L17_AcidBase4_115b08

L17_AcidBase4_115b08 - L17 7.(c) (5 points) You can solve...

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7.(c) (5 points) You can solve the questions in parts (a) and (b) also by defining the fraction that is dissociated as α so that the pressure of N 2 O 5 at equilibrium is n (1- α ), where n is the initial pressure. Derive an equation for K p that expresses it only as a function of α and the total pressure P t at equilibrium. N 2 O 5 (g) NO 3 (g) + NO 2 (g) Initial pp: n 0 0 Equilibrium: n (1- α ) α n α n P t = n (1- α ) + 2 α n = n + α n = n (1+ α )- > n = P t / (1+ α ) 22 2 () ( ) (1 ) ) 1 ) ) t p t P n K P n n αα ααα α == = = + −−− 32 25 NO NO p NO PP K P = L17
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(a) In the experiment n atm of N 2 O 5 (g) were used to prepare NO 3 . Calculate the partial pressure of NO 3 , p NO3 , in a mixture that has a total pressure at equilibrium of P t = 1.00 atm . K p = 4.96 (b) What was the initial pressure, n ? K p = P t α 2 /(1 - α 2 ); n = P t / (1+ α ); N 2 O 5 (g) NO 3 (g) + NO 2 (g) K p and P t are given; can find α K p (1 - α 2 ) = P t α 2 K p -K p α 2 = P t α 2 K p = α 2 ( K p + P t ) --> α 2 = K p / ( K p + P t ) α = {K p / (K p + P t )} 1/2 = {4.96 / (4.96 + 1.00} 1/2 = 0.912 = α n = P t / (1+ α ) = 1.00/1.912 = 0.523 atm; initial pressure of N 2 O 5 α n = 0.523x0.912 = 0.477 atm = pressure of NO
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This note was uploaded on 07/22/2008 for the course CHEM 115b taught by Professor Reisler during the Spring '08 term at USC.

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L17_AcidBase4_115b08 - L17 7.(c) (5 points) You can solve...

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