Nitric oxide and nitrogen dioxide are found in photochemical smog. Nitrogen dioxide is formed from nitrogen monoxide in the exhaust of automobile engines. A possible mechanism for this reaction is given below. What is the rate law predicted by the mechanism?

Reaction:
2NO(g) + O2(g) → 2NO2(g)
Step 1 (fast): NO + NO → N2O2
Step 2 (slow): N2O2 + O2 → 2NO2

A. Rate = k[NO]2
B. Rate = k[NO2]2
C. Rate = k[NO][O2]
D. Rate = k[NO]2[O2]
E. Rate = k[No2]^2/[No]^2[o2]

(D) Rate = k[NO]²[O₂]

Explanation:

The rate law predicted by the mechanism

     Reaction:

                                                  2 NO(g) + O₂(g) →2 NO₂(g)

                                                 Step 1 (fast): NO + NO → N₂O₂

                                               Step 2 (slow): N₂O₂ + O₂ → 2 NO₂

According to rate law

                        rate of reaction is measured from slowest step

                                    ------------------  (i)

Since N₂O₂ is an intermediate

   So we replace intermediate from rate law equation

                                         From step 1 we can write

From equation (i)      

                        ∴   Rate = k[NO]²[O₂]

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

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