Part A: For a different reaction, K c = 9.93, k f=421s1, and k r= 42.4 s1 . Addi
ID: 492862 • Letter: P
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