Part A: For a certain reaction, Kc= 9.33×108 and kf= 4.89×105 M2s1 . Calculate t

Question

Part A: For a certain reaction, Kc= 9.33×108 and kf= 4.89×105 M2s1 . Calculate the value of the reverse rate constant, kr. Express your answer numerically in inverse seconds.

Part B: For a different reaction, Kc=44.8, kf=314s1, and kr= 7.01 s1 . Adding a catalyst increases the forward rate constant to 1.08×105 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst? Express your answer numerically in inverse seconds.

Part C: t 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?

Explanation / Answer

Part-A;

kc = kf /kr

9.33 x10-8 = 4.89×105 M2s1 / kr

kr = 4.89×105 M2s1 / 9.33 x10-8

= 0.524 x 10-13 M2s1

PART-B:

Kc =44.8

kf = 314 s-1

Kc value doesnot change with adding catalyst.

Kc =Kf /Kr

44.8 = 1.08 x 105 s-1 / Kr

Kr = 1.08 x 105 s-1 / 44.8

= 2.41 x 103 s-1

Part-C:

For exothermic reaction with increase in temperature, the equilibrium constant value decreases.

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