Question

QUESTION 9

A reaction is studied and found to proceed in the following two elementary steps:

A + B → C + Q

Q + A → D + B

What is the catalyst for the reaction?

		Q
		D
		C
		B
		A

QUESTION 10

2A → B + C

The above reaction is studied and found to proceed in the following two elementary steps:

A → Q

Q + A → B + C

If the second step is the rate determining step (slow step), what is the rate law for the overall reaction?

rate = k

rate = k[A]/[B][Q]

rate = k[A]²

rate = k[A]

QUESTION 11

2A → B + C

Two trials of the above reaction are run. The concentration of A in the second trial is three times that in the first. It is found that the initial rate of the reaction in the second trial is nine times the initial rate in the first trial. This indicates that the reaction is ...

		zero order in [A]
		first order in [A]
		second order in [A]
		third order in [A]

Answer 1

Q9 solution

B is the catalyst in the reaction because the species which remain unchanged at the end of reaction is called catalyst.

Since B is recovered at the product side in the second reaction therefore B is the catalyst in the reaction.

Q10 Solution

For the given elementary reactions

A --- > Q

Q + A ----- > B+C     slow step

Overall reaction is

2A ----- > B+ C

Q is the intermediate therefore cancelled out

Therefore rate law for the reaction is

Rate = k[A]²

Q11 Solution

Since the rate of reaction increased by factor 9 after tripleing the reactant concentration that means reaction is second order with A

9 = 3^m

log 9 = m * log 3

log 9 / log 3 = m

2 = m

Hence order with respect to A is second order