Given a reaction between an organic molecule, denoted as A, and NaSH, we observe

Question

Given a reaction between an organic molecule, denoted as A, and NaSH, we observe the following observations. Using the observations, write a rate law for the reaction.

(a) The rate triples when the concentration of [A] is tripled and the concentration of [NaSH] is held constant.
(b) The rate is decreased when the concentration of [A] is doubled and the concentration of [NaSH] is cut by a factor of 3.
(c) The rate doubles when the concentration of [A] is cut in half and the concentration of [NaSH] is quadrupled.
(d) The rate increases with an increase in temperature.

Rate=

Explanation / Answer

let the rate law be

rate = k [A]^a [NaSH]^b

now

given [NaSH] is held constant ,

rate tripled when [A] is tripled

so

r2 = 3r1 and A2 =3A1

now

r2/r1 = ( A2/A1)^a

3 = (3)^a

a = 1

2)

now rate doubles when A is halved and NaSH in increased by 4 times

r2 = 2r1

A2 = 0.5 A1

Let NaSH be B

B2 = 4 B1

now

r2/r1 = (A2/A1)^a (B2/B1)^b

2 = (0.5)^1 (4)^b

2 = 0.5 (4)^b

4 = (4)^b

b = 1

so

the rate law is given by

rate = k [A] [NaSH]

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