# eq - This is in response to the following Thinkwell...

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This is in response to the following Thinkwell question: Given the fol- lowing information, ﬁnd the activation energy for the decomposition of ac- etaldehyde. k1 = 1.05 103 M1/2s1 at 759 K k2 = 2.14 102 M1/2s1 at 836 K The student who asked the question was getting the correct answer but with a diﬀerent sign, hence the focus on the sign of the activation energy. So we start with the Arrhenius equation. k = Ae - Ea RT (1) Now we take the log of both sides ln ( k ) = ln ± Ae - Ea RT ² (2) using the rules of logorithmic functions ln ( xy ) = ln ( x ) + ln ( y ) (3) ln ( e z ) = z (4) we are left with ln ( k ) = ln ( A ) - E a RT (5) Now we make an assumption, we assume that when we change T, the tem- perature, A does not change. We know k changes and we know T changes so we will write two equations for the two temperatures ln ( k 1 ) = ln ( A ) - E a RT 1 (6) ln ( k 2 ) = ln ( A ) - E a RT 2 (7) In class Professor Scerri subtracted equation 6 from equation 7 this is com- pletely legal. if you are not comfortable with that you can just solve both

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## This note was uploaded on 02/11/2011 for the course CHEM 14B taught by Professor Lavelle during the Spring '01 term at UCLA.

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eq - This is in response to the following Thinkwell...

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