1. For a certain reaction, K c = 4.47×10 7 and k f= 2.75×10 5 M 2s1 . Calculate
ID: 890741 • Letter: 1
Question
1. For a certain reaction, Kc = 4.47×107 and kf= 2.75×105M2s1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction.
2. For a different reaction, Kc = 5.89×108, kf=4.88×105s1, and kr= 8.28×104 s1 . Adding a catalyst increases the forward rate constant to 3.51×107 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?
3.
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?
increase. decrease. not change.Explanation / Answer
(1) Kc = 4.47×107 and kf= 2.75×105 M2s1
Kc = Kf / Kb
Kr = Kf / Kc = 2.75 x 10^5 / 4.47 x 10^7
Kr = 6.15 x 10^-3
(2)
Kc = 5.89×108, kf= 3.51×107 s1
Kr = Kf / Kc =3.51×107 /5.89×108 = 0.0596
Kr = 0.0596
(note : if Kf increases Kr decreases vice versa . but Kc constant . )
(3) decreases
if the reaction is exothermic reaction equilibrium favours at low temperatures . so Kc value decreases
if the reaction is endothermic reaction equilibrium favours at high temperatures .
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