Chem 170 1st Edition Lecture 22Outline of Last Lecture I. Equilibrium Constant (K) examplesII. Rules of KIII. Reaction Coefficient Quotient (Q)IV.∆ G=∆ G0+RTln(Q)Outline of Current Lecture V.∆ G=∆ G0+RTln(Q)VI. Equilibrium Constants (Kconcentration and Kpartial pressures)VII.∆ G=RTln(QK)Current LectureChapter 15 – Chemical Equilibrium ∆ G=∆ G0+RTln(Q)If the reaction is at equilibrium (ΔG=0 and Q=K)0= ∆ G0+RTln(K)∆ G0=−RTln(K )ex. H2O(l)↔ H2O(g) at 25°C∆ G0system=∑(n Gproducts0)−∑(nGreactants0)=8.6 KJ (use the back of the book)The vapor pressure of H2O(l)↔ H2O(g) is 0.03126 at. When the vapor pressure is 0.03126 atm ΔG=0.Equilibrium Constants (Kconcentration and Kpartial pressures)ex. 2 SO2(g)+O2(g)↔2 SO3(g)Kp=PSO2PSO32PO2 which may not be equal to SOO[SO3]2[¿¿2][¿¿2]¿Kc=¿note that from PV =nRT we get nV=PRT=M thereforeThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.SOO[SO3]2[¿¿2]=PSO2RTPSO32RT×PO2RT=PSO2(RT )PSO32PO2[¿¿2]¿Kc=¿which shows why Kp does not equal Kc Kp=Kc(RT)∆ ngas where R=0.0821L × atmmol × Kex. 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?∆ 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 reactantsKbackward reaction=1Kforward reaction=5.6 ×10−19If the forward reaction was multiplied by a factor of 2 (Kforward reaction)2=3.24 ×1036ex. What is the reaction coefficient quotient of the reaction H2(g)+I2(g)↔ 2 HI(g) where PH2=5.25 atm and PHI=1.75 atm when ∆ G0=26KJmol at 25°C?Q=(PHI)2PH2=(1.75)25.25=0.583∆ G=(2.6KJmol)(1000 J1 KJ)+(8.314Jmol × K)(25+273 K)ln(0.583)¿1263 KJ /mol>0there fore the reaction is non-spontaneous∆ G=−RTln(K)−−RTln(Q)∆ G=RTln(Q)−RTln(K)∆ G=RTln(QK)ex. 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 equilibriumconstant (K) is 1.8×1018.Q=[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-spontaneousex. Find the total vapor pressure of the reactionNH2HS(g)↔ NH3(g)+H2S(g) where 1.08 ×10−1 at 25°C.K=PNH3PH2S=x2=1.08× 10−1, where x=PNH3=PH2Sx=3.29 ×10−1Ptotal=PNH3+PH2S=2 x =2(3.29 × 10−1)=0.658
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