Chapters 15 &amp; 16 - Gas phase &amp; acid-base equilibrium

# Chapters 15 &amp; 16 - Gas phase &amp; acid-base...

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Equilibrium Mostly, we’ve considered reactions which go essentially to completion. However, there are reactions (some of them commercially significant such as the production of ammonia from nitrogen and hydrogen) for which significant quantities of reactant remain , together with products, when no further reaction is evident – i.e. when equilibrium is attained. For all reactions, both forward and reverse reactions proceed at some rate which depends on the amount of reactants (for forward reaction) and products (for reverse reaction). When the forward and reverse rates do not match, there is net accumulation of either products (for net forward reaction) or reactants (for net reverse reaction). One side of the reaction depletes until the rates balance – this is called dynamic equilibrium. Because the rate of forward reaction depends on the amount of reactants & the rate of reverse reaction depends on the amount of products, the equilibrium condition results when a certain ratio of product and reactant amounts equals a specific value. e.g. 2 NO 2 (g) N 2 O 4 (g) The following ratio – the reaction quotient p(N 2 O 4 )/(1 atm) Q = ----------------------- [p(NO 2 )/(1 atm)] 2 determines the direction of net reaction. If Q < K reaction proceeds to right – i.e. forward direction If Q > K reaction proceeds to lef t – i.e. back ward direction

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p eq (N 2 O 4 )/(1 atm) K = ------------------------- [p eq (NO 2 )/(1 atm)] 2 is termed the equilibrium constant . It is the value of reaction quotient under equilibrium conditions. At 25°C, K = 3.10 for this reaction. Suppose we have an equilibrium mixture of NO 2 & N 2 O 4 at 25°C. What happens if the volume of the container is reduced to half its original size? Both p(N 2 O 4 ) & p(NO 2 ) are doubled, and Q is no longer equal to K Q = K/2 Reaction proceeds in forward direction (direction of fewest moles of gas ), increasing Q until it again equals K 2 NO 2 (g) N 2 O 4 (g) prev eq 0.10 0.031 Initial 0.20 0.062 Change 2 x x Equilibrium 0.20 2 x 0.062 + x 0.062 + x 3.10 = -------------- (0.20 2 x ) 2 12.40 x 2 3.48 x + 0.062 = 0 x = [ 3.48 ± ( 3.48 2 − 4× 12.40 × 0.062 ) 1/2 ]/( 12.40) = 0.262 or 0.019
In general, [a(A ΄ )] n(A ΄ ) [a(B ΄ )] n(B ΄ ) Q = ------------------------------- [a(A)] n(A) [a(B)] n(B) for the reaction, n(A) A + n(B) B + … n(A ΄ ) A ΄ + n(B ΄ ) B ΄ + … a(A) is the activity of species A. ---------------------------------------------------------- For gases, a(A) = p(A)/(1 atm) For dissolved species, a(A) = [A]/(1 M) For pure solids or liquids, a(A) = 1 ---------------------------------------------------------- e.g. H 2 O(l) H 2 O(g) Q = p(H 2 O)/(1 atm) K, in this case, is just the vapor pressure of water. At 25°C, K = 0.031

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Le Châtelier’s Principle If a system at equilibrium is changed in some way, it responds by shifting to new equilibrium conditions wherein the change is partially offset.
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Chapters 15 &amp; 16 - Gas phase &amp; acid-base...

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