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CAS CH 131 (61)
Lecture

# Equilibrium Constant and Reaction Quotient

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School
Department
Chemistry
Course
CAS CH 131
Professor
Andrei Lapets
Semester
Fall

Description
Equilibrium Constant and Reaction Quotient • Equilibrium Constant c d a b 1. If aA + bB  cC + dD, k = ([C] [D] )/([A] [B] ) 2. Ex. Calculate the [] s eq a 0.8 M aqueous solution at 25 C and K = 1.2x10 -2 + - 1. HClO 2(aq) H 2 (l) O 3 (aq)+ ClO 2 (aq) 2. [] i 0.8 --- 0 0 3. Δ -x +x +x 4. [] eq 0.8-x x x 5. K = ([H O][ClO ])/([HClO ]) = (x(x))/(0.8-x) = x /(0.8-x) 2 2 3 2 -3 2 6. x +0.012x – 9.6x10 = 0 7. x = (-0.012 +/- sqrt(0.012 -4(1)(-9.6x10 )))/(2(1)) 1. = 0.092 or -0.10 2. negative makes no sense since can’t have negative concentration, so answer is… 3. 0.71 M 0.092 M 0.092 M • Reaction Quotient Q c d a b c d a b 1. If aA + bB  cC + dD, Q = ([C] [D] )/([A] [B] ) or (P P )/(P Pc) d a b 2. Q < K, Q increases and more products would be formed 3. Q = k, reaction at equilibrium 4. Q > K, Q decreases and more reactant would remain 5. Ex. Determine in which direction the reaction will proceed. 1. N 2(g) 3H 2(g) 2NH 3(g) 2. K = 1.9x10 , T = 400 C, V = 1 L 3. n 0.1 mol 0.04 mol 0.02 mol 4. P = nRT/V 1. P =N2.5 atm 2. P =H2.2 atm 3. P = 1.1 atm NH3 5. P 5i5 2.2 1.1 6. Q = (1.1) /((5.5)(2.2) ) = 2.1x10 -2 7. Q > K, so it goes to the left • Le Châtelier’s Principle 1. If an external stress is applied to a system at equilibrium, the system adjusts to partially offset that stress 2. stress can be [], P, V, T, etc.
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