Question

To understand the relationship between the equilibrium constant and rate constants. For a general chemical equation...

To understand the relationship between the equilibrium constant and rate constants.

For a general chemical equation

A+B⇌C+D

the equilibrium constant can be expressed as a ratio of the concentrations:

Kc=[C][D][A][B]

If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows:

forward ratereverse rate==kf[A][B]kr[C][D]

where kf and kr are the forward and reverse rate constants, respectively. When equilibrium is reached, the forward and reverse rates are equal:

kf[A][B]=kr[C][D]

Thus, the rate constants are related to the equilibrium constant in the following manner:

Kc=kfkr=[C][D][A][B]

Part A

For a certain reaction, Kc = 5.79×108 and kf= 634 M−2⋅s−1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction.

Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M−2⋅s−1include ⋅ (multiplication dot) between each measurement.

kr =

SubmitHintsMy AnswersGive UpReview Part

Part B

For a different reaction, Kc = 7.92×105, kf=6.77×105s−1, and kr= 0.855 s−1 . Adding a catalyst increases the forward rate constant to 2.24×108 s−1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M−2⋅s−1include ⋅ (multiplication dot) between each measurement.

kr =

SubmitHintsMy AnswersGive UpReview Part

Part C

Yet another reaction has an equilibrium constant Kc=4.32×105 at 25 ∘C. It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 ∘C , what will happen to the equilibrium constant?

The equilibrium constant will

Yet another reaction has an equilibrium constant  at 25 . It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200  , what will happen to the equilibrium constant?

increase.
decrease.
not change.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Chemical Equilibrium and Chemical Kinetics Part A For a certain reaction, Kc = 8.85×1010 and kf=...
Chemical Equilibrium and Chemical Kinetics Part A For a certain reaction, Kc = 8.85×1010 and kf= 7.52×10−2 M−2⋅s−1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction. Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M−2⋅s−1 include ⋅ (multiplication dot) between each measurement. Part B For a different reaction, Kc = 1.70×1010, kf=6.63×105s−1, and kr= 3.91×10−5 s−1...
For a general chemical equation A+B⇌C+D the equilibrium constant can be expressed as a ratio of...
For a general chemical equation A+B⇌C+D the equilibrium constant can be expressed as a ratio of the concentrations: Kc=[C][D]/[A][B] If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows: forward rate=kf[A][B] reverse rate=kr[C][D] where kf and kr are the forward and reverse rate constants, respectively. When equilibrium is reached, the forward and reverse rates are equal: kf[A][B]=kr[C][D] Thus, the rate constants are...
1. For a certain reaction, Kc = 4.70×10−2 and kf= 95.6 M−2⋅s−1 . Calculate the value...
1. For a certain reaction, Kc = 4.70×10−2 and kf= 95.6 M−2⋅s−1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction. Express your answer with the appropriate units. Include explicit multiplication within units, for example, to enter M−2⋅s−1 include ⋅ (multiplication dot) between each measurement 2. For a different reaction, Kc = 1.51×105, kf=8.12×103s−1 , and kr= 5.39×10−2 s−1 . Adding a catalyst increases the...
a. For a certain reaction, Kc= 1.53×107 and kf= 22.1 M?2?s?1 . Calculate the b. For...
a. For a certain reaction, Kc= 1.53×107 and kf= 22.1 M?2?s?1 . Calculate the b. For a different reaction, Kc=6.70×103, kf=4.58×103s?1, and kr= 0.684 s?1 . Adding a catalyst increases the forward rate constant to 1.04×106 s?1 . What is the new value of the reverse reaction constant, kr, after adding catalyst? Express your answer numerically in inverse seconds. c. Yet another reaction has an equilibrium constant Kc=4.32×105 at 25 ?C. It is an exothermic reaction, giving off quite a...
Learning Goal: To understand reaction order and rate constants. For the general equation aA+bB?cC+dD, the rate...
Learning Goal: To understand reaction order and rate constants. For the general equation aA+bB?cC+dD, the rate law is expressed as follows: rate=k[A]m[B]n where m and n indicate the order of the reaction with respect to each reactant and must be determined experimentally and k is the rate constant, which is specific to each reaction. Order For a particular reaction, aA+bB+cC?dD, the rate law was experimentally determined to be rate=k[A]0[B]1[C]2=k[B][C]2 This equation is zero order with respect to A. Therefore, changing...
The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D][A][B]=3.3 Part A Initially, only A...
The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D][A][B]=3.3 Part A Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached? Express your answer to two significant figures and include the appropriate units. [A] = SubmitHintsMy AnswersGive UpReview Part Part B What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ? Express...
For a different reaction, Kc = 1.92×103, kf=6.92×104s−1, and kr= 36.1 s−1 . Adding a catalyst...
For a different reaction, Kc = 1.92×103, kf=6.92×104s−1, and kr= 36.1 s−1 . Adding a catalyst increases the forward rate constant to 1.09×107 s−1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?
For a different reaction, Kc = 7.22×106, kf=4.13×105s−1, and kr= 5.72×10−2 s−1 . Adding a catalyst...
For a different reaction, Kc = 7.22×106, kf=4.13×105s−1, and kr= 5.72×10−2 s−1 . Adding a catalyst increases the forward rate constant to 7.35×107 s−1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?
For a certain reaction, Kc = 3.33×10−2 and k f = 1.30×10−2 M−2⋅s−1 M − 2...
For a certain reaction, Kc = 3.33×10−2 and k f = 1.30×10−2 M−2⋅s−1 M − 2 ⋅ s − 1 . Calculate the value of the reverse rate constant, kr , given that the reverse reaction is of the same molecularity as the forward reaction. For a different reaction, Kc = 2.66×104, kf=9.40×105s−1kf=9.40×105s−1 , and kr= 35.3 s−1s−1 . Adding a catalyst increases the forward rate constant to 2.25×108 s−1s−1 . What is the new value of the reverse reaction...
Nitric oxide emitted from the engines of supersonic aircraft can contribute to the destruction of stratospheric...
Nitric oxide emitted from the engines of supersonic aircraft can contribute to the destruction of stratospheric ozone: This reaction is highly exothermic (ΔE=−201kJ), and its equilibrium constant Kc is 3.4×1034 at 300 K. Part A Which rate constant is larger, kf or kr? kf kr Part B The value of kf at 300 K is 8.2×106 M−1s−1 . What is the value of kr at the same temperature? Express your answer to two significant figures and include the appropriate units....
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT