Question

A certain reaction has the form aA → bB. At a particular temperature and [A]0 = 2.80 × 10-3 molar, data was collected of concentration versus time for this reaction. A plot of 1/[A]t versus time resulted in a straight line with a slope value of 3.60 × 10-2 M-1s-1. What is the reaction order and rate law for this reaction?

A.
second order, rate = k[A]2
B.
first order, rate = k[A]2
C.
second order, rate = k[A]
D.
first order, rate = k[A]

Answer 1

Hey There!

aA => bB

[A]o = 2.80*10⁻³ M

When drew a plot 1 / [A]t versus time resulted a straight line inidicates second order reation .

Therefore , Rate = K[A]²

Answer A

Hope that helps!

Answer 2

The reaction is second order.

The rate law is:

`Rate=K[A]^(2)`

Step-by-step explanation:

The following plots and line shows the order of reaction

a) if we are getting a straight line in a plot with concentration versus time, the order of reaction is zero.

b) if we are getting a straight line in a plot with ln(concentration) versus time, the order of reaction is one.

c) if we are getting a straight line in a plot with inverse of concentration versus time, the order of reaction is two.

The reaction is second order.

The rate law is:

`Rate=K[A]^(2)`