ch302 notes ch6.2 - CH302 Chapter 6 notes part 2 K and...

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CH302 Chapter 6 notes part 2 K and Reaction Extent K > 1 Equilibrium lies to the right : System will contain more product than reactant. (K >>> 1: reaction goes to completion) K < 1 Equilibrium lies to the left : System will contain more reactant than product. (K <<< 1: very little reaction occurs) Reaction Quotient Looks very similar to the expression for K, BUT the concentrations are those of the INITIAL setup. .. For a simple reversible reaction: j A + k B « l C + m D The reaction quotient is: Q = [C] l [D] m [X] = INITIAL concentration of X [A] j [B] k Q vs. K: Shift in System Q = K Reaction IS already at equilibrium: no shift. Q > K Product concentration is too high: Reaction must shift to the left to reach equilibrium Q < K Product concentration is too low: Reaction must shift to the right to reach equilibrium Example K = 49 at 450 o C for this reaction: H 2(g) + I 2(g) 2HI (g) If 0.22 mole of I 2 , 0.22 mole of H 2 , and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium? Equilibrium Calculations We are now ready for the tricky stuff! Two main types: 1. Given K and initial concentrations, find equilibrium concentrations. 2. Given initial concentrations and one of the equilibrium concentrations, find K BOTH involve setting up ICE tables (Initial, Change, Equilibrium) What follows are several examples to be done in class. We had to get to this point to be able to 'put it all together'
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Use of K c to find Equilibrium Concentrations (1) K c = 3.00 for the following reaction at a given temperature. If 1.00 mole of SO 2 and 1.00 mole of NO 2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?
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ch302 notes ch6.2 - CH302 Chapter 6 notes part 2 K and...

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