A gaseous reaction X _2+ Y _2 ; ;2 , XY has an experimentally determined rate law…

CHEMISTRY · Multi-part · A Level · Chemistry · Rate equations · Rate equations

Problem

A gaseous reaction X2+Y2  ⟶  2 XY\text{X}_2+\text{Y}_2\;\longrightarrow\;2\,\text{XY} has an experimentally determined rate law rate=k[X2]1[Y2]0\text{rate}=k[\text{X}_2]^1[\text{Y}_2]^0 at 400 K. Two steps have been suggested for the mechanism. Step 1 X2_2 ⟶\longrightarrow 2 X (slow) Step 2 X + Y2_2 ⟶\longrightarrow XY + Y (fast) Step 1 is unimolecular, whereas the overall balanced equation shows bimolecular reactants. A student claims that because the overall reaction involves both X2_2 and Y2_2, the rate law must depend on both concentrations and therefore the mechanism is invalid. Address the student’s claim. (a) Describe how the experimental rate law is consistent with the two-step mechanism, despite the absence of Y2_2 in the slow step’s molecularity. (b) Explain why the concentration of Y2_2 does not appear in the rate law even though it is consumed overall. (c) Describe one additional piece of kinetic or spectroscopic evidence that could support the proposed mechanism over an alternative mechanism in which Y2_2 appears in the rate-limiting step.

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