the calculated value is Ea is 18.2 KJ and A is 12.27.
According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.
At 500K, K=0.02s−1
At 700K, k=0.07s −1
The Arrhenius equation can be used to calculate Ea and A.
RT=k=Ae Ea
lnk=lnA+(RT−Ea)
At 500 K,
ln0.02=lnA+500R−Ea
500R Ea (1) At 700K lnA=ln (0.02) + 500R
lnA = ln (0.07) + 700REa (2)
Adding (1) to (2)
700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.
=ln [0.02/0 .07]
Ea= 2/35×100×8.314×1.2528
Ea =18227.6J
Ea =18.2KJ
Changing the value of E an in (1),
lnA=0.02) + 500×8.314/18227.6
= (−3.9120) +4.3848
lnA=0.4728
logA=1.0889
A=antilog (1.0889)
A=12.27
Consequently, Ea is 18.2 KJ and A is 12.27.
Learn more about Arrhenius equation here-
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The kinetic energy of gas particles depends on temperature. Greater the temperature higher will be the average kinetic energy
Kinetic energy is related to the temperature as:
KE = 3/2 kT
where k = Boltzmann constant
T = temperature
In the given example, since the temperature of O2 gas is maintained at room temperature, the average KE will also remain constant.
The fifth postulate of the kinetic molecular theory which states that the temperature of the gas depends on the average KE of the particles of the gas explains the above observation.
I say the Correct answer is
B) There will be a shift toward the products.
I believe that you gain a day
