Question

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.

Answer #1

**we know that**

**K = Ae^(-Ea/RT)**

**ln K = lnA - (Ea/RT)**

**now**

**at two different conditions**

**ln K1 = lnA - (Ea1/RT)**

**lnA = lnK1 + (Ea1/RT)**

**and**

**ln K2 = ln A - (Ea2/RT)**

**ln A = lnK2 + (Ea2/RT)**

**so**

**we get**

**lnK1 + (Ea1/RT) = lnK2 + (Ea2/RT)**

**ln K2 - ln K1 = (Ea1/RT) - (Ea2/RT)**

**ln (K2/K1) = (Ea1 - Ea2)/RT**

**given**

**Ea1 = 122**

**Ea2 = 57**

**T = 25 C = 298 kelvin**

**so**

**ln (K2/K1) = (122-57) x 1000 / 8.314 x 298**

**(K2/K1) = 2.4767 x 10^11**

**so**

**the rate rate will increase by a factor of 2.4767 x
10^11**

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.

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.

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)

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.

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.

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?

The activation energy for the decomposition of hydrogen peroxide
is 55.0 kJ/mol. When the reaction is catalyzed by the enzyme
catalase, it is 11.00 kJ/mol. 2H2O2(aq) → 2H2O(l) + O2(g) Calculate
the temperature that would cause the nonenzymatic catalysis to
proceed as rapidly as the enzyme-catalyzed decomposition at 20.0°C.
Assume the frequency factor, A, to be the same in both cases.
Report your answer to 3 significant figures.

The activation energy of a reaction is 55.6 kJ/mol and the
frequency factor is 1.5×1011/s. Calculate the rate constant of the
reaction at 23 ∘C. Express your answer using two significant
figures.

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