Suppose the decomposition of dinitrogen monoxide proceeds by the following mechanism:
step. elementary reaction. rate constant
1. n2o(g)=n2(g)+o(g). k1
2. n2o(g)+o(g)=n2(g)+o2(g). k2
suppose also k1<
write the balance chemical question for the overall chemical reaction:

write the experimental observable rate law for the overall chemical reaction. know your answer should not contain the concentrations of any intermediates.

express the rate constant k for the overall chemical reaction in terms of k1, k2 and if necessary the reconnaissance k -1 and k -2 for the reverse of the two elementary reactions in the mechanism.

Part A: 2N₂O(g) ⇄ 2N₂(g) + O₂(g)

Part B: -r = K*[N₂O]²

Part C: K= k1*k2

Explanation:

Part A

To do the balance chemical question for the overall chemical reaction, we must sum the reaction of the steps, eliminating the intermediaries, which are the compounds that have the same amount both at reactants and products (bolded).

N₂O(g) ⇄ N₂(g) + O(g)

N₂O(g) + O(g) ⇄ N₂(g) + O₂(g)

2N₂O(g) + O(g) ⇄ 2N₂(g) + O(g) + O₂(g)

2N₂O(g) ⇄ 2N₂(g) + O₂(g)

Part B

The velocity of the reaction (r) can be calculated based on the reactants or based on the products. Let's do it based on the disappearing of the reactant. Because it is disappearing, the variation at its concentration must be negative, so the rate will be negative.

Let's suppose its an elementary reaction, so, the concentration of the reactant must be elevated by its coefficient. And let's call the overall rate constant as K:

-r = K*[N₂O]²

Part C

Because the steps were summed, and the reactions were not multiplied by a constant or inverted, the constant K is just the multiplication of the constants of the steps:

K= k1*k2

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.