Chem 104 1st Edition Lecture 33Outline of Last Lecture I. The equilibrium constantII. Continuation on the ICE methoda. Involving gasesIII. Heterogeneous equilbriaOutline of Current LectureCurrent LectureI. The reaction quotient tells which way the equilibrium constant favors. Remember:K > 1 (the reaction is moving to the product)K < 1 (reaction is moving towards the reactant)K = 1 (equilibrium is established)However, what is the limit? What is happening to the reactant? How can you tell? The reaction quotient is represented just as the equilibrium constant is represented:D ¿d¿B ¿bA ¿a¿¿C ¿c¿¿QC=¿QP=PccPDdPAaPBbWhat’s the difference between Q&K? K is for equilibrium and Q is for concentration at any given time. The only time Q and K are equal is at equilibrium. The same rules above apply between Q and K. What is Q=0? The reaction is only reactants and proceeds forward. What ifQ=∞? Then there are only products. As the products move closer to 0, the numerator will more father away from 0.Example: Which direction is the reaction going to favor if PI2=0.114 atm, PCI2=0.102 atm, and PICI=0.355 atm?Given: I2 and CI2↔2ICl, KP=0.102 atm, 81.9 = KQP=PICl2PI2PCI2QP=(0.355)2(0.114)(0.102)=10.8Since 10.8<81.9 the reaction favors the product and moves to the rightII. Finding the equilibrium concentration:Example: I2+Cl2↔2IClI2Cl2IClI .100 .100 .100CE ? ? ?Which direction is the reaction moving?K=[ICl]2[I2][C l2]Q=[ICl]2[I2][C l2]Q=[.100]2[.100][.100]=1Since Q<K (1<81.9) the reaction moves to the productNow define this in the terms of x:I2Cl2IClI .100 .100 .100C -x -x +2xE ? ? ?So….I2Cl2IClI .100 .100 .100C -x -x +2xE .100-x .100-x .100+2xNow solve for x:Kc=(.100+2 x)2(.100−x)(.100−x)=81.981.9=(.100+2 x)2(.100− x)(.100−x)I2Cl2IClI .100 .100 .100C -.0729 -.0729 (.0729)E .027 .027
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