valence
Practice
KineticsNational 2018

An irreversible reaction A + B → products is studied under conditions where the initial concentrations of A and B are equal. Under these circumstances, a graph of ln⁡([A])\ln([A]) as a function of time is linear. What is the order in A?

7% of national qualifiers got this right

Explanation

draft

Since [A]0=[B]0[A]_0 = [B]_0 and the stoichiometry is 1:1, we have [A]=[B][A] = [B] at all times during the reaction. This means the rate law

rate=k[A]m[B]n\text{rate} = k[A]^m[B]^n

can be rewritten as

rate=k[A]m+n\text{rate} = k[A]^{m+n}

because [B]n=[A]n[B]^n = [A]^n at every instant. A plot of ln⁡[A]\ln[A] vs. time being linear tells us only that the overall kinetics observed for A behaves like a first-order decay, i.e., m+n=1m+n = 1 overall (pseudo-first-order behavior). But this does not tell us the individual value of mm (the order in A alone).

For example, m=1,n=0m=1, n=0 (first order in A, zero in B) gives rate=k[A]\text{rate}=k[A], linear ln⁡[A]\ln[A] vs t. But also m=0.5,n=0.5m=0.5, n=0.5 gives rate =k[A]0.5[B]0.5=k[A]=k[A]^{0.5}[B]^{0.5}=k[A], since [A]=[B][A]=[B], also producing a linear ln⁡[A]\ln[A] plot. Many combinations of mm and nn summing to 1 are indistinguishable under the condition [A]0=[B]0[A]_0=[B]_0.

Therefore, from this experiment alone, the individual order in A cannot be determined — only the sum m+nm+n can be inferred.

A

Incorrectly assumes that because the plot is linear in ln⁡[A]\ln[A], the rate is somehow independent of concentration; but zero order would give a linear plot of [A][A] itself vs t, not ln⁡[A]\ln[A], so this is inconsistent even superficially.

B

Assumes that a linear ln⁡[A]\ln[A] vs t plot always implies first order in A alone, forgetting that because [A]=[B][A]=[B] throughout, the coupling of A and B means the overall exponent (m+n) is what's actually being measured, not just m.

C

Confuses second order behavior (which requires a linear 1/[A]1/[A] vs t plot, not ln⁡[A]\ln[A]) with the given linear ln⁡[A]\ln[A] data, misapplying the integrated rate law form.

D

Correct: because [A]=[B][A]=[B] at all times, the data only reveal the sum of orders m+n=1m+n=1; the individual order in A cannot be isolated without varying the concentrations independently.

Concept · Rate law order determination with coupled initial concentrations

USNCO 2018 National Part I exam, question 28. © American Chemical Society; reproduced for non-commercial practice.