CH131_14A

CH131_14A - Chemical Equilibrium Consider the following gas...

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Chemical Equilibrium Consider the following gas phase chemical reaction: 2HI(g) H 2 (g) + I 2 (g) We’ve written chemical reactions before and the meaning has been that the reaction goes to completion, i.e. all HI reacts to form product. Often not the case; reactions often proceeds until no more chemical change observed => we say chemical equilibrium reached. We’ve discussed equilibrium before => phase equilibrium e.g. liquid vapor Same considerations apply to chemical systems Consider the simple chemical dissociation reaction above
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2HI(g) H 2 (g) + I 2 (g) 1. Start with just HI in container => HI molecules collide; some 2 , I 2 2. Some of the resulting H 2 2 then collide and react to form HI again via the back reaction: H 2 (g) + I 2 (g) 2HI(g) We summarize this chemical situation w/ double arrows: 2HI(g) H 2 (g) + I 2 (g) (like the phase equilibrium) When the rate of the forward rxn = rate of backward rxn => chemical equilibrium achieved & no further changes observed 1. All reactions are reversible, in general. 2. Sometimes a single direction dominates (goes to completion) 3. Equilibrium can be reached from an infinite # of directions
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Say for the HI dissociation reaction: 2HI(g) H 2 (g) + I 2 (g) V = 10.0 L, T = 520 o C initially 2.00 mol HI, 0 H 2 , 0 I 2 => What happens? Plot of HI, I 2 concentration (P) vs. time HI decomposes quickly at first forming H 2 + I 2 , then more slowly (less HI around to collide) I 2 conc. (P) increases quickly then slows due to back reaction H 2 (g) + I 2 (g) 2HI(g) At some time the rates of the forward reaction and back reaction the same and no change in concentrations observed (equil.) At equilibrium, we find 0.20 mol I 2 . How much H 2 ? 0.20 mol How much HI? If 0.20 mol I 2 formed, 0.40 mol HI reacts, so 2.00 - 0.40 = 1.60 mol HI left unreacted
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We could use partial pressures as well as mols to keep track of concentrations: T and V constant P i = n i RT V P HI = n HI RT V where RT V = 0.0821 L atm / K mol 793 K ( ) 10.0 L = 6.51 atm / mol P HI = 10.4 atm P H2 = 1.30 atm P I2 = 1.30 atm at equilibrium
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Consider another initial condition => start with 1.0 mol H
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CH131_14A - Chemical Equilibrium Consider the following gas...

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