Part A For a certain reaction, K c = 8.24×10 2 and k f= 5.47×10 5 M 2s1 . Calcul

Question

Part A

For a certain reaction, Kc = 8.24×102 and kf= 5.47×105M2s1 . 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.

Part B

For a different reaction, Kc = 1.46×104, kf=6.08×103s1, and kr= 0.416 s1 . Adding a catalyst increases the forward rate constant to 2.51×106 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.

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

A. increase, B. decrease, C. not change

Explanation / Answer

A) In a reversible reaction   Kc = kf / kr

8.24 x 10^ -2 = 5.47 x 10^ -5 / kb

kr ( rate of reverse reaction)   = 6.64 x 10^-4 M-2.s-1

B) Kc will not change if catalyst is added ( kr and kf will change but not kc)

hence Kc = kf/kr

        1.46 x 10^4 = 2.51 x 10^6 / kr

kr = 5.82 x 10^ -3 s-1

C) For exothermic reactions high temperature is unfavorable , hence Kc decreases

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects