a. For a certain reaction, K c = 1.53×10 7 and k f= 22.1 M ?2?s?1 . Calculate th

Question

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?

Explanation / Answer

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.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.