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Sagot :
Therefore the activation barrier of this reaction is EA=84lkJ⋅mol−1, If the frequency factor is 1.2*10^13/s, The rate constant of a reaction at 33 Degrees Celsius was measured to be 5.8×10^-2 /s.
The Arrhenius equation states that
k=A⋅e−EA/R⋅T
Taking logarithm of both sides gives
lnk = lnA−EA/R⋅T
Where, the rate constant of this particular reaction
k=0.055ls−1;
The frequency factor (a temperature-dependent constant A=1.2×1013ls−1as given in the question;
The ideal gas constant R=8.314lJ⋅mol−1⋅K−1;
Absolute temperature (T=32+273.15=305.15lK) at which the reaction take place;
EA the activation barrier (a.k.a. activation energy ) the question is asking for
Solve the second equation for EA:
EA/R⋅T=lnA−lnk
EA=(R⋅T)⋅(lnA−lnk)
=(R⋅T)⋅lnA/k
=8.314lJ⋅mol−1⋅K−1⋅305.15lK⋅ln(1.2×1013s−10/055s−1)
EA=8.4⋅104lJ⋅mo−1
Therefore the activation barrier of this reaction is 84lkJ⋅mol−1
Learn more about activation barrier here:
https://brainly.com/question/11296426
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