Question

In: Chemistry

Part A: For a different reaction, Kc = 9.93, kf=421s−1, and kr= 42.4 s−1 . Adding...

Part A:

For a different reaction, Kc = 9.93, kf=421s−1, and kr= 42.4 s−1 . Adding a catalyst increases the forward rate constant to 1.35×10⁵ s−1 . 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

Expert Solution

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

queen_honey_blossom answered 2 years ago

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