Chemical Kinetics: Rate Laws | General Chemistry 2

Reaction rate of a reactant:

Reaction rate of a product:

a A + b B → c C
reaction rate = - $\frac{1}{\mathrm{a}} \frac{∆\left[\mathrm{A}\right]}{∆\mathrm{t}}$ = - $\frac{1}{\mathrm{b}} \frac{∆\left[\mathrm{B}\right]}{∆\mathrm{t}}$ = $\frac{1}{\mathrm{c}} \frac{∆\left[\mathrm{C}\right]}{∆\mathrm{t}}$

NB: reaction rate is in mol.L^-1.s^-1 (or M ^-1.s^-1)

Reaction rate can also be expressed as a product of the concentration of reactants:
 

Reaction rate of first-order reactions: k [A]
Units of rate constant k: s^-1

Dependence of [A] on time: - $\frac{∆\left[\mathrm{A}\right]}{∆\mathrm{t}}$ = k [A]

Test plot of first-order reactions: ln[A] versus t

Half-time t_1/2:  time it takes for one-half of the reactant to react

At t = t_1/2:  [A] = $\frac{{\left[\mathrm{A}\right]}_0}{2}$ 
 

Reaction rate of second-order reactions: k [A]²
Units of rate constant k: M ^-1.s^-1 (mol.L^-1.s^-1)

Dependence of [A] on time: - $\frac{∆\left[\mathrm{A}\right]}{∆\mathrm{t}}$ = k [A]²
After integrating:

Test plot of second-order reactions: $\frac{1}{\left[\mathrm{A}\right]}$ versus t

At t = t_1/2: [A] = $\frac{{\left[\mathrm{A}\right]}_0}{2}$

Radioactive decay: process by which an unstable atomic nucleus loses energy by emitting small particles (α-particles, β-particles, γ-rays)

Radioactive decay is a first-order process:

ln $\left(\frac{\left[\mathrm{A}\right]}{{\left[\mathrm{A}\right]}_0}\right)$ = -kt and t_1/2 = $\frac{\ln 2}{\mathrm{k}}$

⇒ k = $\frac{\ln 2}{{\mathrm{t}}_{1/2}}$ 

⇒ ln $\left(\frac{\left[\mathrm{A}\right]}{{\left[\mathrm{A}\right]}_0}\right)$ = - $\frac{\ln 2}{{\mathrm{t}}_{1/2}}$ x t

Relationship between half-life and number of particles in first-order nuclear decay:
Number of radioactive nuclei N is proportional to the concentration of the radioactive species

 $\frac{\left[\mathrm{A}\right]}{{\left[\mathrm{A}\right]}_0}$ = $\frac{\mathrm{N}}{{\mathrm{N}}_0}$ 

⇒ ln $\left(\frac{\mathrm{N}}{{\mathrm{N}}_0}\right)$ = ln $\left(\frac{\left[\mathrm{A}\right]}{{\left[\mathrm{A}\right]}_0}\right)$ = - $\frac{\ln 2}{{\mathrm{t}}_{1/2}}$ x t