122ch15b_001

122ch15b_001 - 15 15.23) Find K for a reaction that is the...

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15 15.23) Find K for a reaction that is the sum of two reactions. A (aq) + B (aq) W C (aq) K 1 = 1.9 x 10 -4 C (aq) + D (aq) W E (aq) + A (aq) K 2 = 8.5 x 10 2 ----------------------------------------------------- B (aq) + D (aq) W E (aq) K 3 If two rxn’s (eqn’s) can be added to give a 3 rd rxn (eqn) then K 3 is the product of the K 1 and K 2 . R x1 + R x2 = R x3 K 1 * K 2 = K 3 [C] [E] [A] [E] ---------- * ---------- = ---------- ([C] and [A] cancel out) [A] [B] [C] [D] [B] [D] K 3 = K 1 * K 2 = (1.9 x 10 -4 ) (8.5 x 10 2 ) = 0.1 6 15 = 0.16
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15.25) 2 Hg 2 O (s) º 4 Hg ( R ) + O 2 (g) What would K c look like if you included everything (including pure solid and liq)? [Hg] 4 [O 2 ] K c = -------------- [Hg 2 O] 2 However, we normally exclude pure solids and liquids. Their molar conc. are essentially constant so their conc. are incorporated into the equilibrium constant. K c [Hg 2 O] 2 K c = --------------- = [O 2 ] [Hg] 4 K c = [O 2 ] a) The same thing is done for K p (ignore solids and liquids). K p = P O 2 b) As noted above, the molar concentrations for solids and liquids are constant. Their concentration is a ratio of moles to volume occupied (which is related to their density). This doesn’t change whether you have a small or large amounts. NOTE: PURE SOLIDS and LIQUIDS do NOT appear in the expression for K or Q . (Aqueous and gaseous substances should be in K or Q.)
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122ch15b_001 - 15 15.23) Find K for a reaction that is the...

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