Question 25 of 36 Incon irect XIncorrectIncorrect Incorrect Incorrect General Ch

Question

Question 25 of 36 Incon irect XIncorrectIncorrect Incorrect Incorrect General Chemistry 4th Edition University Science Books presenled by Sapling Leaming Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the following reaction equation. The activation energy for this reaction is 75 kJ molr1 The enzyme catalase (found in blood) lowers the activation energy to 8.0 kJ-mo At what temperature would the non-catalyzed reaction need to be run to have a rate equal to that of the enzyme-catalyzed reaction at 25? Number Previous Gare Up & View Solution 9 Check Answer 0 Next Ex.

Explanation / Answer

we have relation k = A exp ( -Ea/RT)

where Ea = 8 KJ/mol = 8000 J/mol , T = 25C =25+273 = 298K , let as assume A = 1 , now we get k value as

k = 1 exp ( -8000J/mol / 8.314J/molK x 298K)

= 0.0396

now non catalysed reaction need to have same rate constant 0.0369 , we have Ea = 75000 J/mol , we find T

0.0396 = 1 exp ( - 75000 /8.314 x T)

0.0396 = exp ( -9021/T)

taking ln both sides we get

ln ( 0.0396) = ln ( exp ( -9021 /T)

-3.2289 = -9021 / T

T = 2794 K is answer

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