c.)At a given temperature, the elementary reaction A<=> B in the forward directi

Question

c.)At a given temperature, the elementary reaction A<=> B in the forward direction is the first order in A with a rate constant of 3.40 × 10-2 s–1. The reverse reaction is first order in B and the rate constant is 5.80 × 10-2 s–1.What is the value of the equilibrium constant for the reaction A<=>B at this temperature. What is the value of the equilibrium constant for the reaction B<=> A at this temperature. d.) consider reaction mechanism: step 1: A<=> B+C equilibrium step 2: C+D-> E slow overall: A+D-> B+E and determine the rate law overall. (wrong answer would be =k[A][D]) Why?

Explanation / Answer

1) we know that

equilibrium constant is given by

Keq = Kf / Kr

given

Kf = 3.4 x 10-2

Kr = 5.8 x 10-2

so

Keq = 3.4 x 10-2 / 5.8 x 10-2

Keq = 0.5862

now

for the equilibrium

B ---->

K`eq = 1 / Keq

K`eq = 1 / 0.5862

K`eq = 1.706

so

the equilibrium constant for B --> A is 1.706

2)

the rate is determined by the slow step

so

consider the slow step

C + D ---> E

so

rate = k [C] [D]

now

conisder the equilibrium step

A ---> B + C

Keq = [B] [C] / [A]

[C] = Keq [A] / [B]

so

rate = k x keq x [A] [D] / [B]

let

k x keq = k1

so

rate = k1 [A] [D] / [B]

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

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