Part A Part B Express your answer numerically ininverse seconds. Part A The acti

Question

Part A Part B Express your answer numerically ininverse seconds.



Part A The activation energy of a certain reactionis 38.4 kJ/mol. At 20 ^{circ}rm C, the rate constant isrm 0.0130~s^{-1}. At whattemperature would this reaction go twice as fast? Express your answer numerically indegrees Celsius Part B Given that the initial rate constant isrm 0.0130~s^{-1} at an initialtemperature of 20 ^circ rm C, what would the rateconstant be at a temperature of 100 rm ^{circ}C? Express your answer numerically ininverse seconds.

Explanation / Answer

         Arrheniusequation , k = Ae-Ea /RT                             ln k1 / k2 = Ea / R [ T1-T2 /T1T2]    Part A                 Ea =38.4 KJ /mol          T1 =20oC       k1 =0.0130sec-1 , k2 = 2k1 = 2 x0.0130sec-1          ln 0.013 / 0.026 = 38.4   [ 20 -T2 / 20T2]             T2 = 23.71oC    part B       ln 0.0130 / k2 = 38.4 [20-100/ 20 x 100]                       k2 = 0.06 sec-          ln 0.013 / 0.026 = 38.4   [ 20 -T2 / 20T2]             T2 = 23.71oC    part B       ln 0.0130 / k2 = 38.4 [20-100/ 20 x 100]                       k2 = 0.06 sec-

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