An example:
Rate law: rate = k[NO2]2
Step 1: NO2(g) + NO2 (g) NO3(g) + NO(g)
Step 2: NO3(g) + CO(g) NO2(g) + CO2(g)
Step 1 is slow and rate-determining; Step 2 is fast.
NO3 is an intermediate!
Rate1 = k1[NO2]2
Rate2 = k2[NO3][CO]
Thus, if k1 = k, the rate law for Step 1 (rate-determining) is the
same as the experimental rate law!
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