For The reaction 2NO2->4NO2 + O2 the experimental rate law is r=k[N2O5].A possib

Question

For The reaction 2NO2->4NO2 + O2 the experimental rate law is r=k[N2O5].A possible mechanism for the reaction is: Step 1: N2O5<->NO2 + NO3 fast step and equilibrium Step 2: NO2+NO3-> NO +NO2 +O2 slow step Step 3 NO3+NO-> 2NO2 Fast Step Demonstrate of this mechanism is consistent with the Experimental Rate Law and What is the molecular its of the slow step For The reaction 2NO2->4NO2 + O2 the experimental rate law is r=k[N2O5].A possible mechanism for the reaction is: Step 1: N2O5<->NO2 + NO3 fast step and equilibrium Step 2: NO2+NO3-> NO +NO2 +O2 slow step Step 3 NO3+NO-> 2NO2 Fast Step Demonstrate of this mechanism is consistent with the Experimental Rate Law and What is the molecular its of the slow step For The reaction 2NO2->4NO2 + O2 the experimental rate law is r=k[N2O5].A possible mechanism for the reaction is: Step 1: N2O5<->NO2 + NO3 fast step and equilibrium Step 2: NO2+NO3-> NO +NO2 +O2 slow step Step 3 NO3+NO-> 2NO2 Fast Step Demonstrate of this mechanism is consistent with the Experimental Rate Law and What is the molecular its of the slow step

Explanation / Answer

Slow step is rate determining step

Rate=k[NO2][NO3]

From fast steps we can infer that

r=k[N2O5]

Molecularity of slow step=2

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