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

## General Chemistry 2 - Quiz 3 - Chemical Kinetics: Rate Laws

The rate of decomposition of a certain compound in solution is first order. If the concentration of the compound is doubled, what happens to the reaction's half-life?

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.

Consider the reaction: 2 ICl (g) + H_{2} (g) → 2 HCl (g) + I_{2} (g). At a certain temperature, the rate constant is found to be 1.63 x 10^{-6} L.mol^{-1}.s^{-1}. What is the overall order of the reaction?

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.

A compound decomposes with a first-order rate constant of 0.00854 s^{-1}. Calculate the concentration after 5.0 minutes for an initial concentration of 1.2 M.

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

[A]_{t} = [A]_{0} e^{-kt}

[A]_{t} = concentration of A at time t (in mol.L^{-1})

[A]_{0} = initial concentration of A (in mol.L^{-1}) = 1.2 M

k = rate constant (in s^{-1}) = 0.00854 s^{-1}

t = time (in s) = 5.0 min = 300 s

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

In a study of the reaction below, the concentration of O_{2} (g) is found to be decreasing by 0.042 M min^{-1}. At what rate is the concentration of nitrogen dioxide gas changing?

2 NO (g) + O_{2} (g) → 2 NO_{2 }(g)

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

Rate of NO_{2} = 2 x 0.042 M.min^{−1} = 0.084 M.min^{−1}

Tert-butyl chloride reacts with hydroxide ion in a process that is first order in both tert-butyl chloride and hydroxide. If both reactants are doubled in concentration, how does the reaction rate change?

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.

The thermal decomposition of NOCl is a second-order process, and the rate constant k for the disappearance of NOCl at 160^{ o}C is 0.0037 M^{-1}.s^{-1}. What is the concentration of NOCl, initially at 0.043 M, after 20.0 minutes at 160 ^{o}C?

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

$\frac{1}{{\left[\mathrm{NOCl}\right]}_{\mathrm{t}}}$ = $\frac{1}{{\left[\mathrm{NOCl}\right]}_{0}}$ + kt

[NOCl]_{t} = concentration of NOCl at time t (in mol.L^{-1})

[NOCl]_{0} = initial concentration of NOCl (in mol.L^{-1}) = 0.043 M

k = rate constant (in M^{-1}.s^{-1}) = 0.0037 M^{-1}.s^{-1}

t = time (in s) = 20.0 min = 1200 s

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

How is the rate of change of the ClF_{3} concentration related to the rate of change of the F_{2} concentration?

Cl_{2} (g) + 3 F_{2} (g) → 2 ClF_{3} (g)

The reaction rate of a reaction is expressed as:

Rate = - $\frac{1}{\mathrm{\alpha}}$ $\frac{\u2206\left[\mathrm{reactant}\right]}{\u2206\mathrm{t}}$ = $\frac{1}{\mathrm{\alpha}}$ $\frac{\u2206\left[\mathrm{reactant}\right]}{\u2206\mathrm{t}}$

Therefore, the reaction rate of Cl_{2} (g) + 3 F_{2} (g) → 2 ClF_{3} (g) is:

Rate = - $\frac{1}{3}\frac{\u2206\left[{\mathrm{F}}_{2}\right]}{\u2206\mathrm{t}}$ = $\frac{1}{2}$ $\frac{\u2206\left[{\mathrm{ClF}}_{3}\right]}{\u2206\mathrm{t}}$

⇒ $\frac{\u2206\left[{\mathrm{ClF}}_{3}\right]}{\u2206\mathrm{t}}$ = - $\frac{2}{3}$ $\frac{\u2206\left[{\mathrm{F}}_{2}\right]}{\u2206\mathrm{t}}$

At a given temperature, a first-order reaction has a rate constant of 3.33 x 10^{-3} s^{-1}. How much time is required for the reaction to be 75% complete?

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:

ln $\left(\frac{{\left[A\right]}_{0}}{{\left[A\right]}_{t}}\right)$ = kt

[A]_{t} = concentration of A at time t (in mol.L^{-1})

[A]_{0} = initial concentration of A (in mol.L^{-1})

k = rate constant (in s^{-1})

t = time (in s)

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