Chemical Kinetics

Integrated rate equation for a first-order reaction

For $A \rightarrow \text{Products}$, let $a$ = initial concentration of A and $x$ = amount reacted in time $t$, so the concentration left is $(a - x)$.

For a first-order reaction the rate is proportional to the concentration of A remaining:

$$\frac{dx}{dt} = k(a - x)$$

Separate the variables and integrate:

$$\int \frac{dx}{a - x} = k\int dt \;\Rightarrow\; -\ln(a - x) = kt + C$$

At $t = 0$, $x = 0$, so $C = -\ln a$:

$$-\ln(a - x) = kt - \ln a \;\Rightarrow\; kt = \ln\frac{a}{a - x}$$ $$k = \frac{2.303}{t}\log\frac{a}{a - x}$$

Explanation:

• This equation lets us find $k$ (or the time for a given extent of reaction) from concentrations at two times.
• In exponential form, $[A] = [A]_0e^{-kt}$: the concentration falls exponentially with time.
• A plot of $\log(a - x)$ against $t$ is a straight line with slope $= -\dfrac{k}{2.303}$ and intercept $\log a$. This plot is used to test for first-order kinetics.
• The unit of $k$ is time⁻¹ (s⁻¹ or min⁻¹). It does not depend on the concentration unit.

Half-life ($t_{1/2}$)

Half-life is the time needed for half the initial reactant to react. At $t = t_{1/2}$, $x = a/2$:

$$t_{1/2} = \frac{2.303}{k}\log\frac{a}{a/2} = \frac{2.303\log 2}{k} = \frac{2.303\times0.3010}{k} = \frac{0.693}{k}$$

The half-life of a first-order reaction is constant and independent of the initial concentration. For example, 100% → 50% → 25% → 12.5% each takes the same time $t_{1/2}$. Radioactive decay is an example.

Numerical

Given: 60% transformed in $t = 100$ min. If $a = 100$, then $a - x = 40$.

$$k = \frac{2.303}{100}\log\frac{100}{40} = \frac{2.303}{100}\times0.3979 = 9.16\times10^{-3}\ \text{min}^{-1}$$

For 90% transformation, $a - x = 10$:

$$t = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303\times1}{9.16\times10^{-3}} = 251.3\ \text{min}$$

(Equivalently, $\dfrac{t_{90}}{t_{60}} = \dfrac{\log 10}{\log 2.5} = \dfrac{1}{0.3979} = 2.513$, so $t_{90} = 2.513\times100$ min.)

Time for 90% transformation ≈ 251 minutes

(b) it lowers the activation energy

A catalyst provides an alternative pathway (mechanism) with a lower activation energy. More reactant molecules then have energy at least equal to this lower $E_a$, so more collisions are effective and the reaction goes faster.

a)

i) Rate law: The rate law is the experimentally found expression that relates the rate of a reaction to the molar concentrations of the reactants, each raised to some power. For $mP + nQ \rightarrow z$: rate $= K[P]^m[Q]^n$.

ii) Why the rate law is experimental: The powers in the rate law (the orders) do not have to equal the stoichiometric coefficients of the balanced equation. They depend on the reaction mechanism, which is usually unknown. So the rate law cannot be written from the balanced equation. It has to be found by measuring rates at different concentrations.

iii) Rate constant: The rate constant ($K$) is the rate of the reaction when the concentration of each reactant is unity (1 mol L⁻¹). It is constant for a given reaction at a given temperature and changes with temperature.

iv) Order vs molecularity (one difference): Order is the sum of the powers of the concentration terms in the experimental rate law. It is found by experiment and can be zero, fractional or a whole number. Molecularity is the number of reacting species taking part in one elementary step. It is a theoretical quantity and is always a whole number (1, 2 or 3), never zero or fractional.

b) The reaction is first order in P and zero order in Q, so

$$\text{rate} = k[P]^1[Q]^0 = k[P]$$

The rate depends only on [P].

ii) Rate constant (from Expt I):

$$k = \frac{\text{rate}}{[P]} = \frac{2\times10^{-2}\ \text{M s}^{-1}}{0.1\ \text{M}} = 0.2\ \text{s}^{-1}$$

i) Missing values

• (A): $[P] = \dfrac{\text{rate}}{k} = \dfrac{4\times10^{-2}}{0.2} = 0.2$ M. The rate is double that of Expt I, so [P] is doubled.
• (B): $[P] = \dfrac{2\times10^{-2}}{0.2} = 0.1$ M. The rate is the same as in Expt I, so [P] is the same. The change in [Q] has no effect because the reaction is zero order in Q.
• (C): rate $= k[P] = 0.2 \times 0.4 = 8\times10^{-2}$ M s⁻¹

Rate constant, $k = 0.2\ \text{s}^{-1}$ (the unit s⁻¹ fits an overall first-order reaction).

(a) increase 2 times

Taking the reaction as elementary, rate $= k[A]^2[B]$. The new rate is $k(2[A])^2\left(\tfrac{[B]}{2}\right) = 4 \times \tfrac{1}{2}\,k[A]^2[B] = 2 \times$ the original rate.

a) Factors affecting the rate of reaction

1. Concentration of reactants: A higher concentration gives more collisions per unit time, so the rate is higher (rate ∝ concentration terms in the rate law).
2. Temperature: Raising the temperature increases the kinetic energy of the molecules, so more molecules have energy ≥ $E_a$. The rate roughly doubles for every 10 °C rise (temperature coefficient ≈ 2–3).
3. Catalyst: A catalyst gives an alternative path of lower activation energy, so the rate increases. It is not used up in the reaction.
4. Surface area (particle size): Finely divided solids expose more surface, so there are more collisions and a faster reaction.
5. Nature of reactants: Ionic reactions are fast. Reactions that break strong covalent bonds are slow.
6. Light (radiation): Photochemical reactions, such as $H_2 + Cl_2$, go faster when light is absorbed.
7. Pressure: For gaseous reactions, higher pressure means higher concentration, so the rate is higher.

b) Let rate $= k[A]^x[B]^y$.

i) Order with respect to A and B

Experiments 1 and 3 ([B] is constant):

$$\frac{r_3}{r_1} = \frac{1.0\times10^{-2}}{4.0\times10^{-4}} = 25 = \left(\frac{0.5}{0.1}\right)^x = 5^x \Rightarrow x = 2$$

Experiments 3 and 4 ([A] is constant):

$$\frac{r_4}{r_3} = \frac{1.0\times10^{-2}}{1.0\times10^{-2}} = 1 = \left(\frac{0.5}{0.1}\right)^y = 5^y \Rightarrow y = 0$$

The reaction is second order with respect to A and zero order with respect to B, so the overall order is 2.

(Check with Expt 2: $k(0.2)^2 = 0.04 \times 0.04 = 1.6\times10^{-3}$, which agrees with the table.)

ii) Rate constant (Expt 1):

$$k = \frac{\text{rate}}{[A]^2} = \frac{4.0\times10^{-4}}{(0.1)^2} = 4.0\times10^{-2}\ \text{L mol}^{-1}\text{s}^{-1}$$

iii) Rate law:

$$\text{rate} = k[A]^2[B]^0 = k[A]^2$$

iv) Rate when [A] = 0.2 mol L⁻¹ and [B] = 0.35 mol L⁻¹

$$\text{rate} = 4.0\times10^{-2} \times (0.2)^2 = 4.0\times10^{-2}\times0.04 = 1.6\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$$

The concentration of B does not affect the rate because the reaction is zero order in B.

a) Pseudo first order vs first order reaction

b) Energy profile for an endothermic reaction

Potential
energy
  ^           Transition state (activated complex)
  |               ___
  |              /   \
  |             /     \_______ Products
  |        Ea  /               dH (+ve)
  | Reactants_/
  |  ------
  +-----------------------------------> Reaction progress

In an endothermic reaction the products lie at a higher energy than the reactants, so $\Delta H = E_{products} - E_{reactants}$ is positive. $E_a$ is the height of the energy barrier, measured from the reactants up to the transition state (activated complex).

c) Let rate $= k[A]^x[B]^y$.

Order with respect to B (Expts 2 and 3, [A] constant):

$$\frac{r_3}{r_2} = \frac{2.88\times10^{-1}}{7.2\times10^{-2}} = 4 = \left(\frac{0.4}{0.2}\right)^y = 2^y \Rightarrow y = 2$$

Order with respect to A (Expts 1 and 4, [B] constant):

$$\frac{r_4}{r_1} = \frac{2.40\times10^{-2}}{6\times10^{-3}} = 4 = \left(\frac{0.4}{0.1}\right)^x = 4^x \Rightarrow x = 1$$

Rate law: rate $= k[A][B]^2$ (overall order = 3)

Rate constant (Expt 1):

$$k = \frac{6\times10^{-3}}{(0.1)(0.1)^2} = \frac{6\times10^{-3}}{1\times10^{-3}} = 6\ \text{L}^2\text{mol}^{-2}\text{s}^{-1}$$

(Check with Expt 2: $6 \times 0.3 \times (0.2)^2 = 7.2\times10^{-2}$, which agrees.)

Rate of formation of C when [A] = 0.6 and [B] = 0.3 mol L⁻¹:

$$\text{rate} = 6 \times 0.6 \times (0.3)^2 = 6 \times 0.6 \times 0.09 = 0.324\ \text{mol L}^{-1}\text{s}^{-1}$$

Rate of formation of C = $3.24\times10^{-1}$ mol L⁻¹ s⁻¹

(b) First order reaction

Only for a first-order reaction is the half-life independent of the initial concentration ($t_{1/2} = 0.693/k$). Doubling the starting concentration therefore leaves it unchanged at 50 s.

Instantaneous rate of reaction: The instantaneous rate is the rate of a reaction at a particular moment of time. It is the rate of change of concentration over an infinitesimally small time interval, i.e. $-\dfrac{d[R]}{dt}$ or $+\dfrac{d[P]}{dt}$. It equals the slope of the tangent drawn to the concentration–time curve at that instant.

Rate expressions for $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$:

$$\text{Rate} = -\frac{1}{2}\frac{d[N_2O_5]}{dt} = +\frac{1}{4}\frac{d[NO_2]}{dt} = +\frac{d[O_2]}{dt}$$

The negative sign shows that the reactant is decreasing. Each term is divided by its stoichiometric coefficient so that the rate has a single value.

Numerical

Given: volume of $O_2$ formed at NTP = 2.24 L, time = 30 min

Moles of $O_2$ formed:

$$n_{O_2} = \frac{2.24\ \text{L}}{22.4\ \text{L mol}^{-1}} = 0.1\ \text{mol}$$

From the equation, 2 mol $N_2O_5$ gives 1 mol $O_2$. So the $N_2O_5$ used up is

$$n_{N_2O_5} = 2 \times 0.1 = 0.2\ \text{mol}$$

Rate of disappearance of $N_2O_5$:

$$-\frac{\Delta n_{N_2O_5}}{\Delta t} = \frac{0.2\ \text{mol}}{30\ \text{min}} = 6.67\times10^{-3}\ \text{mol min}^{-1}$$ $$= \frac{0.2}{30\times60}\ \text{mol s}^{-1} = 1.11\times10^{-4}\ \text{mol s}^{-1}$$

Rate of disappearance of $N_2O_5$ = $6.67\times10^{-3}$ mol min⁻¹ (= $1.11\times10^{-4}$ mol s⁻¹)

(The rate of formation of $O_2$ is $0.1/30 = 3.33\times10^{-3}$ mol min⁻¹, which is half the rate of disappearance of $N_2O_5$, as the rate expression requires.)

(c) independent of initial concentration of the reactants

For a first-order reaction $t_{1/2} = \dfrac{0.693}{k}$. It depends only on the rate constant and not on the initial concentration.

(a) (i) Rate of reaction: The rate of reaction is the change in concentration of a reactant or product per unit time. Its unit is mol L⁻¹ s⁻¹.

$$\text{Rate} = -\frac{\Delta[\text{Reactant}]}{\Delta t} = +\frac{\Delta[\text{Product}]}{\Delta t}$$

(ii) Let rate $= k[X]^a[Y]^b$.

• When [X] is doubled ([Y] constant), the rate doubles: $2^a = 2 \Rightarrow a = 1$.
• When both [X] and [Y] are doubled, the rate becomes 8 times: $2^a \times 2^b = 8 \Rightarrow 2 \times 2^b = 8 \Rightarrow 2^b = 4 \Rightarrow b = 2$.

Order: first order in X, second order in Y, so the overall order is $1 + 2 = 3$ (third order).

Rate law:

$$\text{Rate} = k[X][Y]^2$$

(b) Let rate $= k[NO_2]^x[F_2]^y$.

Order in $NO_2$ (Expts 1 and 2, [F₂] constant):

$$\frac{r_2}{r_1} = \frac{1.2\times10^{-2}}{6\times10^{-3}} = 2 = \left(\frac{0.40}{0.20}\right)^x = 2^x \Rightarrow x = 1$$

Order in $F_2$ (Expts 2 and 3, both concentrations doubled):

$$\frac{r_3}{r_2} = \frac{4.8\times10^{-2}}{1.2\times10^{-2}} = 4 = 2^x \cdot 2^y = 2 \times 2^y \Rightarrow y = 1$$

Rate law: rate $= k[NO_2][F_2]$ (overall second order)

Rate constant (Expt 1):

$$k = \frac{6\times10^{-3}}{(0.20)(0.05)} = \frac{6\times10^{-3}}{1\times10^{-2}} = 0.6\ \text{L mol}^{-1}\text{s}^{-1}$$

Rate when $[NO_2] = 0.50$ and $[F_2] = 0.60$ mol L⁻¹:

$$\text{Rate} = 0.6 \times 0.50 \times 0.60 = 0.18\ \text{mol L}^{-1}\text{s}^{-1}$$

Rate = $1.8\times10^{-1}$ mol L⁻¹ s⁻¹

(a) Zero order reaction

A straight line with constant slope means the concentration changes by equal amounts in equal times. The rate is therefore constant and independent of concentration, which is the mark of a zero-order reaction. A line rising from the origin is the product concentration, $[P] = kt$. The reactant concentration, $[R] = [R]_0 - kt$, would be a falling straight line. A first-order plot would be a curve.

(b) Pseudo-first order

Acid hydrolysis of an ester is really bimolecular (it needs both ester and water). Water is present in large excess, however, so [H₂O] stays practically constant and the rate $= k[\text{RCOOR}]$. The reaction therefore behaves as first order.

a) Differences between order and molecularity

b) Concentration vs time for a zero-order reaction

[R]
  ^
[R]0 *
  |    \
  |      \      slope = -k
  |        \
  |          \
  |            \
  +--------------*---------> Time

For a zero-order reaction $[R] = [R]_0 - kt$. A plot of the reactant concentration against time is a straight line with intercept $[R]_0$ and slope $-k$.

c) Let rate $= k[X]^a[Y]^b$.

Order in Y (Expts I and II, [X] constant):

$$\frac{r_{II}}{r_I} = \frac{3.2\times10^{-4}}{1.6\times10^{-4}} = 2 = \left(\frac{1.0}{0.5}\right)^b = 2^b \Rightarrow b = 1$$

Order in X (Expts II and III, [Y] constant):

$$\frac{r_{III}}{r_{II}} = \frac{3.2\times10^{-4}}{3.2\times10^{-4}} = 1 = \left(\frac{1.0}{0.5}\right)^a = 2^a \Rightarrow a = 0$$

Rate law:

$$\text{Rate} = k[X]^0[Y]^1 = k[Y]$$

Rate constant (Expt I):

$$k = \frac{\text{rate}}{[Y]} = \frac{1.6\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}}{0.5\ \text{mol L}^{-1}} = 3.2\times10^{-4}\ \text{s}^{-1}$$

d) The reaction is zero order with respect to X. Doubling [X] (Expts II and III) does not change the rate.

(a) Zero order reaction

For a zero order reaction the rate equation is $r = k[A]^0 = k$. The rate is constant and does not depend on the concentration of the reactant.

a) Rate law

$$\text{Rate} = k[M]^x[N]^y$$

b) Order with respect to each reactant

Order in M: compare experiments 1 and 4 ($[N]$ constant at 0.1 M). $[M]$ changes from 0.1 to 0.4 (four times) and the rate changes from $6.0\times10^{-3}$ to $2.4\times10^{-2}$ (also four times):

$$\frac{2.4\times10^{-2}}{6.0\times10^{-3}} = \left(\frac{0.4}{0.1}\right)^x \Rightarrow 4 = 4^x \Rightarrow x = 1$$

Order in N: compare experiments 2 and 3 ($[M]$ constant at 0.3 M). $[N]$ doubles (0.2 to 0.4) and the rate rises from $7.2\times10^{-2}$ to $2.88\times10^{-1}$ (four times):

$$\frac{2.88\times10^{-1}}{7.2\times10^{-2}} = \left(\frac{0.4}{0.2}\right)^y \Rightarrow 4 = 2^y \Rightarrow y = 2$$

(Check with experiments 1 and 2: $3^1 \times 2^2 = 12$, and $6.0\times10^{-3}\times12 = 7.2\times10^{-2}$, which agrees.)

The reaction is first order in M, second order in N, and third order overall ($1 + 2 = 3$). Rate $= k[M][N]^2$.

c) Rate constant

Using experiment 1:

$$k = \frac{\text{Rate}}{[M][N]^2} = \frac{6.0\times10^{-3}}{(0.1)(0.1)^2} = \frac{6.0\times10^{-3}}{1.0\times10^{-3}} = 6.0$$ $$k = 6.0\ L^2\,mol^{-2}\,s^{-1}$$

d) Unit of k for the overall reaction

For overall order $n$, $k = \dfrac{\text{rate}}{(\text{conc})^n}$, so the unit is $(mol\,L^{-1})^{1-n}\,s^{-1}$. For $n = 3$:

$$\frac{mol\,L^{-1}s^{-1}}{(mol\,L^{-1})^3} = L^2\,mol^{-2}\,s^{-1}$$

Zero order reaction: a reaction whose rate does not depend on the concentration of the reactant. For $A \rightarrow$ products,

$$\text{Rate} = -\frac{d[A]}{dt} = k[A]^0 = k$$

Integrated rate equation

Rearrange: $d[A] = -k\,dt$. Integrate from $t = 0$ (concentration $[A]_0$) to time $t$ (concentration $[A]$):

$$\int_{[A]_0}^{[A]} d[A] = -k\int_0^t dt$$ $$[A] - [A]_0 = -kt \quad\Rightarrow\quad [A] = [A]_0 - kt$$

or

$$k = \frac{[A]_0 - [A]}{t}$$

A graph of $[A]$ against $t$ is a straight line of slope $-k$ and intercept $[A]_0$. The unit of $k$ is $mol\,L^{-1}s^{-1}$.

Half-life

At $t = t_{1/2}$, $[A] = [A]_0/2$:

$$\frac{[A]_0}{2} = [A]_0 - k\,t_{1/2} \quad\Rightarrow\quad t_{1/2} = \frac{[A]_0}{2k}$$

Explanation. The half-life of a zero order reaction is directly proportional to the initial concentration (unlike a first order reaction, where it is independent of concentration). Examples are the decomposition of $NH_3$ on a hot platinum surface and of $HI$ on gold, where the surface is saturated and so the rate is constant.

Rate of reaction [1]

The rate of a reaction is the change in concentration of a reactant or product per unit time. Its unit is $mol\,L^{-1}s^{-1}$.

$$\text{Rate} = -\frac{\Delta[R]}{\Delta t} = +\frac{\Delta[P]}{\Delta t}$$

Factors affecting the rate [4]

• Concentration: increasing the concentration of reactants gives more molecules per unit volume, more collisions per second, and so a faster reaction (rate law: rate $= k[A]^n$).
• Temperature: a rise of about 10 °C roughly doubles or triples the rate. More molecules have energy equal to or above the activation energy and collisions are more frequent (Arrhenius equation $k = Ae^{-E_a/RT}$).
• Catalyst: a catalyst gives the reaction a new path with a lower activation energy, so more collisions are effective and the reaction is faster. It is not used up.
• Surface area: for solid reactants, a finer powder exposes more surface, so more collisions occur and the reaction is faster (powdered marble reacts faster with acid than lumps).

Calculation [3]

First order, $t_{1/2} = 3$ h:

$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{3} = 0.231\ h^{-1}$$

87.5% complete: if $[A]_0 = 100$, $[A] = 100 - 87.5 = 12.5$.

$$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{0.231}\log\frac{100}{12.5} = \frac{2.303}{0.231}\log 8$$ $$t = 9.97 \times 0.9031 \approx 9.0\ \text{h}$$

Check: 87.5% completion means three half-lives ($100 \to 50 \to 25 \to 12.5$), and $3 \times 3 = 9$ h. Time required = 9 hours.

(a) The rate of a reaction is the change in concentration of a reactant or product per unit time, e.g. mol L$^{-1}$ s$^{-1}$.

Catalyst: it provides an alternative path with a lower activation energy. More colliding molecules then have energy at least equal to the (lower) activation energy, so the rate increases, though the catalyst is not consumed.

Temperature: a rise in temperature increases the average kinetic energy of molecules, so the fraction of molecules with energy at or above the activation energy increases greatly (as in $k = Ae^{-E_a/RT}$), and collisions are also more frequent. The rate usually doubles or triples for each $10^\circ C$ rise.

(b) The half-life is the time required for the concentration of a reactant to fall to half of its initial value.

For a first-order reaction:

$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{5\ \text{s}^{-1}} = 0.1386\ \text{s}$$

Average rate vs instantaneous rate

Rate data

i) Order with respect to A: in runs 1 and 2, $[B]$ is constant and $[A]$ doubles, and the rate doubles ($0.005 \to 0.010$). So the rate is proportional to $[A]^1$; order = 1.

Order with respect to B: in runs 1 and 3, $[A]$ is constant and $[B]$ doubles, but the rate is unchanged. So the rate is independent of $[B]$; order = 0.

ii) Overall order $= 1 + 0 =$ 1.

iii) Rate law: $\text{Rate} = k[A]$

iv) Using run 1:

$$k = \frac{\text{Rate}}{[A]} = \frac{0.005}{0.01} = 0.5\ \text{s}^{-1}$$

v) For a first-order reaction:

$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.5} = 1.386\ \text{s}$$

(b) 0.8

For a first-order reaction, rate $= k[N_2O_5]$, so

$$[N_2O_5] = \frac{\text{rate}}{k} = \frac{2.4\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}}{3.0\times10^{-5}\ \text{s}^{-1}} = 0.8\ \text{mol L}^{-1}$$

A) Integrated rate equation for a first-order reaction

Consider the first-order reaction $A \rightarrow \text{Products}$.

Let $a$ = initial concentration of A (at $t = 0$) and $x$ = amount of A that has reacted after time $t$. The concentration of A left at time $t$ is $(a - x)$.

Rate of reaction (the rate of formation of products):

$$\frac{dx}{dt} = k(a - x)$$

Separate the variables:

$$\frac{dx}{a - x} = k\,dt$$

Integrate both sides:

$$-\ln(a - x) = kt + C \qquad \text{...(i)}$$

At $t = 0$, $x = 0$, so $C = -\ln a$. Put this into (i):

$$-\ln(a - x) = kt - \ln a$$ $$kt = \ln a - \ln(a - x) = \ln\frac{a}{a - x}$$ $$k = \frac{1}{t}\ln\frac{a}{a - x} = \frac{2.303}{t}\log\frac{a}{a - x}$$

This is the integrated rate equation for a first-order reaction.

Half-life period

At $t = t_{1/2}$, half of the reactant has reacted, so $x = \dfrac{a}{2}$ and $a - x = \dfrac{a}{2}$:

$$k = \frac{2.303}{t_{1/2}}\log\frac{a}{a/2} = \frac{2.303}{t_{1/2}}\log 2$$ $$t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k}$$

The expression contains only the rate constant $k$ and no term for the initial concentration $a$. So for a first-order reaction the half-life is independent of the initial concentration: the same time is needed to halve the reactant, whatever the starting amount.

B) Time for 90% completion

(The time for 50% completion is read as $t_{1/2} = 1.26\times10^{14}$ s.)

Given: $t_{1/2} = 1.26 \times 10^{14}$ s

Rate constant:

$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1.26\times10^{14}} = 5.5\times10^{-15}\ \text{s}^{-1}$$

For 90% completion, if $a = 100$ then $a - x = 10$:

$$t_{90\%} = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303 \times 1}{5.5\times10^{-15}} = 4.19\times10^{14}\ \text{s}$$

Required time ≈ $4.19 \times 10^{14}$ s (that is, $t_{90\%} = 3.32\,t_{1/2}$).

(If the given time is 10 s, the same method gives $t_{90\%} = 3.32 \times 10 = 33.2$ s.)

(c) 512 times

The rise from 10 °C to 100 °C is 90 °C, which is 9 intervals of 10 °C. Since the rate doubles in each interval, it increases $2^9 = 512$ times.

a) Rate of reaction vs rate constant

b) Let rate $= k[A]^x[B]^y$.

i) Order with respect to A and B

Expts 1 and 2 ([A] constant, [B] doubled):

$$\frac{r_2}{r_1} = \frac{3.2\times10^{-4}}{1.6\times10^{-4}} = 2 = 2^y \Rightarrow y = 1$$

Expts 2 and 3 ([B] constant, [A] doubled):

$$\frac{r_3}{r_2} = \frac{3.2\times10^{-4}}{3.2\times10^{-4}} = 1 = 2^x \Rightarrow x = 0$$

The reaction is zero order with respect to A and first order with respect to B. Rate $= k[B]$, and the overall order is 1.

ii) Rate constant (Expt 1):

$$k = \frac{1.6\times10^{-4}}{0.50} = 3.2\times10^{-4}\ \text{s}^{-1}$$

iii) Rate when [A] = 0.5 M and [B] = 0.4 M:

$$\text{Rate} = k[B] = 3.2\times10^{-4}\times0.4 = 1.28\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}$$

a) Order of reaction

The order of a reaction is the sum of the powers to which the concentration terms are raised in the experimentally determined rate law. It is $1 + 0 = 1$ here.

b) Rate law

$$\text{Rate} = k[A]^1[B]^0 = k[A]$$

c) Effect of doubling the concentrations

Initially $r_1 = k[A]$. After doubling both, $[A]' = 2[A]$ and $[B]' = 2[B]$:

$$r_2 = k(2[A])(2[B])^0 = 2k[A] = 2r_1$$

The rate increases 2 times, because B has zero order and does not affect the rate.

d) Effect of temperature

For about every $10^\circ C$ rise the rate becomes roughly 2 to 3 times (the temperature coefficient). A rise of $20^\circ C$ therefore increases the rate about $2^2$ to $3^2$ = 4 to 9 times, because more molecules have energy equal to or above the activation energy.

(a) zero order

For a reaction of order $n$, $t_{1/2} \propto [A]_0^{\,1-n}$. If $t_{1/2} \propto [A]_0$, then $1-n = 1$, so $n = 0$. Directly, for zero order $t_{1/2} = \dfrac{[A]_0}{2k}$.

(a) Let the rate law be $r = k[X]^a[Y]^b$.

Experiments 1 and 2: $[X]$ is constant and $[Y]$ is multiplied by 4, and the rate also becomes 4 times. So $b = 1$.

Experiments 2 and 3: $[Y]$ is constant and $[X]$ is doubled, and the rate also doubles. So $a = 1$.

(iii) Rate law: $r = k[X][Y]$. The reaction is second order overall.

(i) Using experiment 1:

$$k = \frac{r}{[X][Y]} = \frac{1 \times 10^{-3}}{0.1 \times 0.1} = 0.1 \ L\ mol^{-1}min^{-1}$$

(ii) Using experiment 4:

$$1.8 \times 10^{-2} = 0.1 \times 0.5 \times A \Rightarrow A = \frac{1.8 \times 10^{-2}}{0.05} = 0.36 \ mol\ L^{-1}$$

(b) A pseudo first order reaction is one that is truly of higher order (usually second) but follows first-order kinetics because one reactant is present in such large excess that its concentration stays practically constant. Example: acid hydrolysis of ethyl acetate, $CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$. Water is in large excess, so the rate depends only on the ester concentration.

(c) Effective collision. Diagram description: two molecules (e.g. $A_2$ and $B_2$) approach each other with the proper orientation, so that the reacting atoms face each other. They collide with energy equal to or greater than the activation energy, the bonds break and form at the point of impact, and two $AB$ molecules separate. A collision with the wrong orientation, or too little energy, simply makes the molecules bounce apart unchanged.