Quiz 3 - Chemical Kinetics: Rate Laws | Chemical Kinetics: Rate Laws

For a first-order reaction, the half-life (t_1/2​) is independent of the initial concentration of the reactant. The half-life is given by the equation: t_1/2 = $\frac{0.693}{\mathrm{k}}$, where k is the rate constant. Since the half-life depends only on the rate constant and not on the concentration, doubling the concentration does not affect the half-life.

The units of the rate constant k (L.mol^-1.s^-1) indicate a second-order reaction. For a reaction with these units, the overall order is two.

For a first-order reaction, the concentration at any time t is given by the equation:

[A]_t = [A]₀ e^-kt = 1.2 x e^{-0.00854 x 300} = 0.093 M

The stoichiometry of the reaction shows that for every 1 mole of O₂ consumed, 2 moles of NO₂ are produced. The rate of change of concentration of NO₂ is therefore twice the rate of change of O₂. Given that O₂ is decreasing at a rate of 0.042 M min⁻¹, the rate of increase of NO₂ is:

Rate of NO₂ = 2 x 0.042 M.min⁻¹ = 0.084 M.min⁻¹

The reaction rate for a reaction that is first order in both tert-butyl chloride and hydroxide ion can be expressed as: Rate = k [tert-butyl chloride][OH⁻]

If the concentration of both reactants is doubled, the new rate becomes:

New Rate = k (2 [tert-butyl chloride] x 2 [OH^-]) = k ( 4 [tert-butyl chloride][OH⁻]) = 4 x Rate

Thus, the reaction rate quadruples when both reactants are doubled in concentration.

For a second-order reaction, the relationship between concentration and time is given by the equation:

Thus,

$\frac{1}{{\left[\mathrm{NOCl}\right]}_{\mathrm{t}}}$ = $\frac{1}{0.043 \mathrm{M}}$ + (0.0037 M^-1.s^-1) (1200 s) = 27.7 M^-1 

⇒ [NOCl]_t = $\frac{1}{27.7 {\mathrm{M}}^{-1}}$ = 0.036 M

The reaction rate of a reaction is expressed as:

Rate = - $\frac{1}{\mathrm{\alpha }}$ $\frac{∆\left[\mathrm{reactant}\right]}{∆\mathrm{t}}$ = $\frac{1}{\mathrm{\alpha }}$ $\frac{∆\left[\mathrm{reactant}\right]}{∆\mathrm{t}}$

Therefore, the reaction rate of Cl₂ (g) + 3 F₂ (g) → 2 ClF₃ (g) is:

Rate = - $\frac{1}{3} \frac{∆\left[{\mathrm{F}}_2\right]}{∆\mathrm{t}}$ = $\frac{1}{2}$ $\frac{∆\left[{\mathrm{ClF}}_3\right]}{∆\mathrm{t}}$

⇒ $\frac{∆\left[{\mathrm{ClF}}_3\right]}{∆\mathrm{t}}$ = - $\frac{2}{3}$ $\frac{∆\left[{\mathrm{F}}_2\right]}{∆\mathrm{t}}$

For a first-order reaction, the time required for a certain percentage of the reaction to be complete can be calculated using the first-order rate law equation:

Since the reaction is 75% complete, 25% of the original concentration remains, so $\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[\mathrm{A}\right]}_0}$ = 0.25. Thus:

ln $\left(\frac{1}{0.25}\right)$ = kt ⇒ t = $\frac{1}{\mathrm{k}}$ ln $\left(\frac{1}{0.25}\right)$ = $\frac{1}{3.33 \times {10}^{-3}}$ ln(4) = 416 s