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H 2(g) + I 2(g) 2HI (g) Write the K expression [H 2 ] = 0.106 M [I 2 ] = 0.022 M [HI] = 1.29 M

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H 2(g) + I 2(g) 2HI (g) What is the equilibrium conc of [HI] if: [H 2 ] = 0.81 M [I 2 ] = 0.035 M

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Solve for K if: It is the reverse rxn The eqn is 1/3 of the original The eqn is reversed and is 4 times the original.

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Equilibria with pressures? So far, we’ve only used concentration gases can be described in terms of pressures

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PV = nRT P =(n/V]RT = CRT C= molar conc of gas K p is the equilibrium in terms of partial pressures of the gases

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N 2(g) + 3H 2(g) 2NH 3(g) Write the K p expression How do K and K p relate? Let’s try it for this rxn… K= K p (RT) 2

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H 2(g) + F 2(g) 2HF (g) Try it for this rxn… K = K p b/c sum of coefficients on either side is equal

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General relationship K p = K(RT) n n= the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants

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CH 3 OH (g) CO (g) + 2H 2(g) At 25 o C P CH3OH = 6.10 x 10 -4 atm P CO = 0.387 atm P H2 = 1.34 atm What is K?

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So far... Our systems have only involved gases! Homogeneous equilibria- where phases are all the same

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Heterogeneous equilibria Equilibria involving more than one phase Take for instance… CaCO 3(s) CaO (s) + CO 2(g) Write the K

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However... Experiments show that the position of heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present!!

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Why?? Concentrations of pure solids and liquids can NOT change!! So… LEAVE THEM OUT of K!! Let’s try…

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Homework P 640 (WOW!) 22, 25, 28, 29, 31

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To do now… Learn how to use “assignments” folder set groups for presentations set up chem olympics teams start homework!

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Ch. 14: Chemical Equilibrium I.Introduction II.The Equilibrium Constant (K) III.Values of Equilibrium Constants IV.The Reaction Quotient (Q) V.Equilibrium.

Ch. 14: Chemical Equilibrium I.Introduction II.The Equilibrium Constant (K) III.Values of Equilibrium Constants IV.The Reaction Quotient (Q) V.Equilibrium.

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