Chemical Kinetics

Chemical Kinetics

Question

Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to-
(R = 8.314 J mol–1K–1)

  • 8

  • 12

  • 6

  • 4

Answer

D.

4

From Arrhenius equation
straight K space equals space straight A. straight e to the power of fraction numerator negative Ea over denominator RT end fraction end exponent
So comma space straight K subscript 1 space equals space straight A. straight e to the power of bevelled fraction numerator negative straight E subscript straight a 1 end subscript over denominator RT end fraction end exponent space..... space left parenthesis 1 right parenthesis
straight K subscript 2 space equals space straight A. straight e to the power of bevelled fraction numerator negative straight E subscript straight a 2 end subscript over denominator RT end fraction end exponent space..... space left parenthesis 2 right parenthesis
so space divide space equation space 2 space by space 1 space we space get
rightwards double arrow space straight K subscript 1 over straight K subscript 2 space equals space straight e to the power of fraction numerator left parenthesis straight E subscript straight a 1 end subscript minus straight E subscript straight a 2 end subscript over denominator RT end fraction end exponent
Taking space log space on space both space side
ln space open parentheses straight K subscript 2 over straight K subscript 1 close parentheses space equals space fraction numerator straight E subscript straight a 1 end subscript minus straight E subscript straight a 2 end subscript over denominator RT end fraction space equals space fraction numerator 10 comma 000 over denominator 8.314 space straight x space 300 end fraction space equals space 4

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