U5.9 · Steady-State Approximation (Pre-Equilibrium)Practice · 稳态近似（预平衡）

The rate-determining step is step 2: rate = $k_2$[$\ce{NOBr2}$][NO]. $\ce{NOBr2}$ is an intermediate, so use the fast equilibrium in step 1: $K_1 = \dfrac{[\ce{NOBr2}]}{[\ce{NO}][\ce{Br2}]}$, so $[\ce{NOBr2}] = K_1[\ce{NO}][\ce{Br2}]$. Substituting gives rate = $k_2K_1$[NO]²[$\ce{Br2}$].

Rate-determining step: rate = $k_2$[X][$\ce{Y2}$]. X is an intermediate; the fast equilibrium $K_1 = \dfrac{[\ce{X}]^2}{[\ce{X2}]}$ gives [X] = $K_1^{1/2}[\ce{X2}]^{1/2}$. Substituting: rate = $k[\ce{X2}]^{1/2}[\ce{Y2}]$, which is order 1/2 in $\ce{X2}$.

Derivation: rate = $k_2$[Z][X]; Z is an intermediate with $[\ce{Z}] = K_1\dfrac{[\ce{Y}]}{[\ce{W}]}$, so rate = $k\dfrac{[\ce{Y}][\ce{X}]}{[\ce{W}]}$, with W in the denominator (order −1). Adding W pushes the first equilibrium back to the left, leaving less Z, so the slow step slows down.

$k_{-1} = \dfrac{k_2 k_1}{k}$ = 80 × $1.5 \times 10^{3}$ ÷ 2.4 = $5.0 \times 10^{4}$. So $K_1$ = $k_1/k_{-1}$ = 0.030: the equilibrium lies toward the reactants and the intermediate stays scarce.

In the fast equilibrium, the forward and reverse rates are large and equal; the slow step only occasionally drains a little intermediate and barely disturbs it, so $K_1 = k_1/k_{-1}$ can be used to express the intermediate. The overall reaction is not at equilibrium; product keeps forming.