step for reaction). converted to products. - The species present at the transition Collision Theory state is called the activated complex. - In a chemical reaction, bonds are - The energy gap between the reactants broken and new bonds are formed. and the activated complex is the activation energy barrier. ℗jisuedo  k: the rate constant RATE LAW x: order with respect to A y: order with respect to B Rate Law - an equation that relates the rate of reaction to the concentrations of reactants Goodluck sa exam guys !! Figure 1: EXO Figure 2: ENDO FORMULA: rate = k[A]^x [B]^y PROBLEMS AND SOLUTIONS (credits kay ada) INTERNAL ENERGY Question 1: A system loses 354 J of heat and at the same time it does 94 J of work to its surroundings. What is the change in internal energy of the system in the same unit? Identify the given values: The system loses heat, so heat q is negative: q = -354 J The system does work on its surroundings, so work w is negative: w = -94 J Solve: ΔE = q + w ΔE = (-354 J ) + (-94 J ) ΔE = -448 J Interpretation: The change in internal energy is -448 J, which means the system lost 448 J of internal energy. Question 2: Calculate the work if the system has a net change of 0.500 kJ of internal energy after it gained 143 J of heat. What does the sign of work indicate about its direction? Identify the given values: The change in Internal Energy: ΔE = 0.500 kJ [Convert to Joules] ΔE= 0.500 kJ x 1000 = 500 J The heat is gained by the system, so q is positive: ℗jisuedo q = 143 J Solve: (Rearrange the formula ΔE = q + w) w = ΔE - q w = 500 J - 143 J w = 357 J; The work is +357 J. A positive value for work means that work is done on the system. ENTHALPY Q1. Ag⁺(aq) + Cl⁻(aq) 🡪 AgCl(s) Enthalpy of Formation Values: ○ ΔH°f (Ag⁺(aq)) = +105.6 kJ/mol ○ ΔH°f (Cl⁻(aq)) = -167.6 kJ/mol ○ ΔH°f (AgCl(s)) = -127.0 kJ/mol So, ΔH°rxn = -65.0 kJ/mol Q2. CaCO₃(s) 🡪 CaO(s) + CO₂(g) Enthalpy of Formation Values: ○ ΔH°f (CaCO₃(s)) = -1207.0 kJ/mol ○ ΔH°f (CaO(s)) = -635.1 kJ/mol ○ ΔH°f (CO₂(g)) = -393.5 kJ/mol So, ΔH°rxn = 178.4 kJ/mol ℗jisuedo ENTROPY 1. 2 SO2(g) + O2(g) 🡪 2 SO3(g) Find the standard molar entropy values (S°): [Refer to the standard thermodynamics value] SO2(g) = 248.1 O2(g) = 205.0 SO3(g) = 256.66 ΔS°rxn = [2 (256.66)] - [2(248.1) + (205.0)] ΔS°rxn = -187.88 J/K 2. CO2(g) + 4 H2(g) 🡪 CH4(g) + 2 H2O(g) Find the standard molar entropy values (S°): [Refer to the standard thermodynamics value] CO2(g) = 213.7 H2(g) = 130.6 CH4(g) = 186.2 H2O(g) = 188.72 ΔS°rxn = [(186.2) + 2(188.72)] - [(213.7) + 4(130.6)] ΔS°rxn = -172.46 J/K HESS’S LAW ℗jisuedo GIBBS FREE ENERGY 4 KClO3 (s) 🡪 3 KClO4(s) + KCl(s) ΔH°rxn ΔS°rxn ΔG°rxn KClO3 (s) = -397.7 KClO3 (s) = 1431 KClO3 (s) = -303.2 KClO4(s) = -432.75 KClO4(s) = 151.0 KClO4(s) = -296.3 KCl(s) = -436.7 KCl(s) = 82.59 KCl(s) = -409.2 Solving for ΔH°rxn: ΔH°rxn = [3(-432.75) + (-436.7)] - [4(-397.7)] ΔH°rxn = -144.15 kJ Solving for ΔS°rxn: ΔS°rxn = [3(151.0) + (82.59)] - [4(143.1)] ΔS°rxn = -36.8 J/K Solving ΔG°rxn: ΔG°rxn = [3(-303.2) + (-409.2)] - [4(-296.3)] ΔG°rxn = -133.6 kJ ΔG°rxn = ΔH° - T ΔS° (298𝑘)(−36.8𝐽/𝐾) ΔG°rxn = (-144.15 kJ) - 1000 ΔG°rxn = -133.1836 ΔG°rxn = -133.18 kJ ℗jisuedo RATE LAW ℗jisuedo