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

Answer:The value of rate constant is $2.7\times 10^{-3} Ls^{-1} mol^{-1}$.

Explanation:

Concentration of [A]=0.10 mol/L

The rate of the reaction = $2.7\times 10^{-5} M/s$

Rate constant = k

The given rate law:

$R=k[A]^2$

$2.7\times 10^{-5} M/s=k\times (0.10 mol/L)^2$

$k=\frac{2.7\times 10^{-5} mol/L s}{0.10 mol/L\times 0.10 mol/L}=2.7\times 10^{-3} Ls^{-1} mol^{-1}$

The value of rate constant is $2.7\times 10^{-3} Ls^{-1} mol^{-1}$.

Answer 2

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