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 as a function of time is linear. What is the order in A?
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Explanation
draftSince and the stoichiometry is 1:1, we have at all times during the reaction. This means the rate law
can be rewritten as
because at every instant. A plot of vs. time being linear tells us only that the overall kinetics observed for A behaves like a first-order decay, i.e., overall (pseudo-first-order behavior). But this does not tell us the individual value of (the order in A alone).
For example, (first order in A, zero in B) gives , linear vs t. But also gives rate , since , also producing a linear plot. Many combinations of and summing to 1 are indistinguishable under the condition .
Therefore, from this experiment alone, the individual order in A cannot be determined — only the sum can be inferred.
Incorrectly assumes that because the plot is linear in , the rate is somehow independent of concentration; but zero order would give a linear plot of itself vs t, not , so this is inconsistent even superficially.
Assumes that a linear vs t plot always implies first order in A alone, forgetting that because throughout, the coupling of A and B means the overall exponent (m+n) is what's actually being measured, not just m.
Confuses second order behavior (which requires a linear vs t plot, not ) with the given linear data, misapplying the integrated rate law form.
Correct: because at all times, the data only reveal the sum of orders ; the individual order in A cannot be isolated without varying the concentrations independently.
USNCO 2018 National Part I exam, question 28. © American Chemical Society; reproduced for non-commercial practice.