Question

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 constant, kr

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
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...
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?
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...
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...
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 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...
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...
The rate constant for a certain reaction is k = 1.30×10−3 s−1 . If the initial...
The rate constant for a certain reaction is k = 1.30×10−3 s−1 . If the initial reactant concentration was 0.900 M, what will the concentration be after 10.0 minutes?
Consider the gas-phase hydration of hexafluoroacetone, (CF3)2CO: (CF3)2CO(g)+H2O(g)⇌krkf(CF3)2C(OH)2(g) At 76 ∘C, the forward and reverse rate...
Consider the gas-phase hydration of hexafluoroacetone, (CF3)2CO: (CF3)2CO(g)+H2O(g)⇌krkf(CF3)2C(OH)2(g) At 76 ∘C, the forward and reverse rate constants are kf=0.13M−1s−1 and kr=6.2×10−4s−1. Part A What is the value of the equilibrium constant Kc?
CO (g) + Cl2 (g) <--> COCl (g) + Cl (g) Rate constant for the forward...
CO (g) + Cl2 (g) <--> COCl (g) + Cl (g) Rate constant for the forward reaction (kf) = 1.38 x 10^-28 M/sec. Rate constant for the reverse reaction (kr) = 9.3 x 10^10 M/sec Reaction was run at 25 degrees celsius 1) Calc equilibirum constant's value (kc) 2) What would be the kc value for 2Cl(g) + 2COCl (g) --> <-- 2Cl2 (g) + 2CO (g) 3) In which direction is this reaction favored?