a. For a certain reaction, Kc= 1.53×107 and kf= 22.1 M?2?s?1 . Calculate the
b. For a different reaction, Kc=6.70×103, kf=4.58×103s?1, and kr= 0.684 s?1 . Adding a catalyst increases the forward rate constant to 1.04×106 s?1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?
Express your answer numerically in inverse seconds.
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?
Part a
Equilibrium constant
Kc= 1.53×10^7 = kf/kr = forward constant / reverse constant
Forward rate constant
kf= 22.1 M-2 s-1
Reverse rate constant
kr = kf/Kc
= 22.1/1.53×10^7
= 1.44 x 10^6 M-2 s-1
Part b
Kc=6.70×10^3
kf=4.58×10^3 s-1
kr= 0.684 s-1
After adding catalyst
kf = 1.04×10^6 s-1
New value of reverse constant
kr = kf/Kc
= (1.04×10^6) / (6.70×10^3)
= 155.22 s-1
Part c
Kc=4.32×10^5 at T1 = 25+273 = 298 K
T2 = 200 + 273 = 473 K
From Arrhenius equation
Kc = A exp (-Ea/RT)
For exothermic reaction, increase in the temperature equilibrium rate constant decreases.
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