Chem 170 1st Edition Lecture 23Outline of Last Lecture I.∆ G=∆ G0+RTln(Q)II. Equilibrium Constants (Kconcentration and Kpartial pressures)III.∆ G=RTln(QK)Outline of Current Lecture IV. Examples of Chemical EquilibriumCurrent LectureChapter 15 – Chemical Equilibriumex. Consider the reaction 2 NO(g)+O2(g)↔2 NO2(g) whose Kc is 375 at temperature T. Write the Kc and find [NO2][NO] when there is 0.0148 mol O2 in 0.755 L.Kc=[NO2]2[NO]2[O2]375=[NO2]2[NO]2(0.0148 m ol0.755 L) [NO2][NO]=√(375)(0.0148mol0.755 L)=0.369ex. Consider the reaction I2(g)↔2 I(g) whose Kc=1.1 ×10−2 at 1260°C. When this reaction is at equilibrium there is 1.0 mol of I2 and 0.5 mol I what is the volume of the flask?1.1× 10−2=(0.5L)21L=0.52LL=0.51.1× 10−2=22.7 Lex. Consider the reaction SO2Cl2(g)↔ SO2(g)+Cl2(g) whose KP=2.9 ×10−2 at 303 K. What is the Kc?These 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.KP=Kc(RT)∆ nKc=KP(RT)∆n=2.9 ×10−2[(0.0821) (303)]2−1=1.2× 10−3ex. Consider the reaction H2(g)+I2(g)↔ 2 HI (g)whose Kc is 50.2. What is the KP?KP=KC(RT )∆n=50.2(RT)1−1=50.2ex. Consider the reaction PCl5(g)↔ PCl3(g)+Cl2(g) whose KP is 78.3 at 250°C. If the initial partial pressure of PCl5 is 1.30 bar what the final pressure of all components?PCl5PCl3Cl2Initial 1.30 0 0Reaction -x +x +xEquilibrium 1.3-x x xKp=PPCl3PCl2PPCl5=(x) (x)1.3−x0=x2+78.3 x−102x=−78.3 ±√78.32−4(1) (102)(2) (1)=1.28at equilibrium PPCl5=1.30−1.28=0.02¯¿PCl2and PPCl3 is 1.28
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