Question

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.

Answer #1

So, the ratio of rate constants of catalyzed and uncatalyzed
reactions will be equal to exp(-Ea_{1}/RT) /
exp(-Ea_{2}/RT) = exp((Ea_{2}-Ea_{1})/RT),
where Ea_{2} - activation energy of catalyzed reaction -
15.3 * 4184 J = 64015.2 J; and Ea_{1} - activation energy
of uncatalyzed reaction - 8.7 * 4184 J = 36400.8 J

Ratio = exp ( ( 64015.2-36400.8 ) / 8.314 * 298 ) = 69269

Catalyzed reaction will be 69269 times faster than uncatalyzed.

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 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.

An uncatalyzed reaction has a rate of 0.0000048 sec–1 at room
temperature (25 °C). When an enzyme is added the rate is 32955
sec–1. If the height of the activation barrier for the uncatalyzed
rate is 27.9 kcal/mol, what is the height of the activation barrier
for the catalyzed rate? Report your answer in terms of kcal/mol to
the nearest tenths. Also, assume the pre-exponential terms for the
uncatalyzed and catalyzed reactions are the same.

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?

An uncatalyzed reaction has a rate of 4.8 x 10-7 s–1 at room
temperature (25°C). When an enzyme is added the rate is 3.3 x 104
s–1. If the height of the activation barrier for the uncatalyzed
rate is 28.2 kcal/mol, what is the height of the activation barrier
for the catalyzed rate? Report your answer in terms of kcal/mol to
the nearest tenths. Also, assume the pre-exponential terms for the
uncatalyzed and catalyzed reactions are the same.

The standard free energy of activation of a reaction A is 88.2
kJ mol–1 (21.1 kcal mol–1) at 298 K. Reaction B is ten million
times faster than reaction A at the same temperature. The products
of each reaction are 10.0 kJ mol–1 (2.39 kcal mol–1) more stable
than the reactants.
(a) What is the standard free energy of activation of reaction
B?
(b) What is the standard free energy of activation of the
reverse of reaction A?
(c) What...

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.

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 ratio of an enzyme catalyzed reaction rate to the
uncatalyzed rate (i.e. kcat / kuncat) is
equal to 10,000. Compute the amount, in kj/mol, by which the
activation energy for the reaction is lowered by the enzyme. Assume
the free energy of the reactants is the same in both cases. The
temperature is 300 K and the gas constant R is 8.314 J / (mol
K).

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