Can you get this USNCO question?

KineticsNational 2020

A reaction $\ce{A <=> B}$ is reversible, with $K_\text{eq} = 1.00$ . The forward reaction is first-order in A. In a reaction initially containing only A at an initial concentration of $[A]_0$ , which plot will be linear over the course of the reaction?

26% of national qualifiers got this right

Explanation

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For $\ce{A <=> B}$ with $K_{eq}=1.00$ and only A initially at $[A]_0$ , mass balance gives $[B]=[A]_0-[A]$ .

The net rate law (forward and reverse both first-order, since $K_{eq}=k_f/k_r$ ) is:

$-\dfrac{d[A]}{dt}=k_f[A]-k_r[B]=k_f[A]-k_r([A]_0-[A])$

Since $K_{eq}=k_f/k_r=1$ , we have $k_f=k_r=k$ , so

$-\dfrac{d[A]}{dt}=k[A]-k([A]_0-[A])=k(2[A]-[A]_0)=2k\left([A]-\tfrac12[A]_0\right)$

Let $x=[A]-\tfrac12[A]_0$ (the displacement from the equilibrium concentration, since at equilibrium $[A]_{eq}=[B]_{eq}=\tfrac12[A]_0$ ). Then

$\dfrac{dx}{dt}=-2kx$

This is a standard first-order decay in $x$ , so integrating gives

$\ln\!\left([A]-\tfrac12[A]_0\right)=\ln\!\left([A]_0-\tfrac12[A]_0\right)-2kt$

which is linear in time with slope $-2k$ . Thus the correct linear plot is $\ln([A]-0.5[A]_0)$ vs. time — option (B).

This would be correct only for an irreversible first-order reaction ( $\ce{A->B}$ ) where $[A]\to0$ ; it ignores that the reverse reaction stops decay at $[A]_{eq}=0.5[A]_0$ , not at zero.

Correct: accounting for the reverse reaction shows the true decaying quantity is the displacement from equilibrium, $[A]-0.5[A]_0$ , which decays exponentially, making its natural log linear in time.

This is the integrated rate law form for a second-order irreversible reaction; it wrongly assumes the reaction is second-order in A rather than reversible first-order.

This mixes the second-order linear form with the equilibrium shift correction, but the underlying kinetics are first-order (linear in $x$ via its log, not via $1/x$ ), so this combination is not linear in time.

Concept · Reversible first-order kinetics approaching equilibrium