The kinetics of the reaction 3 A → B was studied by monitoring the concentration of A as a function of time.

Given the following concentrations, determine the specific rate constant for the disappearance of [A] at the temperature of the experiment. (In the future another blank will be added asking about the rate constant of the REACTION, in which case one would divide by the coefficient in front of A.)

t 0 1000. 10000. s
[A] 0.100 0.0878 0.0273 M

What are the units of the rate constant?

Rate constant = 0.0000422 M⁰•⁵/s

The units of the rate constant = M⁰•⁵/s

Explanation:

Rate of reaction = rate of change in the concentration of reactant A per unit time

Rate = | {[A]ₙ - [A]ₙ₋₁} ÷ (tₙ - tₙ₋₁) |

The rate of the reaction at any time is also proportional to the concentration of the reactant raised to the power of the order of the reaction.

Rate = | k [A]ˢ |

where k = rate constant

s = order of the reaction

In [Rate] = In K + s In [A]

t | 0 | 1000 | 10000 s

[A] | 0.100 | 0.0878 | 0.0273 M

t₀ = 0 s

[A]₀ = 0.100 M

t₁ = 1000 s

[A]₁ = 0.0878 M

t₂ = 10000 s

[A]₂ = 0.0273 M

Rate at t₁ = {[A]₁ - [A]₀} ÷ (t₁ - t₀)

= | (0.0878 - 0.100)/1000 | = 0.0000122 M/s

Rate at t₂ = {[A]₂ - [A]₁} ÷ (t₂ - t₁)

= | (0.0273 - 0.0878)/(10000-1000) | = 0.0000067222 M/s

In [Rate] = s In [A] + In k

Rate₁ = 0.0000122 M/s

[A]₁ = 0.0878 M

In [Rate₁] = -11.314

In [A]₁ = -2.433

Rate₂ = 0.0000067222 M/s

[A]₂ = 0.0273 M

In [Rate₂] = -11.91

In [A]₂ = -3.601

We can then obtain a simultaneous equation from

In [Rate] = s In [A] + In k

-11.314 = -2.433s + In k

-11.91 = -3.601s + In k

Solving the simultaneous equation,

s = 0.51 ≈ 0.50

In k = -10.0725

k = 0.0000422249

Rate = | k [A]ˢ |

Rate = k [A]⁰•⁵

Hence the units for the rate constant to keep the units of the rate of reaction as M/s would be M⁰•⁵/s.

Hope this Helps!!!