For a certain reaction. Kc 6.25x 108 and kr 8.33 x 105 M 2-S-I . Calculate the v

Question

For a certain reaction. Kc 6.25x 108 and kr 8.33 x 105 M 2-S-I . Calculate the value of the reverse rate constant, that the reverse reaction is of the same molecularity as the forward reaction. given Learning Goal Express your answer with the appropriate units. To understand the relationship between the equilibrium View Available Hintis) constant and rate constants. For a general chemical equation the equilibrium constant can be expressed as a ratio of A+B C+D the concentrations: kValue AJB If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows: Submit ard rate kr A B reverse ratekD Part B where kr and k are the forward and reverse rate constants, respectively. When equilibrium is reached, the forward and reverse rates are equal: For a different reaction. Kc 1.40 x 103 . kf 7.98 x 103s 1 , and kr 5.68s 1. Adding a catalyst increases the forward rate constant to 2.59x10 s 1.What is the new value of the reverse reaction constant, k,, after adding catalyst? Express your answer with the appropriate units Thus, the rate constants are related to the equilibrium constant in the following manner: View Available Hint(s) CAB kValue Units Submit Part C Yet another reaction has an equilibrium constant Kc 4.32 x 105 at 25 °C. lt 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? View Available Hint(s) increase The equilibrium constant will O decrease not change Submit

Explanation / Answer

Part A

Kc = Kf/Kr

Kr = kf/kc

substituting Kc and Kf value

kr = 8.33×105M-2 s-1/6.25×108

      = 1.33×10-3 M-2 s-1

Part B

By adding equillibrium constantant will not change

kr= kf/Kc

after addition of catalyst, kr = 2.59×106s-1

   kr = 2.59×106s-1/1.40×103

   = 1.85×103s-1

Part C

The equillibrium constant will decrease

Exothermic reaction will not favour farward reaction and exothermic reaction will favour reverse reaction as the reverse reaction as the reverse is endothermic reaction.So, rate constant for forward direction will decrease and rate constant for reverse direction will increase, thus Kc will decrease at 200°C.

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