The generic reaction 2AB has the following rate laws: forward reaction:reverse r

Question

The generic reaction 2AB has the following rate laws: forward reaction:reverse reaction:rate=kf[A]2rate=kr[B] where kf is the rate constant for the forward reaction and kr is the rate constant for the reverse reaction. At equilibrium, the two rates are equal and so kf[A]2=kr[B]. The equilibrium constant for a reaction is related by the law of mass action to the rate constants for the forward and reverse reactions: Kc=[B][A]2=kfkr

PART B

At this temperature the rate constant for the reverse reaction is 360. M2s1 . What is kf for the reaction?

Express your answer to three significant figures and include the appropriate units. Include an asterisk to indicate mulitplication in compound units, for example to write the units for a second order rate constant type M-1*s-1.

Explanation / Answer

For the above shown reaction,

2A <==> B

the equilibrium constant is related to the rate constants for the forward and the reverse reaction by the relation as given above,

Kc = Kf.Kr

with,

Kc = 19.8

Kr = 360 M-2.s-1

we get,

rate constant for the forward reaction Kf = Kc/Kr

                                                                = 19.8/360 = 0.055

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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