1
answer
0
watching
129
views
12 Dec 2019

Part A For a certain reaction, Kc = 9.58×104 and kf= 0.564 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. kr =

Part B For a different reaction, Kc = 73.6, kf=513s−1, and kr= 6.97 s−1 . Adding a catalyst increases the forward rate constant to 3.85×104 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−1 include ⋠(multiplication dot) between each measurement. kr =

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.

For unlimited access to Homework Help, a Homework+ subscription is required.

Reid Wolff
Reid WolffLv2
13 Dec 2019

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
discord banner image
Join us on Discord
Chemistry Study Group
Join now

Related textbook solutions

Related questions

Weekly leaderboard

Start filling in the gaps now
Log in