Question

How much does an enzyme need to reduce the activation energy of a reaction at 25 ºC in order to increase the rate 100-fold over the uncatalyzed reaction? R = 8.3145 J/K mol. When answering, convert your response to kJ/mol and input ONLY a number, rounded to one decimal place.

Hint: the rate constant for an enzymatic reaction divided by the uncatalyzed reaction = 10

Answer #1

Since the rate is to be increased 100 times,

k_{enzyme}/ k_{uncatalyzed} = 100 ( Unlike the
100 given in the hint)

k_{enzyme} = k_{o} exp(-E_{enzyme} /
RT)

k_{uncatalyzed} =
k_{o}exp(-E_{uncat}/RT)

Divide the two,

k_{enzyme} / k_{uncat} = exp((E_{uncat}
- E_{cat})/RT) = 100

Substituting values for R = 8.314 and T = 298

We get

100 = exp(E_{red} / 8.314 / 298)

Take log on both sides to get

log (100) = 4.6052 = E_{red} / (2477.57)

E_{red} = 11,409.7 J/mol = **11.41
kJ/mol**

**Required ans - 11.4**

( Since its not clear if its 10 fold or 100 fold, for 10 fold,
the log value will be half of this, **and for 10 fold the
answer will be 5.7** accordingly)

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).

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 60.0 kJ/mol and a
frequency factor of A1 = 2.80×1012 M−1s−1 . What is the rate
constant, k, of this reaction at 30.0 ∘C ?
I tried k = Ae^(-Ea/RT) but it's giving me 1.26E2 M^-1s^-1 which
is wrong on my online homework, I did convert C to K and kJ/mol to
J.

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 energy, Ea for a particular reaction is
13.6 kj/mol. If the rate constant at 754 degrees celsius is
24.5/min at egat temperature in celsius will the rate constant be
12.7/min? r= 8.314j/mol • K

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?

If an enzyme rate increases 10-fold for every 5.7 kJ/mol
decrease in the energy of activation, what rate increase would be
observed if three H-bonds were stabilizing the transition state
with each H-bond having a strength of 21 kJ/mol

The hydrolysis of urea to ammonia plus CO2 normally has an
activation energy of 125 kJ/mol at 294 K, but when the urease
enzyme catalyzes the reaction it lowers the activation energy to 46
kJ/mol. By what factor does this increase the velocity of the
reaction? The rate constant equation might be helpful here, where k
= A·e(-ΔG°‡/RT) and A = 6.12 x 1012 s-1. Look at both kcat and
knon, finding the ratio between the two

The frequency factor and activation energy for a chemical
reaction are A = 8.08 x 10–12 cm3/(molecule·s) and Ea = 15.0 kJ/mol
at 368.4 K, respectively. Determine the rate constant for this
reaction at 368.4 K.

Calculate the rate constant, k, for a reaction at 65.0 °C that
has an activation energy of 87.1 kJ/mol and a frequency factor of
8.62 × 1011 s–1.

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