Part A For a certain reaction, Kc = 9.58×104 and kf= 0.564 M2s1 . Calculate the

Question

Part A For a certain reaction, Kc = 9.58×104 and kf= 0.564 M2s1 . 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 M2s1 include (multiplication dot) between each measurement. kr =

Part B For a different reaction, Kc = 73.6, kf=513s1, and kr= 6.97 s1 . Adding a catalyst increases the forward rate constant to 3.85×104 s1 . 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 M2s1 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.

Explanation / Answer

the rate constant for the forward reaction indicates the reaction to be third order

hence 3A<---> 3B

Kf= forward reaction rate constant and Kr= backward reaction rate constant

Kf[A]3= Kr[B]3 at equilibrium

KC= [B]3/[A]3 = Kf/Kr = 9.58*104, Kf=0.564/M2.s

Kr= Kf/KC= 0.564/(9.58*104)= 5.88*10-6 /M2.s

2. The addition of catalyst does not affect the rate constant in the forward direction

Kc = 73.6, kf=513s1, and kr= 6.97 s1 . Adding a catalyst increases the forward rate constant to 3.85×104 s1

Kc= 73.6 = 3.85*104/Kr ( upon addition of catalyst)

Kr= 3.85*104/ 73.6=523.1/s

3. since d(lnK)= -deltaH/RT2

since deltaH is -ve for exothermic reaction

ln(K2/K1)= deltaH/R*(1/T1-1/T2)

since deltaH is -ve, the value of K2/K1 will be less than 1.

K2 decreases. So equilibrium constant decreaes for exothermic reaction with an increase in temperature

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