U5.8 · Reaction Mechanism and Rate LawPractice · 反应机理与速率定律

The slowest step sets the rate. Step 1 is slow and elementary, so its rate is k[$\ce{H2}$][ICl]; it contains no intermediate, so it is the overall rate law. Using the overall coefficients, k[$\ce{H2}$][ICl]², is wrong, and HI is an intermediate, so it can’t appear.

All four add up to the overall reaction, so compare rate laws. From each rate-determining step: A rate = $k[\ce{A}][\ce{B}]$; B rate = $k[\ce{A}]^{2}[\ce{B}]$; C rate = $k[\ce{A}]^{2}$; D rate = $k[\ce{B}]$. In B, the slow step contains the intermediate $\ce{A2}$; substituting $[\ce{A2}] = K[\ce{A}]^2$ from the fast equilibrium gives rate = k[A]²[B], which matches.

The rate law is that of the slow step, and it contains only X, second order, so the slow step must be two X colliding: $\ce{2X -> X2}$. The fast step 2, $\ce{X2 + Y -> X2Y}$, then completes the overall reaction. Y enters only after the rate-determining step, so it isn’t in the rate law; a single step ($\ce{2X + Y -> X2Y}$) would put [Y] in it.

An acceptable mechanism must add up to the overall reaction and predict the experimental rate law. This one does both, so it is plausible; but other mechanisms might too, so we can only say it is consistent with the data, not proven. The mismatch with the overall coefficients just shows the reaction isn’t a single step.

The steps do add up to the overall reaction, but that is only one requirement. The slow step is step 1, $\ce{2NO -> N2O2}$, which predicts rate = k[NO]²; $\ce{O2}$ enters after the rate-determining step and drops out. That disagrees with the experimental k[NO]²[$\ce{O2}$], so the mechanism is not acceptable.