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 =
1.70×10⁻² and kf= 80.0 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 =
1.29×10¹⁰, kf=6.76×105s−1, and kr=
5.25×10⁻⁵ s−1 . Adding a catalyst increases the forward
rate constant to 7.91×10⁷ 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.

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?