HW7 - Ch8-5 a. Michaelis-Menten equation shows that d[P ]...

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Ch8-5 a. Michaelis-Menten equation shows that d [ P ] dt = - k 2 [ E ] 0 [ S ] K M + [ S ] When [ S ] = 0 . 1 M and [ E ] 0 = 1 . 0 × 10 - 5 M at 280 K, d [ P ] dt = 100 × 1 × 10 - 5 × 0 . 1 1 × 10 - 4 + 0 . 1 = - 1 × 10 - 3 Ms - 1 . b. E a = RTT 0 T - T 0 ln k 2 k 0 2 (see Example 7.5 in textbook) = 8 . 314 × 280 × 300 300 - 280 ln 200 100 = 24 . 2 kJ/mol . c. We need to find K eq = k 1 k - 1 . K M = k 1 + k 2 k 1 . Since k 1 ,k - 1 ± k 2 , K M = 1 × 10 - 4 k - 1 /k 1 . Hence, K eq = 1 K M = 1 × 10 4 M - 1 . d. By using Van Hoff equation, 1
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ln K eq 2 K eq 1 = - Δ H 0 R ± 1 T 2 - 1 T 1 ² ln K M 1 K M 2 = - Δ H 0 R ± 1 T 2 - 1 T 1 ² ln 1 × 10 - 4 1 . 5 × 10 - 4 = - Δ H 0 8 . 314 ± 1 300 - 1 280 ² Δ H 0 = - 14 . 16 kJ/mol Ch8-10 a. From equation 8.9 in textbook, [ ES ] = [ E ] 0 1 + k - 1 + k 2 k 1 [ S ] [ ES ] [ E ] 0 = ± 1 + k - 1 + k 2 k 1 [ S ] ² - 1 = ± 1 + K M [ S ] ² - 1 = 0 . 99 . b. From equation on page 409 in textbook, K M ln [ S ] [ S ] 0 + [ S ] - [ S ] 0 = - V max t K M ln [ S ] [ S ] 0 + [ S ] - [ S ] 0 = - k 2 [ E ] 0 t 5 × 10 - 5 ln 1 2 + 2
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This note was uploaded on 08/03/2009 for the course BME 100 taught by Professor Yuan during the Spring '07 term at Duke.

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HW7 - Ch8-5 a. Michaelis-Menten equation shows that d[P ]...

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