Q2(c). Opposing first-order reactions A ⇌ B with k₁/k₂=2

Integrated rate law

Take [A]=[A]₀ and [B]=0 at t=0; then at every instant [A]+[B]=[A]₀.

1. Both steps are elementary and first order: -(d[A])/(dt)=k₁[A]-k₂[B]=k₁[A]-k₂([A]₀-[A])=(k₁+k₂)[A]-k₂[A]₀
2. At equilibrium the net rate is zero: k₁[A]_(eq)=k₂[B]_(eq), so [A]_(eq)=(k₂[A]₀)/(k₁+k₂) and the rate law becomes -(d[A])/(dt)=(k₁+k₂)([A]-[A]_(eq))
3. Integrating: ln[A]₀-[A]_(eq)[A]-[A]_(eq)=(k₁+k₂) t, that is [A]=[A]₀ k₂+k₁e^(-(k₁+k₂)t)k₁+k₂, [B]=[A]₀ k₁(1-e^(-(k₁+k₂)t))k₁+k₂

The plots for k₁/k₂=2

1. Equilibrium values: K=[B]_(eq)[A]_(eq)=(k₁)/(k₂)=2, so [A]_(eq)=13[A]₀, [B]_(eq)=23[A]₀
2. Shape: [A] decays exponentially, not to zero but to [A]₀/3; [B] rises as its mirror image to 2[A]₀/3. The sum stays [A]₀ at every time.
3. Start: the initial slopes are -k₁[A]₀ and +k₁[A]₀; at first the back reaction is absent and the reaction looks like a simple first-order one.
4. Crossing point: [A]=[B]=[A]₀/2 when e^(-(k₁+k₂)t)=14, that is at (k₁+k₂)t=4, or t=0.462/k₂. A crossing occurs only because K>1; for K<1 the curves would never meet.
5. Speed of approach: the distance from equilibrium decays with the sum k₁+k₂ (not k₁). It halves every 2/(k₁+k₂)=0.231/k₂; after (k₁+k₂)t=4 the curves are within 2 % of their final values.
6. Explanation: as A is used up the forward rate k₁[A] falls and the backward rate k₂[B] rises; they become equal at equilibrium, a dynamic equilibrium, and the concentrations stop changing although both reactions continue.

Test of the mechanism: a plot of ln([A]-[A]_(eq)) against t is a straight line of slope -(k₁+k₂); with K=k₁/k₂ from the end concentrations the two rate constants are found separately: here k₁=23(k₁+k₂) and k₂=13(k₁+k₂).

Application. interconversions such as cis–trans isomerisation and the mutarotation of glucose (α⇌β, followed by optical rotation) obey this rate law; relaxation methods (temperature jump) for fast reactions use the same result, 1/τ=k₁+k₂.