Kc=kf/kr=[C][D]/[A][B] Part A For a certain reaction, Kc = 6.56×1010 and kf= 7.2

Question

Kc=kf/kr=[C][D]/[A][B]

Part A For a certain reaction, Kc = 6.56×1010 and kf= 7.29×105 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.

Part B For a different reaction, Kc = 1.99×104, kf=6.76×105s1, and kr= 33.9 s1 . Adding a catalyst increases the forward rate constant to 1.49×108 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

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

A) Given Kc = [C][D]/[A][B] and the molecularity of the forward and reverse reactions are equal, we can approximate

Kc = kf/kr

Plug in values and obtain

6.56*1010 = (7.29*105 M-2.s-1)/kr

====> kr = (7.29*105 M-2.s-1)/(6.56*1010) = 1.11*10-5 M-2.s-1 (ans).

B) We shall assume the molecularities are same and approximate the equilibrium constant as

Kc = kf/kr

Kc is an equilibrium constant and remains constant at a particular temperature; hence, Kc must stay constant even when kf = 1.49*108 s-1. Plug in values and obtain

1.99*104 = (1.49*108 s-1)/kr

=====> kr = (1.49*108 s-1)/(1.99*104) = 7.49*103 s-1 (ans).

C) The equilibrium constant for the reaction is Kc = 4.32*105 at 25°C. The reaction is exothermic, i.e, releases heat. Hypothetically, we can assume heat to be one of the products of the reaction. Increasing the temperature increases the heat in the system and we can hypothesize that one of the products in the reaction has increased. Therefore, the equilibrium favors the reverse reaction to counteract the effect of increased temperature.

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