Question

The rate law for a hypothetical reaction is rate = k [A]2. When the concentration is 0.10 moles/liter, the rate is 2.7 × 10-5 M*s-1. What is the value of k?

Answer 1

Given:

rate = k [A]2

concentration is 0.10 moles/liter

rate is 2.7 × 10-5 M*s-1

Required:

Value of k

Solution:

rate = k [A]2

2.7 × 10-5 M*s-1 = k (0.10 moles/liter)^2

k = 2.7 x 10^-3 liter per mole per second

Answer 2

Answer:The value of rate constant is
`2.7* 10^(-3) Ls^(-1) mol^(-1)`.

Step-by-step explanation:

Concentration of [A]=0.10 mol/L

The rate of the reaction =
`2.7* 10^(-5) M/s`

Rate constant = k

The given rate law:

`R=k[A]^2`

`2.7* 10^(-5) M/s=k* (0.10 mol/L)^2`

`k=(2.7* 10^(-5) mol/L s)/(0.10 mol/L* 0.10 mol/L)=2.7* 10^(-3) Ls^(-1) mol^(-1)`

The value of rate constant is
`2.7* 10^(-3) Ls^(-1) mol^(-1)`.