Koratalyst = e . = e 55-76 8.314 R (2૧૪) 3.3 The reaction will be 3.3 times faster c. As the temperature increases, will the catalyst be equally effective, or will it lose or gain effectiveness? Show this in a semi-quantitative way. kinetic energy is available for As the temperature increases, more the weactants to overcome the activation energy barrier, which reduces the contribution of the catalyst, so the catalyst will it lose effectiveness. In a semi-quantitative way we can say: waction rate with catalyst a reaction rate without catalyst +T d. The rate law for this reaction is: rate = k [NH₂). Give a reason why this reaction [H₂] rate is inverse in H₂. The reaction rate is inversely proportional to the [H₂] because the rale-determining step of the reaction involver the collation of 2 molecdes to N₂ to 1₂. The H₂ affects the probability prevent this collision, the role is proportional to [NH₂)² and inversely proportiond to [H.]. that will you
Koratalyst = e . = e 55-76 8.314 R (2૧૪) 3.3 The reaction will be 3.3 times faster c. As the temperature increases, will the catalyst be equally effective, or will it lose or gain effectiveness? Show this in a semi-quantitative way. kinetic energy is available for As the temperature increases, more the weactants to overcome the activation energy barrier, which reduces the contribution of the catalyst, so the catalyst will it lose effectiveness. In a semi-quantitative way we can say: waction rate with catalyst a reaction rate without catalyst +T d. The rate law for this reaction is: rate = k [NH₂). Give a reason why this reaction [H₂] rate is inverse in H₂. The reaction rate is inversely proportional to the [H₂] because the rale-determining step of the reaction involver the collation of 2 molecdes to N₂ to 1₂. The H₂ affects the probability prevent this collision, the role is proportional to [NH₂)² and inversely proportiond to [H.]. that will you
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Hello, Can you help me with this correcting me this questions below.
Here I leave the comments to correct: In 3b, you didn't convert the Ea's to J/mol (also, why "45" and "76"? I don't see where those numbers came from). Also, the 298 should be in the denominator. It should end up that it's roughly 10^30 times faster. In 3c, NO. Try a higher temperature with the same Ea you used in 3b. The k should go down at a higher temperature. In 3d, NO. Mention that H2 is a product, and that this is an example of product inhibition.
![3. For the reaction 2 NH3 (g) → N2 (g) + 3 H₂ (g), the following information is known
about the activation energies.
Catalyst
Activation energy
(kJ/mol)
Kunsten
Kuoratalyst
None
335
W (tungsten)
163
b. How many times faster is the reaction when tungsten catalyst is used compared to no
catalyst? Assume T-298K.
T
= e
1. Edtungates - Edmorotalyst,
R
Os (osmium)
197
55-76
= e 8.314 (298)
= 3.3
The reaction will be 3.3 times faster
c. As the temperature increases, will the catalyst be equally effective, or will it lose or
gain effectiveness? Show this in a semi-quantitative way.
kinetic
more
energy
As the temperature increases,
is available for
the seactants to overcome the activation energy barrier,
which reduces the contribution of the catalyst, so the
catalyst will it lose effectiveness. In a semi-quantitative way
we can say: reaction rate with catalyst ≈ reaction rate without catalyst +T
d. The rate law for this reaction is: rate = k
(NH₂)
[H₂]
. Give a reason why this reaction
rate is inverse in H₂.
the rate-determining step of
The reaction rate is inversely proportional to the [H₂] because
the reaction involuer the collision
of 2 molecdes to N₂ to H₂. The H₂ affects the probability
that will prevent this collision, the role is proportional
to [NH₂)² and inversely proportiond to [Hh₂].
you](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F603d5899-1291-47a0-990b-5ec94451be91%2F35c0d3fb-feb1-4b9f-9e67-0ca223effcc4%2F30w6m2_processed.png&w=3840&q=75)
Transcribed Image Text:3. For the reaction 2 NH3 (g) → N2 (g) + 3 H₂ (g), the following information is known
about the activation energies.
Catalyst
Activation energy
(kJ/mol)
Kunsten
Kuoratalyst
None
335
W (tungsten)
163
b. How many times faster is the reaction when tungsten catalyst is used compared to no
catalyst? Assume T-298K.
T
= e
1. Edtungates - Edmorotalyst,
R
Os (osmium)
197
55-76
= e 8.314 (298)
= 3.3
The reaction will be 3.3 times faster
c. As the temperature increases, will the catalyst be equally effective, or will it lose or
gain effectiveness? Show this in a semi-quantitative way.
kinetic
more
energy
As the temperature increases,
is available for
the seactants to overcome the activation energy barrier,
which reduces the contribution of the catalyst, so the
catalyst will it lose effectiveness. In a semi-quantitative way
we can say: reaction rate with catalyst ≈ reaction rate without catalyst +T
d. The rate law for this reaction is: rate = k
(NH₂)
[H₂]
. Give a reason why this reaction
rate is inverse in H₂.
the rate-determining step of
The reaction rate is inversely proportional to the [H₂] because
the reaction involuer the collision
of 2 molecdes to N₂ to H₂. The H₂ affects the probability
that will prevent this collision, the role is proportional
to [NH₂)² and inversely proportiond to [Hh₂].
you
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