Question

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.

Answer #1

I'm not quite sure if the height of activation barrier is the same of activation energy Ea. I'm looking in different sources and I can't fin another equation or formula refering to height of activation, so, I'll use instead that formula to give you the idea of how to do it. If we have to use another formula then put it, and I'll do it for you.

the general equation for the rate would be:

K = Ae^{-Ea/RT}

for the uncatalized reaction:

K1 = Ae^{-Ea1/RT}

For the catalized reaction:

K2 = Ae^{-Ea2/RT}

so:

K1/K2 = A/A e^{Ea2-Ea1/RT}

ln K1/K2 = Ea2 - Ea1/RT

ln K1/K2 x RT + Ea1 = Ea2

Ea2 = ln (0.0000048/32955) x 8.3143 x 298 + (27.9 x 4.184)

Ea2 = 60.62 kJ/mol

Ea2 = 60.64 / 4.184

**Ea2 = 14.48 kcal/mol**

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 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.

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).

calculate at room temperature: a) the ratio of the rate
constants for two reactions that have the same pre-exponential
value but have activation energies that differ by 40.0 kJ/mole b)
the difference in the activation energies (Ea1-Ea2) when the first
reaction has a pre-exponential value that is 5 times the
pre-exponential value of the second reaction and where the rate
constant of the first reaction is 100 times larger than the rate
constant of the second reaction.

If the uncatalyzed reaction proceeds at a rate of 2.6 x 10-5 s-1
and the enzymatic reaction rate is 50 s-1 , what is the difference
in the height of the activation barrier for the enzymatic reaction?
If the average H-bond in the chorismate mutase active site releases
18.5 kJ/mol of energy, how many H-bonds would have to form to
account for the rate enhancement?

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.

It is often stated in many introductory chemistry texts that
near room temperature, a reaction rate doubles if the temperature
increases by 10°. Calculate the activation energy of a reaction
that obeys this rule exactly. Would you expect to find this rule
violated frequently? [R=1.987 cal/mol•K=8.314 J/mol•K]?
2) An endothermic reaction A → B has a positive internal energy
change, ΔE, or enthalpy change,
ΔH. In such a case, what is the minimum value that the
activation energy can have?...

The activation energy of a particular reaction is 245 kJ/mol.
How many degrees above room temperature (25°C) would the reaction
need to be heated in order to see a 10 fold increase in the rate
constant?

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