Derive a rate law equation for 1) a first order reaction close to equilibrium A

Question

Derive a rate law equation for 1) a first order reaction close to equilibrium A --> 2B d[A]/dt = ? [A] eq = ?
2) a second order reaction close to equilibrium A --> B d[A]/dt = ? [A] eq = ?
Derive a rate law equation for 1) a first order reaction close to equilibrium A --> 2B d[A]/dt = ? [A] eq = ?
2) a second order reaction close to equilibrium A --> B d[A]/dt = ? [A] eq = ?
1) a first order reaction close to equilibrium A --> 2B d[A]/dt = ? [A] eq = ? A --> 2B d[A]/dt = ? [A] eq = ?
2) a second order reaction close to equilibrium A --> B d[A]/dt = ? [A] eq = ? d[A]/dt = ? [A] eq = ? d[A]/dt = ? [A] eq = ?

Explanation / Answer

1) d[A]/dt = -Kf*[A] + Kb[B]

dx/dt = -Kf*x + Kb*([A]o-x), x is the concentration A at t, and hence concentration of B is Ao - x.

Rearranging it yields,

dx/dt = Kb[A]o - (Kf + Kb)x (rate law)

at equlibrium, K = Kf/Kb = [A]o -x(eq)/xeq,

rearranging and simplifying,

dx/dt = Kb[A]o - Kb([A]o/x(eq))x

dx/dt = Kb[A]o/x(eq){x(eq) - x}

integrating this gives expression for [A] (eq)

this applies for both the questions. Just that expression for x changes as K = Kf/Kb = {(Ao -x)/x}^2

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