Part A: For a different reaction, K c = 9.93, k f=421s1, and k r= 42.4 s1 . Addi

Question

Part A:

For a different reaction, Kc = 9.93, kf=421s1, and kr= 42.4 s1 . Adding a catalyst increases the forward rate constant to 1.35×105 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

Part B:

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?

answers: equilibrium constant will:

a. increase

b. decrease

c. not change

Explanation / Answer

part A

un catalyst                                    catalyst

Kf   = 421sce^-1                           Kf1 = 1.35*10^5 sec^-1

Kr   = 42.4sec^-1                          Kr1   =

Kf/Kr   = Kf1/Kr1

421/42.4   = 1.35*10^5/kr1

Kr1          = 1.35*10^5*42.4/421

                 = 1.35*10^4 sec^-1

The new value of the reverse reaction constant, kr, after adding catalyst is = 1.35*10^4 sec^-1

Part B

Every 10C^0 the equilibrium constant increases two or three times.

The equilibrium constant is directly propotional to temperature.

   The temperature is increases equilibrium constant is increases.

a. increase

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