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CHM 170 Final Study Guide

by: Joseph Notetaker

CHM 170 Final Study Guide 46657 - CHM 170 - 002

Joseph Notetaker

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The study guide for the final
General Chemistry 2
Mark M Richter
Study Guide
General Chemistry
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This 14 page Study Guide was uploaded by Joseph Notetaker on Tuesday May 10, 2016. The Study Guide belongs to 46657 - CHM 170 - 002 at Missouri State University taught by Mark M Richter in Spring 2016. Since its upload, it has received 12 views.


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Date Created: 05/10/16
CHM 170 Final Exam Study Guide Calculating rates: H 2 I -2------- 2HI −∆[H ] 2 −∆[I ]2 +1 ∆[HI ] For H 2 ∆t For I2: ∆t For 2HI: 2 ∆t Δ: The change in []: Molarity -: losing +: gaining ½: corresponds to stoichiometry Rate Laws: Note: Rate laws are always defined by reactants Rate = K[A] [B] m K = constant 0 K[N 2] : Zero order reaction, in this reaction the order is independent of the reactant K[N 2] : First order reaction, in this reaction the order is directly proportional to the reactant K[N 2] : Second order reaction, in this reaction the order is proportional to the square of the reactant Integrated rate laws: [A]t= concentration of A at a particular time Integrated zero order rate law: A¿ 0 A¿ t−kt+¿ ¿ Integrated first order rate law: A¿ 0 A¿t=−kt+ln ?¿ ln¿ Integrated second order rate law: A¿ t ¿ A¿ 0 ¿ ¿ 1 ¿ Half-life equations: Zero order half-life equation: A¿ 0 ¿ ¿ t =¿ 2 First order half-life equation: t = 0.693 2 k Second order half-life equation: A¿ 0 k¿ t = 1 2 ¿ Activation energy: Ea=Rln k1 1 − 1 k2(T 1 T 2) Note: A catalyist is an agent that speeds up a chemical reaction Chemical equilibrium Chemical reactions can travel in both directions In the equation aA + bB -------- cC + dD D¿ d ¿ b B¿ A¿ ¿ ¿ c C¿ ¿ ¿ K=¿ Dynamic equilibrium is when the reaction is going the same rate both forwards and backwards If you add two chemical equations you have to multiply the equilibrium constants +¿ ¿ H 3 ¿ [H3O ]>[OH] then solution is acidic ¿ pH=−log¿ −¿ ¿ OH ¿ [H3O ]<[OH]  ¿ pOH=−log¿ then solution is basic pH+pOH=14 [H 3 ]=[OH] then solution is neutral pKa=−log? [Ka] Strong Acid: +¿ ¿ H 3 ¿ HCl+H O­­­­­H O +Cl        Ka= −¿ ¿    Ka is very large in this case 2 3 Cl ¿ ¿ ¿ [H O ] = [Cl] = [HCl] in the case of strong acids due to complete ionization 3 Weak Acid: +¿ ¿ H O3 ¿ −¿ HC H2O3+H2O+­2­­H O +C 3 O   2 K3= 2 ¿      Ka=1.8*10 ­5 C 2 O3 2 ¿ ¿ ¿ HA represents a weak acid [H 3 ] ≠ [HA] in the case of weak acids due to incomplete ionization Assuming [H O ]3represents a small number then:  +¿ ¿ H 3 ¿   ¿ with HA reiresenting the initial concentration  Check this value with the equation +¿ O H 3HA ∗100 ¿ If the answer exceeds 5% then [H 3 ] is not small + If [H3O ] is large then: +¿ ¿ H 3 −¿ ¿ ¿ 1 A Ka +4Ka H ( ¿ ) 2                   or                     i ¿ ¿ pH=pKa+log¿ −Ka+¿ ¿ Use according to personal preference Solubility: Solubility Product (Ksp) In the equation              2PbCl ­­­­­­­Pb +2Cl Ksp = [Pb ][Cl]­ 2 Ksp ≠ Molar solubility n+m Ksp Molar solubility= n m   with n and m representing products √n ∗m Complex Ion formation In the equation Fe +6CN­­­­­­­­­­­Fe(CN) 6 4−¿ ¿ CN¿ 6 ¿ 2+¿¿ Fe ¿ C N −¿¿ ¿ Fe¿ ¿ Kf =¿ Some other things to remember: BAD: Bronsted Acid Donates (+) LAA: Lewis Acid accepts (­) Use an ICE table to find stoichiometric coefficients  Strong acids and bases ionize completely  Weak acids and bases ionize incompletely The stronger the acid the weaker the conjugate base and vice versa Strong bonds lead to weak acids Polyprotic acids (acids with more than one H ) require multiple calculations Buffers contain an acid and a conjugate base Acidity increases as you go left to right and top to bottom on the periodic table H-Enthalpy S-Entropy G-Gibbs Free Energy Energy changes that occur during a chemical reaction are due to the making and breaking of chemical bonds The first law of thermodynamics states that energy is conserved ΔETotal = Δ SEstem + Δ SurEoundings = 0 If the system absorbs energy it’s represented by a positive value (+) If the system releases energy it’s represented by a negative value (-) The second law of thermodynamics states that energy is most stable in a randomized state This means that a reaction can never be 100% efficient because some energy will always be lost due to entropy Thermodynamics predicts whether a process will occur under given conditions -spontaneous reactions (Thermodynamically favorable) do not require external stimuli -nonspontaneous reactions require external stimuli In all reactions one direction is spontaneous and the other is nonspontaneous The direction of spontaneity can be determined by comparing the potential energy of the system at the start and at the end Spontaneity is determined by comparing the chemical potential before the reaction with the potential energy after the reaction. Spontaneity ≠ speed of the reaction A nonspontaneous process can occur by coupling it to a process that is spontaneous or by supplying energy from an external source Spontaneous processes release energy Most go from higher energy to lower energy (Exothermic) But some go from lower energy to higher energy (Endothermic) Entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases S = klnW -23 k=1.38 * 10 Joules/Kelvin W= # of energetically equivalent ways to arrange the compounds of the system Macrostates are concerned with the entire system Microstates are concerned with the actual arrangements (this is the actual W measurement) ΔS = S final Sinitial ΔS > 0 for spontaneous reactions ΔS > 0 for more disorder ΔS < 0 for more order Entropy can be increased by increasing particles or increasing temperature N 2 3H =22NH 3 This is a system of decreasing entropy because a set of four moles has to be combined into a set of two moles When the entropy change in the system is unfavorable (negative) the entropy change in the surroundings must be positive ΔS surroundingsΔH systemtemperature Gibbs Free energy Free energy refers to energy that is available to do work ΔG=ΔH-TΔS If the answer is negative the reaction is spontaneous In the example ΔS ORXNrefers to the standard state of entropy The third law of thermodynamics states that In a perfect crystal at absolute zero the entropy is zero This means ΔS is always positive Entropy is affected by -State of substance -Gas has higher entropy than liquid -Molar mass of the substance -Larger the molar mass the greater the entropy -The particular Allotrope (Elemental arrangement) -The less constrained the greater the entropy -Molecular complexity More complexity = More vibrations = More entropy -Extent of dissolution -dissolved solids have larger entropy The equation for ΔS RXNis as follows: ∆ SO = n ∆S O(products)− n ∆S (reactants) RXN ∑ p ∑ r N= Stoichiometric Coefficient P= products R= reactants The values of ΔS will be provided for you as the are determined experimentally Electrochemistry is the branch of chemistry that deals with the chemical action of electricity and the production of electricity by chemical reactions. Assigning oxidation numbers: Fe2O 3 3CO ------- 2Fe + 3CO 2 (Fe )3+(These elements balance each other) 2 6- (O3) (Oxygen usually has a charge of -2. 2*3=6) 2- (O) (See above) (3C)2+ (These elements must balance each other out) 0 (2Fe) (Elemental form has no charge) (3C) 4+ (See above) (O 2 4- (See above) The Half Reaction Method in an acidic medium (As demonstrated by this equation) 2- 2- 3+ SO 2 Cr O 2 7 ------------- SO4 + Cr First assign oxidation values: +4 -2 +6 -2 +6 -2 +3 S O 2 +Cr 2 O 7------ S O 4 + Cr Next split the reaction into two half reactions: Oxidation: S O +4 2-2--------- S O 4-2 +6 -2 +3 Reduction: Cr 2 O7 -------- Cr Now we have to balance the atoms do this by adding an H O for every Oxyg2n atom and 2 hydrogens on the opposite side to balance out the equation like so: Oxidation: SO + 22 O ----2------------- SO 4-2+ 4H + -2 + +3 Reduction: Cr O 2 7 + 14H -------------- 2Cr + 7H O2 Add electrons to make the charges even: Oxidation: SO + 22 O ----2--------------- SO 4-2+ 4H + 2e - Reduction: Cr O 2 7-2+ 14H + 6e ---------------- 2Cr +3 + 7H O2 Next we use multiplication to make the electrons equal: 3[SO + 2H O -------------- SO -2 + 4H + 2e ] - 2 2 4 3SO +26H O --2------------ 3SO 4-2+ 12H + 6e - Finally we combine the equations: -2 + - 3SO +26H O --2-- 3SO 4 + 12H + 6e + Cr O -2+ 14H + 6e ------ 2Cr +3 + 7H O 2 7 2 -2 + -2 +3 3SO +2Cr O 2 7 + 2H -------- 3SO 4 + H 2 + 2Cr A reaction in a basic solution uses the same method except in a basic solution you - + must add OH to the H in order to neutralize any acid joule volt= coulomb Electricity is caused by the flow of electrons through a circuit Redox reactions involve the flow of electrons A voltaic cell is an electrochemical cell that produces electrical current from a spontaneous chemical reaction This equation can be used to calculate overall cell potential: E o =E o +E o Cell Cathode Anode You’ll find that the values of the cathodes and anodes have been calculated already o E cell 0 means that the reaction is spontaneous. The amount of material in the cell has no effect on the cell potential You can calculate the free energy of a cell using this equation: o o ∆ G =−nF E cell N = number of moles F = faraday’s constant (96,485) The Nernst equation calculates cell potential outside of standard conditions: o 0.059v E cell cell logQ n Electrolytic cells use electricity to drive a reaction in reverse In electrolysis the anode and cathode are the inverse of voltaic cells Anode: (+) Cathode: (-) The voltage required for electrolysis depends on the specific half reactions In electrolysis the electrons act as reactants and have a stoichiometric coefficient Current is recorded in Amps Amp= coulomb second In order for rusting to occur Moisture must be present Additional electrolytes promote rust Acids promote rust We can stop corrosion by Covering the surface Using sacrificial electrodes (some materials are more easily oxidized than others) Nuclear processes often result in one element changing into another, frequently emitting tremendous amounts of energy. 4 2+ α particle = 2He - β particle = e γ particle = 00γ Proton: 11P Neutron: 10 Electron: -1e Alpha Decay Alpha decay occurs when an unstable nucleus emits a particle at high energy Alpha particles cause atoms to change into other atoms Heres an example of an alpha decay reaction: 224 220 4 88a= 86Rn+ 2e Notice that the numbers on both sides of the equation are equal Alpha radiation has the highest ionizing power but the lowest penetrating power Beta Decay Beta decay occurs when an unstable nucleus emits an electron (A neutron splits apart to form a proton and an electron) Here is an example of a Beta decay reaction: 249 249 0 97k= Cf98 e −1 Beta radiation has intermediate ionizing and penetrating power Gamma Decay Gamma ray emission is a form of electromagnetic radiation they are high energy photons which are given off as a byproduct of alpha or beta emissions Here’s an example of a Gamma decay reaction: 137 137 0 56Ba= Ba56γ 0 Gamma Rays have the highest penetrating power and lowest ionizing power. Radiation is a natural component in our environment and humans are exposed to it every day. Half-life is the time it takes for half of the nuclei to decay Here are a few equations: Rate=KN N = Number of radioactive nuclei T = 0.693 1 K 2 A¿ t ¿ A¿ 0 ¿ ¿ ¿ ln¿


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