chem107f07wk14s

chem107f07wk14s - Chemical Equilibrium Chapter 12.2 N2O4...

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N 2 O 4 2NO 2 at equilibrium: rate fwd = rate rev rate fwd = k fwd [N 2 O 4 ] eq rate rev = k rev [NO 2 ] 2 eq k fwd [N 2 O 4 ] eq = k rev [NO 2 ] 2 eq Chemical Equilibrium Chapter 12.2 eq eq eq rev fwd K ] O N [ ] NO [ k k = = 4 2 2 2
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Chemical Equilibrium Chapter 12.2 Equilibrium is a dynamic state.
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a A + b B c C + d D K c = [C] c [D] d [A] a [B] b equilibrium concentrations Equilibrium Constant 12.3 Expression ignores solids and pure liquids.
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SnO 2 (s) + 2CO(g) Sn(s) + 2CO 2 (g) CaCO 3 (s) CaO(s) + CO(g) Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq) Write the expression for the equilibrium constant, K c
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K p = ( P C ) c ( P D ) d ( P A ) a ( P B ) b K p = K c (RT) Δ n Δ n = (moles of gaseous product) (moles of gaseous reactant) equilibrium pressures Equilibrium Constant K p
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The formation of NO from oxygen and nitrogen might be considered a means of fixing nitrogen. O 2 (g) + N 2 (g) 2NO(g) The value of the equilibrium constant for this reaction at 25 o C is K c = 1 x 10 -30 . Describe the feasibility of fixing nitrogen by forming NO at 25 o C.
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For the reaction H 2 (g) + I 2 (g) 2HI(g) K p = 794 at 298 K and K p = 54 at 700 K. Is the formation of HI favored more at the higher or lower temperatures?
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For the reaction below. K c = 3.6 at 300 o C. What is the K p for this reaction at this temperature? 3H 2 (g) + N 2 (g) 2NH 3 (g)
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A B K B A 2A 2B C D K 1 E F K 2 C + E F + D Manipulating Equilibrium Constant 12.3
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H + (aq) + F - (aq) K c = 6.8 x 10 -4 H 2 C 2 O 4 (aq) 2H + (aq) + C 2 O 4
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This note was uploaded on 02/23/2010 for the course CHEM 107 taught by Professor Generalchemforeng during the Fall '07 term at Texas A&M.

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chem107f07wk14s - Chemical Equilibrium Chapter 12.2 N2O4...

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