Part A For a certain reaction, K c = 8.05×10 9 and k f= 0.777 M 2s1 . Calculate

Question

Part A

For a certain reaction, Kc = 8.05×109 and kf= 0.777 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.

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Part B

For a different reaction, Kc = 31.1, kf=2.59×103s1, and kr= 83.3 s1 . Adding a catalyst increases the forward rate constant to 1.06×105 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.

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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?

Explanation / Answer

Part A: (A) K(c) = k(f) / k(r), so k(r) = K(f) / K(c)

(B) K(c) is unchanged by adding a catalyst. So you just use your new K(f) to get the new K(r) same as in (A).

Part C: It will decrease

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