Exam # 2 Study Guide Lectures: 15-23Lecture 15 (February 23) What are colligative properties?What is the equation for the boiling point elevation and what is the equation for the freezing point depression?ex. Consider the reaction CxHyOz + O2 → CO2 + H2O. If 1.205g of CxHyOz reacted with excess oxygen to make 2.479g CO2 and 0.698g H2O What is the molecular formula of the compound?What is osmotic pressure?ANS: Colligative properties are properties that depends on the amount of particles and does not depend on the identity of the particles.Boiling Point Elevation: ∆ Tb=kbm=Tsolution−TsolventFreezing Point Depression: ∆ Tf=kfm=Tsolvent−Tsolution2.479 gC O2(1 mol C O244 g)(1 mol C1 mol C O2)=0.0563 mol C ×(12.011 g1 mol C)=0.676 g C0.698 g H2O(1mol H2O18 g)(2 mol H1 mol H2O)=0.0775 mol H ×(1.008 g1 mol H)=0.0775 g H1.205 g−(0.0775 g+0.676 g)=0.457 gO ×(1 mol O16 g)=0.0282 mol O C0.676H0.775O0.0282=C8H11O4Osmosis is the selective passage of solvents through a semipermeable membrane. The solute cannot pass through the semipermeable membrane.π =MRTLecture 16 (February 25) ex. if there was 1.08g of serum albumin dissolved in 50cm3 of water when the osmotic pressure was 5.85mmHg at 298K, then what is the molecular weight (MW)?What is Van’t Hoff Factor?What are electrolytes?ANS:(MW)=mRTVπ=(1.08 g)(0.0821L × atmmol × K)(298 K)(50 g ×1 L1000 g)(5.85 mmHg ×1 atm760 mmHg)=68655gmolVan’t-Hoff factor= i=the actual number of particles in the solution after dissolution.Electrolytes are substances that disassociates into ions in water. Solutions with electrolytes are conductive. If you run a current through the solution with an electrolyte and place two wires and a lightbulb Lecture 17 (March 2) Chem 175 1st EditionWhat is a spontaneous reaction?What is a non-spontaneous reaction?How are spontaneous reactions and exothermic reactions related?What is entropy?ANS: Spontaneous describes a process that occurs in a system left to itself. Non-spontaneous describes a process that will not occur unless some external action is continuously applied.In the old times, spontaneity was predicted by looking at whether a reaction was exothermic. For example, combustion reactions (solid +O2→C O2+H2O) are exothermic. People also looked at the potential energy for spontaneity. For example, a ball on a hill would have a lot of potential energy and it can spontaneously roll down the hill. However a ball at the bottom of the hill would not be able to spontaneously roll up the hill. BUT endothermic reactions can also be spontaneous.Entropy (S) is a thermodynamic property related to the degree of disorder. The greater the randomness or disorder the greater the entropyLecture 18 (March 4) Write Boltzmann’s equation.Write the equation for ∆ Ssystem.ex. What is the change in entropy when 0.321 mol of O2 experiences a pressure change form 0.30 atm to 12 atm while temperature is constant? What is standard entropy?Describe some trends of entropy.ANS: Boltzmann’s equation: S=kln(W)∆ Ssystem=Rn ln(VfinalVinitial)V1P1n1T1=V2P2n2T2→V1(0.30 atm)=12 V20.3012=V1V2∆ Ssystem=(0.321mol)(8.314Jmol × K)(0.312)=−9.85JKS0 is entropy at 1 atmTrends:As the mass increase the entropy also increases. (ex. SXe>SNe)For 2 substances (usually molecules) with about the same molecular weight the substance with the more complex structure has a greater entropy than the less complex structure.∆ Sreaction0=∑(n Sproducts0)−∑(n Sreactants0)Lecture 19 (March 6) What is the ∆ Suniverse when the reaction is spontaneous?ex. Is the reaction H2O(l )→ H2O(s) spontaneous or non-spontaneous at -10°C?What is Gibbs Free Energy?ANS: Since all spontaneous processes produce an increase in the entropy of the universe,∆ Suniverse=∆ Ssystem+∆ Ssurroundings>0.S0 of H2O(s)=43.2 J/mol*K and S0 of H2O(l )=65.2 J/mol*K.∆ Ssystem=∑(n Sproducts0)−∑(n Sreactants0)¿(1)(43.2Jmol × K)−(1)(65.2Jmol × K)¿−22 J / K∆ Ssurroundings=−−6000Jmol263 K=23JK∆ Suniverse=∆ Ssystem+∆ Ssurroundings=−22+23=1>0therefore the reaction is spontaneousGibbs Free energy is ∆ G=∆ Hsystem−T ∆ Ssystem.if ∆ G<0, then the reaction is spontaneous.if ∆ G>0, then the reaction is not spontaneous, but the reverse process is spontaneous.if ∆ G=0, then the reaction will be at equilibrium.Lecture 20 (March 9) What are the characteristics of an equilibrium state?What is the empirical law of mass action?ANS: Characteristics of equilibrium state:1. it displays no macroscopic evidence of change2. it is reached through spontaneous processes3. it shows dynamic behavior between forward and reverse processes4. they are the same regardless of direction of approachwhen there is the reaction aA+bB ↔cC +dD, thenK=[C ]eqc[D]eqd[ A ]eqa[B]eqbLecture 21 (March 11) What is the equilibrium constant for −¿(aq)−¿(aq)+Cl O3¿−¿(aq)↔2 C l¿3 Cl O¿?What are the rules of the equilibrium constant?What is K if the following reaction was reversed? 2 A+B ↔ CWhat is K if the following reaction was multiplied by a factor of 2? 2 A+B ↔ CWhat happens if the reaction coefficient quotient (Q) is larger than the equilibrium constant?Write an equation for ΔG including ΔG°.ANS:−¿C l¿¿−¿Cl O3¿¿¿¿K=¿Rules of K: 1. Gasses use partial pressures (atm)2. Dissolved solvents use concentration (M)3. Pure liquids and pure solids do not appear in K even if the solvent participates in the reaction.4. When there is the reaction aA+bB ↔cC +dD, thenK2=1K1=[A]2[B][C]K3=K12=[C]2[A]4[B]2When Q>K the reaction goes to the left (toward the reactants).∆ G=∆ G0+RTln(Q)Lecture 22 (March 13) Write the equation that relates Kp to Kc. What is the equilibrium constant of the reaction 3 NO(s)↔N2(g)+N O2(g)at 25°C, given that∆ GNO(s)0=86.55KJmol, ∆ GN2( g)0=104.18KJmol and ∆ GN O2(g)0=5.92KJmol?What is the ΔG of the reaction 3 NO(s)↔ N2(g)+N O2(g) at 25°C whenNO=0.001 atm, N2O=0.001 atm and NO2=0.001atm ? The equilibrium constant (K) is 1.8×1018.ANS:Kp=Kc(RT)∆ ngas∆ G0=(104.18KJmol+5.92KJmol)−3(86.55KJmol)=−104.18 KJln(K)=−∆ G0RT=−(−104.18 KJ)(1000 J1 KJ)(8.314Jmol × K)(25+273 K)=42.02K=e42.02=1.8 ×1018since K is so large the products are favored over the reactantsQ=[0.001] [0.001][0.001]2=1.0× 103<Kln(1.0 ×1031.8 × 1018)=−87.1KJmol<0∆ G=RTln(QK)=(8.314Jmol × K)(25+273 K )¿therefore the reaction is non-spontaneousLecture 23 (March 23) ex. Consider the reaction 2 NO(g)+O2(g)↔2 NO2(g) whose Kc