Question

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How many times faster is the catalyzed than the uncatalyzed reaction at 400°C? Assume that the frequency factor remains the same.

Answer #1

So, the ratio of rate constants of catalyzed and uncatalyzed
reactions will be equal to exp(-Ea1/RT) / exp(-Ea2/RT) =
exp((Ea2-Ea1)/RT), where Ea1 - activation energy of catalyzed
reaction and Ea1 - activation energy of not catalyzed
reaction.

ratio = exp((64000-55000)/8.314/673.15)=5

Catalyzed reaction will be 5 times faster than uncatalyzed
reaction.

A reaction proceeds with ∆ H = -80 kJ/mol. The energy of
activation of the uncatalyzed reaction is 80 kJ/mol, whereas it is
55 kJ/mol for the catalyzed reaction. How many times faster is the
catalyzed reaction than the uncatalyzed reaction at 25°C?

At 298K, adding a catalyst makes a certain reaction go 250,000
times faster than the uncatalyzed reaction. The activation energy
for the uncatalyzed reaction is 80.0 kJ/mol. What is the activation
energy for the catalyzed reaction? Assume the frequency factor A is
the same for both catalyzed and uncatalyzed reactions.
A. 49.2 kJ/mol
B. 68.4 kJ/mol
C. 34.7 kJ/mol
D. 54.1 kJ/mol
E. 60.8 kJ/mol
Please provide explanation.

The activation energy for a reaction is changed from 184 kJ/mol
to 59.5 kJ/mol at 600. K by the introduction of a catalyst. If the
uncatalyzed reaction takes about 2627 years to occur, about how
long will the catalyzed reaction take? Assume the frequency factor
A is constant and assume the initial concentrations are the
same.

The activation barrier for an uncatalyzed reaction is estimated
to be 15.3 kcal/mol. The activation barrier for the catalyzed
reaction is estimated to be 8.7 kcal/mol. How many times faster is
the catalyzed rate versus the uncatalyzed rate? In other words, by
what factor/coefficient do you have to multiply the uncatalyzed
rate to equal the catalyzed rate? Assume the temperature is 298 K,
and enter your answer to the nearest ones.

The activation barrier for an uncatalyzed reaction is estimated
to be 15.3 kcal/mol. The activation barrier for the catalyzed
reaction is estimated to be 8.7 kcal/mol. How many times faster is
the catalyzed rate versus the uncatalyzed rate? In other words, by
what factor/coefficient do you have to multiply the uncatalyzed
rate to equal the catalyzed rate? Assume the temperature is 298 K,
and enter your answer to the nearest ones.

The activation barrier for an uncatalyzed reaction is estimated
to be 15.3 kcal/mol. The activation barrier for the catalyzed
reaction is estimated to be 8.7 kcal/mol. How many times faster is
the catalyzed rate versus the uncatalyzed rate? In other words, by
what factor/coefficient do you have to multiply the uncatalyzed
rate to equal the catalyzed rate? Assume the temperature is 298 K,
and enter your answer to the nearest ones.

A catalyst decreases the activation energy of a particular
exothermic reaction by 58 kJ/mol, to 24 kJ/mol. Assuming that the
mechanism has only one step, and that the products are 89 kJ lower
in energy than the reactants, sketch approximate energy-level
diagrams for the catalyzed and uncatalyzed reactions.
What is the activation energy for the uncatalyzed reverse reaction?
(Please Show Work)

Suppose that a catalyst lowers the activation barrier of a
reaction from 122 kJ/mol to 57 kJ/mol .
By what factor would you expect the reaction rate to increase at
25 ∘C? (Assume that the frequency factors for the catalyzed and
uncatalyzed reactions are identical.)
Express your answer using two significant figures.

A certain reaction has an activation energy of 31.40 kJ/mol. At
what Kelvin temperature will the reaction proceed 6.50 times faster
than it did at 293 K?
Use the Arrhenius equation
ln (k2/k1) = Ea/R [(1/T1)-(1/T2)]
Where R=8.3145 J/(molxK)

A certain reaction has an activation energy of 47.88 kJ/mol. At
what Kelvin temperature will the reaction proceed 8.00 times faster
than it did at 335K?

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