Find the rate law for the reaction and describe the mathematical relationship between the overall rate constant, k, and the rate constants for the elementary steps.

Question

The two-step mechanism below has been proposed for the reaction 2 NO(g) + O2(g) → 2 NO2(g)

which is believed to be responsible for the conversion of NO in automobile exhaust to NO2.

 

2 NO(g)   N2O2(g)   (fast equilibrium)

 

N2O2(g) + O2(g) → 2 NO2 (slow)

 

Find the rate law for the reaction and describe the mathematical relationship between the overall rate constant, k, and the rate constants for the elementary steps.

Answer 1

The two step proposed mechanisms has N₂O₂ as an intermediate. The rate of the slowest step should be the rate for the overall reaction, and the overall rate would be rate = k[N₂O₂][O₂]. But since N2O2 is an intermediate and does NOT appear in the final reaction, i.e. 2NO + O2 ==> 2NO2, we can replace it as follows:

2NO <==> N₂O₂ and at equilibrium k1[NO]² = k-1[N₂O₂] and [N₂O₂] = (k1/k-1)[NO]²

Now plugging this into the rate law, we have... rate = k(k1/k-1)[NO]²[O₂]

Combining all rate constants into one constant, we now have the final rate law...

rate = k'[NO]²[O₂] and this would be the rate law for the reaction

The relationship between the overall rate constant (k'0 and the rate constants for the elementary steps is that the overall rate constant k' = k1/k-1 x k or in other words it is a combination of the forward and reverse reactions in step one and the forward reaction in step 2