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Unformatted text preview: Homework 6 Mar. 8 21. a. d [ P ] dt = k [ A ][ B ] = 1 × 10 5 × . 1 × . 1 = 10 3 M/s b. d [ P ] dt = k [ A ][ B ] = 1 × 10 5 × 1 × 10 4 × 1 × 10 6 = 10 5 M/s c. For [ A ] = [ B ] = 0 . 1 M, d [ A ] dt = k [ A ] 2 . Then, Z . 1 . 05 [ A ] d [ A ] dt = k Z t dt 1 [ A ]  . 05 . 1 = kt t = 1 × 10 4 s d. E a = RT 2 T 1 T 2 T 1 ln k 2 k 1 where T 1 = 300 K, T 2 = 400 K, and k 2 k 1 = 10 3 . Also, Δ H ‡ = E a RT , and so E a = 68 . 9 kJ Δ H ‡ = 66 . 4 kJ/mol 1 22. a. d [ A ] dt = k 1 [ A ][ B ] Z [ A ] [ A ] d [ A ] [ A ] = k 1 [ B ] Z t dt ln [ A ] [ A ] = k 1 [ B ] t Reaction is first order in A and B . b. d [ A ] dt = k 2 [ B ] Z [ A ] [ A ] d [ A ] = k 2 [ B ] Z t [ A ] [ A ] = k 2 [ B ] t Reaction is zero order in A and first order in B . c. The equation that is consistent with experiments at both high and low [ A ] is d [ A ] dt = k 1 [ A ][ B ] 1 + k 1 [ A ] /k 2 ....
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 Spring '07
 YUAN
 Trigraph, dt, Bicarbonate, k2

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