Chemical Kinetics

Given:
- Initial concentration, $[R]_1 = 0.03\,\text{M}$
- Final concentration, $[R]_2 = 0.02\,\text{M}$
- Time interval, $\Delta t = 25\,\text{min}$

Formula:
$$\text{Average rate} = -\frac{\Delta[R]}{\Delta t} = -\frac{[R]_2 - [R]_1}{\Delta t}$$

In minutes:
$$\text{Average rate} = -\frac{(0.02 - 0.03)\,\text{mol L}^{-1}}{25\,\text{min}} = -\frac{-0.01}{25}\,\text{mol L}^{-1}\text{min}^{-1}$$
$$= 4 \times 10^{-4}\,\text{mol L}^{-1}\text{min}^{-1}$$

In seconds (converting: $25\,\text{min} = 25 \times 60 = 1500\,\text{s}$):
$$\text{Average rate} = -\frac{(0.02 - 0.03)}{1500}\,\text{mol L}^{-1}\text{s}^{-1} = \frac{0.01}{1500}$$
$$= 6.66 \times 10^{-6}\,\text{mol L}^{-1}\text{s}^{-1}$$

Answer: Average rate $= 4 \times 10^{-4}\,\text{mol L}^{-1}\text{min}^{-1} = 6.66 \times 10^{-6}\,\text{mol L}^{-1}\text{s}^{-1}$

Given:
- Reaction: $2\text{A} \rightarrow \text{Products}$
- $[A]_1 = 0.5\,\text{mol L}^{-1}$, $[A]_2 = 0.4\,\text{mol L}^{-1}$
- $\Delta t = 10\,\text{min}$

Formula:
$$\text{Rate of reaction} = -\frac{1}{2}\frac{\Delta[A]}{\Delta t}$$

Calculation:
$$\text{Rate} = -\frac{1}{2} \times \frac{(0.4 - 0.5)\,\text{mol L}^{-1}}{10\,\text{min}}$$
$$= -\frac{1}{2} \times \frac{-0.1}{10}\,\text{mol L}^{-1}\text{min}^{-1}$$
$$= \frac{0.1}{20} = 0.005\,\text{mol L}^{-1}\text{min}^{-1}$$

Answer: Rate of reaction $= 5 \times 10^{-3}\,\text{mol L}^{-1}\text{min}^{-1}$

Given: Rate law: $r = k[A]^{1/2}[B]^2$

Concept: Order of reaction with respect to each reactant is the power of its concentration in the rate law. Overall order is the sum of all powers.

Calculation:
- Order with respect to A $= \dfrac{1}{2}$
- Order with respect to B $= 2$

$$\text{Overall order} = \frac{1}{2} + 2 = \frac{5}{2} = 2.5$$

Answer: The order of the reaction is 2.5.

Given: $X \rightarrow Y$, second order reaction.

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

Initial rate: $r_1 = k[X]^2$

New rate when $[X]$ is tripled, i.e., $[X]_{\text{new}} = 3[X]$:
$$r_2 = k(3[X])^2 = 9k[X]^2 = 9\,r_1$$

Answer: The rate of formation of Y will increase 9 times when the concentration of X is tripled.

Given:
- $k = 1.15 \times 10^{-3}\,\text{s}^{-1}$
- Initial amount, $[R]_0 = 5\,\text{g}$
- Final amount, $[R] = 3\,\text{g}$

Formula for first order reaction:
$$t = \frac{2.303}{k}\log\frac{[R]_0}{[R]}$$

Calculation:
$$t = \frac{2.303}{1.15 \times 10^{-3}}\log\frac{5}{3}$$
$$= \frac{2.303}{1.15 \times 10^{-3}} \times \log(1.667)$$
$$= \frac{2.303}{1.15 \times 10^{-3}} \times 0.2219$$
$$= 2003 \times 0.2219$$
$$\approx 444\,\text{s}$$

Answer: The time required is $t \approx 444\,\text{s}$.

Given:
- Half-life, $t_{1/2} = 60\,\text{min}$
- First order reaction

Formula:
$$t_{1/2} = \frac{0.693}{k}$$

Calculation:
$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{60\,\text{min}}$$
$$= 0.01155\,\text{min}^{-1}$$

Converting to s⁻¹: $k = \dfrac{0.01155}{60}\,\text{s}^{-1} = 1.925 \times 10^{-4}\,\text{s}^{-1}$

Answer: $k = 1.925 \times 10^{-4}\,\text{s}^{-1}$

Answer:

The rate constant of a reaction increases with increase in temperature.

According to the Arrhenius equation:
$$k = A\,e^{-E_a/RT}$$

As temperature $T$ increases, the exponential term $e^{-E_a/RT}$ increases (becomes less negative in exponent), so $k$ increases. It has been found experimentally that for most reactions, the rate constant nearly doubles for every $10\,\text{K}$ rise in temperature. This is because at higher temperatures, more molecules possess energy equal to or greater than the activation energy $E_a$.

Given:
- $T_1 = 298\,\text{K}$, $T_2 = 308\,\text{K}$
- $k_2 = 2k_1$ (rate doubles)

Formula (Arrhenius equation in two-temperature form):
$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$$

Substituting values:
$$\log 2 = \frac{E_a}{2.303 \times 8.314}\left(\frac{308 - 298}{298 \times 308}\right)$$
$$0.3010 = \frac{E_a}{19.147} \times \frac{10}{91784}$$
$$0.3010 = \frac{E_a \times 10}{19.147 \times 91784}$$
$$0.3010 = \frac{E_a}{1.757 \times 10^5}$$
$$E_a = 0.3010 \times 1.757 \times 10^5$$
$$E_a = 52,897\,\text{J mol}^{-1} \approx 52.9\,\text{kJ mol}^{-1}$$

Answer: $E_a \approx 52.9\,\text{kJ mol}^{-1}$

Given:
- $E_a = 209.5\,\text{kJ mol}^{-1} = 209500\,\text{J mol}^{-1}$
- $T = 581\,\text{K}$
- $R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}$

Formula:
The fraction of molecules having energy $\geq E_a$ is:
$$x = e^{-E_a/RT}$$

Taking logarithm:
$$\ln x = -\frac{E_a}{RT} = -\frac{209500}{8.314 \times 581}$$
$$= -\frac{209500}{4830.4} = -43.37$$

$$\log x = \frac{-43.37}{2.303} = -18.83$$

$$x = \text{antilog}(-18.83) = 10^{-18.83}$$
$$= 1.471 \times 10^{-19}$$

Answer: The fraction of molecules having energy equal to or greater than activation energy $= 1.471 \times 10^{-19}$.

Concept: Order of reaction = sum of powers of concentration terms in rate law. Units of $k$ are derived from: $\text{Rate} = k[\text{conc}]^n$, so $k = \dfrac{\text{Rate}}{[\text{conc}]^n} = \dfrac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^n} = \text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}$

(i) Rate $= k[\text{NO}]^2$
- Order = 2 (second order)
- Units of $k$: $$k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^2} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol}^2\text{L}^{-2}} = \text{mol}^{-1}\text{L}\,\text{s}^{-1}$$

(ii) Rate $= k[\text{H}_2\text{O}_2][\text{I}^-]$
- Order = 1 + 1 = 2 (second order)
- Units of $k$: $$k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^2} = \text{mol}^{-1}\text{L}\,\text{s}^{-1}$$

(iii) Rate $= k[\text{CH}_3\text{CHO}]^{3/2}$
- Order = 3/2 = 1.5 (1.5 order)
- Units of $k$: $$k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^{3/2}} = \text{mol}^{1-3/2}\text{L}^{3/2-1}\text{s}^{-1} = \text{mol}^{-1/2}\text{L}^{1/2}\text{s}^{-1}$$

(iv) Rate $= k[\text{C}_2\text{H}_5\text{Cl}]$
- Order = 1 (first order)
- Units of $k$: $$k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}$$

Given:
- Rate $= k[A][B]^2$
- $k = 2.0 \times 10^{-6}\,\text{mol}^{-2}\text{L}^2\text{s}^{-1}$
- Initial: $[A] = 0.1\,\text{mol L}^{-1}$, $[B] = 0.2\,\text{mol L}^{-1}$

Part 1 – Initial rate:
$$r_1 = k[A][B]^2 = 2.0 \times 10^{-6} \times 0.1 \times (0.2)^2$$
$$= 2.0 \times 10^{-6} \times 0.1 \times 0.04$$
$$= 2.0 \times 10^{-6} \times 4 \times 10^{-3}$$
$$= 8.0 \times 10^{-9}\,\text{mol L}^{-1}\text{s}^{-1}$$

Part 2 – Rate when [A] is reduced to 0.06 mol L⁻¹:

The reaction is $2\text{A} + \text{B} \rightarrow \text{A}_2\text{B}$.

Decrease in $[A] = 0.1 - 0.06 = 0.04\,\text{mol L}^{-1}$

Since stoichiometry: 2 mol A reacts with 1 mol B,
$$\text{Decrease in }[B] = \frac{0.04}{2} = 0.02\,\text{mol L}^{-1}$$
$$[B]_{\text{new}} = 0.2 - 0.02 = 0.18\,\text{mol L}^{-1}$$

$$r_2 = k[A]_{\text{new}}[B]_{\text{new}}^2 = 2.0 \times 10^{-6} \times 0.06 \times (0.18)^2$$
$$= 2.0 \times 10^{-6} \times 0.06 \times 0.0324$$
$$= 2.0 \times 10^{-6} \times 1.944 \times 10^{-3}$$
$$= 3.888 \times 10^{-9}\,\text{mol L}^{-1}\text{s}^{-1}$$
$$\approx 3.89 \times 10^{-9}\,\text{mol L}^{-1}\text{s}^{-1}$$

Answer:
- Initial rate $= 8.0 \times 10^{-9}\,\text{mol L}^{-1}\text{s}^{-1}$
- Rate after $[A]$ reduces to $0.06\,\text{mol L}^{-1}$ $= 3.89 \times 10^{-9}\,\text{mol L}^{-1}\text{s}^{-1}$

Given:
- Reaction: $2\text{NH}_3 \xrightarrow{\text{Pt}} \text{N}_2 + 3\text{H}_2$
- Zero order reaction, $k = 2.5 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$

For zero order: Rate $= k = 2.5 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$

This is the rate of disappearance of $\text{NH}_3$:
$$-\frac{1}{2}\frac{d[\text{NH}_3]}{dt} = k$$

Rate of production of N₂:
$$\frac{d[\text{N}_2]}{dt} = \frac{k}{2} \times 2 \cdot \frac{1}{2}$$

Using stoichiometry: $\text{Rate} = -\frac{1}{2}\frac{d[\text{NH}_3]}{dt} = \frac{d[\text{N}_2]}{dt} = \frac{1}{3}\frac{d[\text{H}_2]}{dt}$

So:
$$\frac{d[\text{N}_2]}{dt} = k = 2.5 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$$

Rate of production of H₂:
$$\frac{d[\text{H}_2]}{dt} = 3k = 3 \times 2.5 \times 10^{-4} = 7.5 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$$

Answer:
- Rate of production of $\text{N}_2 = 2.5 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$
- Rate of production of $\text{H}_2 = 7.5 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$

Given:
- Rate $= k\,(p_{\text{CH}_3\text{OCH}_3})^{3/2}$
- Pressure in bar, time in minutes

Units of Rate:
$$\text{Rate} = \frac{\text{change in pressure}}{\text{time}} = \text{bar min}^{-1}$$

Units of rate constant $k$:
$$k = \frac{\text{Rate}}{(\text{pressure})^{3/2}} = \frac{\text{bar min}^{-1}}{(\text{bar})^{3/2}} = \text{bar}^{1-3/2}\,\text{min}^{-1} = \text{bar}^{-1/2}\,\text{min}^{-1}$$

Answer:
- Units of rate $= \text{bar min}^{-1}$
- Units of rate constant $k = \text{bar}^{-1/2}\,\text{min}^{-1}$

The following factors affect the rate of a chemical reaction:

1. Concentration of reactants: Rate generally increases with increase in concentration of reactants (more molecules available for collision).

2. Temperature: Rate increases with increase in temperature. For most reactions, rate nearly doubles for every $10\,\text{K}$ rise in temperature (Arrhenius equation: $k = Ae^{-E_a/RT}$).

3. Presence of a catalyst: A catalyst provides an alternate pathway with lower activation energy, thereby increasing the rate of reaction without being consumed.

4. Nature of reactants: Physical state, surface area (for heterogeneous reactions), and bond types influence the rate.

5. Pressure (for gaseous reactions): Increasing pressure increases concentration of gaseous reactants, thus increasing the rate.

6. Surface area: For heterogeneous reactions, finely divided solids (larger surface area) react faster.

Given: Rate $= k[A]^2$ (second order with respect to reactant A)

(i) When concentration is doubled ($[A]_{\text{new}} = 2[A]$):
$$r_{\text{new}} = k(2[A])^2 = 4k[A]^2 = 4\,r$$
$$\frac{r_{\text{new}}}{r} = 4$$
The rate becomes 4 times the original rate.

(ii) When concentration is reduced to half ($[A]_{\text{new}} = \dfrac{[A]}{2}$):
$$r_{\text{new}} = k\left(\frac{[A]}{2}\right)^2 = \frac{k[A]^2}{4} = \frac{r}{4}$$
$$\frac{r_{\text{new}}}{r} = \frac{1}{4}$$
The rate becomes one-fourth of the original rate.

Effect of temperature on rate constant:

The rate constant of a reaction increases with increase in temperature. Experimentally, it is found that for most reactions, the rate constant approximately doubles for every 10 K rise in temperature.

Quantitative representation — Arrhenius Equation:

The effect of temperature on rate constant is given by the Arrhenius equation:
$$k = A\,e^{-E_a/RT}$$

where:
- $k$ = rate constant
- $A$ = Arrhenius factor (pre-exponential factor or frequency factor)
- $E_a$ = activation energy (J mol⁻¹)
- $R$ = gas constant $= 8.314\,\text{J K}^{-1}\text{mol}^{-1}$
- $T$ = absolute temperature (K)

Taking logarithm:
$$\ln k = \ln A - \frac{E_a}{RT}$$
$$\log k = \log A - \frac{E_a}{2.303\,RT}$$

For two temperatures $T_1$ and $T_2$:
$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$$

As $T$ increases, $e^{-E_a/RT}$ increases, so $k$ increases.

Given:
- At $t = 30\,\text{s}$: $[A] = 0.31\,\text{mol L}^{-1}$
- At $t = 60\,\text{s}$: $[A] = 0.17\,\text{mol L}^{-1}$

Formula:
$$\text{Average rate} = -\frac{\Delta[A]}{\Delta t} = -\frac{[A]_{60} - [A]_{30}}{60 - 30}$$

Calculation:
$$\text{Average rate} = -\frac{(0.17 - 0.31)\,\text{mol L}^{-1}}{(60 - 30)\,\text{s}}$$
$$= -\frac{-0.14}{30}\,\text{mol L}^{-1}\text{s}^{-1}$$
$$= \frac{0.14}{30} = 4.67 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}$$

Answer: Average rate $= 4.67 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}$

Given: First order in A, second order in B.

(i) Differential rate equation:
$$\text{Rate} = -\frac{d[A]}{dt} = k[A]^1[B]^2 = k[A][B]^2$$

(ii) Effect of tripling [B]:

Original rate: $r = k[A][B]^2$

New rate when $[B]_{\text{new}} = 3[B]$:
$$r_{\text{new}} = k[A](3[B])^2 = 9k[A][B]^2 = 9\,r$$

The rate increases 9 times.

(iii) Effect of doubling both [A] and [B]:

New rate when $[A]_{\text{new}} = 2[A]$ and $[B]_{\text{new}} = 2[B]$:
$$r_{\text{new}} = k(2[A])(2[B])^2 = k \times 2[A] \times 4[B]^2 = 8k[A][B]^2 = 8\,r$$

The rate increases 8 times.

Let Rate $= k[A]^m[B]^n$

Finding order with respect to B (comparing experiments 1 and 2):

$[A]$ is same (0.20) in both; $[B]$ changes from 0.30 to 0.10.
$$\frac{r_1}{r_2} = \frac{k(0.20)^m(0.30)^n}{k(0.20)^m(0.10)^n} = \left(\frac{0.30}{0.10}\right)^n = 3^n$$
$$\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}} = 1 = 3^n$$
$$\Rightarrow n = 0$$

Order with respect to B = 0.

Finding order with respect to A (comparing experiments 2 and 3):

$[B]$ changes but since $n = 0$, it doesn't matter. Using experiments 2 and 3:
$$\frac{r_3}{r_2} = \frac{k(0.40)^m(0.05)^0}{k(0.20)^m(0.10)^0} = \left(\frac{0.40}{0.20}\right)^m = 2^m$$
$$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = 2.82 \approx 2^m$$
$$2^m \approx 2.82 \approx 2^{1.5}$$
$$\Rightarrow m \approx 1.5$$

Verification: $2^{1.5} = 2\sqrt{2} = 2.828 \approx 2.82$ ✓

Answer:
- Order with respect to A $= 1.5$ (or $\dfrac{3}{2}$)
- Order with respect to B $= 0$
- Overall order $= 1.5$

Let Rate $= k[A]^m[B]^n$

Finding order with respect to B (comparing experiments II and III):
$[A]$ is same (0.3):
$$\frac{r_{III}}{r_{II}} = \left(\frac{[B]_{III}}{[B]_{II}}\right)^n = \left(\frac{0.4}{0.2}\right)^n = 2^n$$
$$\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = 4 = 2^n \Rightarrow n = 2$$

Order with respect to B = 2.

Finding order with respect to A (comparing experiments I and IV):
$[B]$ is same (0.1):
$$\frac{r_{IV}}{r_I} = \left(\frac{[A]_{IV}}{[A]_I}\right)^m = \left(\frac{0.4}{0.1}\right)^m = 4^m$$
$$\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = 4 = 4^m \Rightarrow m = 1$$

Order with respect to A = 1.

Rate law: $\text{Rate} = k[A][B]^2$

Calculating rate constant $k$ (using Experiment I):
$$k = \frac{\text{Rate}}{[A][B]^2} = \frac{6.0 \times 10^{-3}}{0.1 \times (0.1)^2} = \frac{6.0 \times 10^{-3}}{0.1 \times 0.01} = \frac{6.0 \times 10^{-3}}{10^{-3}} = 6.0\,\text{mol}^{-2}\text{L}^2\text{min}^{-1}$$

Verification with Experiment II:
$$k = \frac{7.2 \times 10^{-2}}{0.3 \times (0.2)^2} = \frac{7.2 \times 10^{-2}}{0.3 \times 0.04} = \frac{7.2 \times 10^{-2}}{0.012} = 6.0\,\text{mol}^{-2}\text{L}^2\text{min}^{-1}$$ ✓

Answer:
- Rate law: $\text{Rate} = k[A][B]^2$
- $k = 6.0\,\text{mol}^{-2}\text{L}^2\text{min}^{-1}$

Rate law: Rate $= k[A]^1[B]^0 = k[A]$

Finding k from Experiment I:
$$k = \frac{\text{Rate}}{[A]} = \frac{2.0 \times 10^{-2}}{0.1} = 0.2\,\text{min}^{-1}$$

Experiment II: Rate $= 4.0 \times 10^{-2}$, $[B] = 0.2$
$$[A] = \frac{\text{Rate}}{k} = \frac{4.0 \times 10^{-2}}{0.2} = 0.2\,\text{mol L}^{-1}$$

Experiment III: $[A] = 0.4$, $[B] = 0.4$
$$\text{Rate} = k[A] = 0.2 \times 0.4 = 8.0 \times 10^{-2}\,\text{mol L}^{-1}\text{min}^{-1}$$

Experiment IV: Rate $= 2.0 \times 10^{-2}$, $[B] = 0.2$
$$[A] = \frac{\text{Rate}}{k} = \frac{2.0 \times 10^{-2}}{0.2} = 0.1\,\text{mol L}^{-1}$$

Completed Table:

| Experiment | [A]/mol L⁻¹ | [B]/mol L⁻¹ | Rate/mol L⁻¹ min⁻¹ |
|---|---|---|---|
| I | 0.1 | 0.1 | $2.0 \times 10^{-2}$ |
| II | 0.2 | 0.2 | $4.0 \times 10^{-2}$ |
| III | 0.4 | 0.4 | $\mathbf{8.0 \times 10^{-2}}$ |
| IV | 0.1 | 0.2 | $2.0 \times 10^{-2}$ |

Formula for first order half-life:
$$t_{1/2} = \frac{0.693}{k}$$

(i) $k = 200\,\text{s}^{-1}$:
$$t_{1/2} = \frac{0.693}{200} = 3.465 \times 10^{-3}\,\text{s} \approx 3.47 \times 10^{-3}\,\text{s}$$

(ii) $k = 2\,\text{min}^{-1}$:
$$t_{1/2} = \frac{0.693}{2} = 0.3465\,\text{min} \approx 0.35\,\text{min}$$

(iii) $k = 4\,\text{years}^{-1}$:
$$t_{1/2} = \frac{0.693}{4} = 0.173\,\text{years}$$

Answers:
- (i) $t_{1/2} = 3.47 \times 10^{-3}\,\text{s}$
- (ii) $t_{1/2} = 0.35\,\text{min}$
- (iii) $t_{1/2} = 0.173\,\text{years}$

Given:
- $t_{1/2} = 5730\,\text{years}$
- Remaining $^{14}\text{C} = 80\%$ of original, so $[R] = 0.80\,[R]_0$

Step 1: Find rate constant $k$:
$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} = 1.209 \times 10^{-4}\,\text{year}^{-1}$$

Step 2: Use first order integrated rate equation:
$$t = \frac{2.303}{k}\log\frac{[R]_0}{[R]} = \frac{2.303}{1.209 \times 10^{-4}}\log\frac{[R]_0}{0.80[R]_0}$$
$$= \frac{2.303}{1.209 \times 10^{-4}} \times \log\left(\frac{1}{0.80}\right)$$
$$= \frac{2.303}{1.209 \times 10^{-4}} \times \log(1.25)$$
$$= \frac{2.303}{1.209 \times 10^{-4}} \times 0.09691$$
$$= 19035 \times 0.09691$$
$$\approx 1845\,\text{years}$$

Answer: The age of the sample is approximately 1845 years.

Data Table:

| $t$/s | $[\text{N}_2\text{O}_5] \times 10^2$/mol L⁻¹ | $\log[\text{N}_2\text{O}_5]$ |
|---|---|---|
| 0 | 1.63 | $-1.788$ |
| 400 | 1.36 | $-1.866$ |
| 800 | 1.14 | $-1.943$ |
| 1200 | 0.93 | $-2.031$ |
| 1600 | 0.78 | $-2.108$ |
| 2000 | 0.64 | $-2.194$ |
| 2400 | 0.53 | $-2.276$ |
| 2800 | 0.43 | $-2.367$ |
| 3200 | 0.35 | $-2.456$ |

(i) Plot [N₂O₅] vs t:
Plot $[\text{N}_2\text{O}_5]$ on y-axis and $t$ on x-axis. The curve is exponential (decreasing), confirming first order kinetics.

(ii) Half-life from graph:
Initial concentration $= 1.63 \times 10^{-2}\,\text{mol L}^{-1}$
Half of initial $= 0.815 \times 10^{-2}\,\text{mol L}^{-1}$
From the graph, at $[\text{N}_2\text{O}_5] = 0.815 \times 10^{-2}$, $t \approx 1440\,\text{s}$
$$t_{1/2} \approx 1440\,\text{s}$$

(iii) Graph of log[N₂O₅] vs t:
The values of $\log[\text{N}_2\text{O}_5]$ decrease linearly with time, confirming first order reaction. Plot $\log[\text{N}_2\text{O}_5]$ on y-axis and $t$ on x-axis — a straight line is obtained.

(iv) Rate law:
Since $\log[\text{N}_2\text{O}_5]$ vs $t$ is linear, the reaction is first order.
$$\text{Rate} = k[\text{N}_2\text{O}_5]$$

(v) Calculating rate constant $k$:
For first order: slope of $\log[\text{N}_2\text{O}_5]$ vs $t$ graph $= -k/2.303$

Using two points (0 s and 3200 s):
$$\text{slope} = \frac{-2.456 - (-1.788)}{3200 - 0} = \frac{-0.668}{3200} = -2.088 \times 10^{-4}\,\text{s}^{-1}$$
$$k = -2.303 \times \text{slope} = 2.303 \times 2.088 \times 10^{-4}$$
$$k = 4.81 \times 10^{-4}\,\text{s}^{-1}$$

(vi) Half-life from k:
$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4.81 \times 10^{-4}} = 1440\,\text{s}$$

Comparison: The half-life calculated from $k$ ($\approx 1440\,\text{s}$) matches the value obtained from the graph in part (ii). This confirms the first order nature of the reaction.

Given:
- $k = 60\,\text{s}^{-1}$
- $[R] = \dfrac{[R]_0}{16}$, i.e., $\dfrac{[R]_0}{[R]} = 16$

Formula:
$$t = \frac{2.303}{k}\log\frac{[R]_0}{[R]}$$

Calculation:
$$t = \frac{2.303}{60}\log 16$$
$$= \frac{2.303}{60} \times \log 2^4$$
$$= \frac{2.303}{60} \times 4 \times \log 2$$
$$= \frac{2.303}{60} \times 4 \times 0.3010$$
$$= \frac{2.303 \times 1.204}{60}$$
$$= \frac{2.773}{60} = 4.62 \times 10^{-2}\,\text{s}$$

Answer: Time required $= 4.62 \times 10^{-2}\,\text{s}$

Given:
- $t_{1/2} = 28.1\,\text{years}$
- Initial amount $= 1\,\mu\text{g}$

Step 1: Find rate constant:
$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{28.1} = 2.466 \times 10^{-2}\,\text{year}^{-1}$$

After 10 years:
$$t = \frac{2.303}{k}\log\frac{[R]_0}{[R]}$$
$$\log\frac{[R]_0}{[R]} = \frac{k \times t}{2.303} = \frac{2.466 \times 10^{-2} \times 10}{2.303} = \frac{0.2466}{2.303} = 0.1071$$
$$\frac{[R]_0}{[R]} = \text{antilog}(0.1071) = 1.28$$
$$[R] = \frac{1}{1.28} = 0.7813\,\mu\text{g}$$

After 60 years:
$$\log\frac{[R]_0}{[R]} = \frac{2.466 \times 10^{-2} \times 60}{2.303} = \frac{1.4796}{2.303} = 0.6425$$
$$\frac{[R]_0}{[R]} = \text{antilog}(0.6425) = 4.39$$
$$[R] = \frac{1}{4.39} = 0.2278\,\mu\text{g}$$

Answer:
- Amount remaining after 10 years $\approx 0.7813\,\mu\text{g}$
- Amount remaining after 60 years $\approx 0.2278\,\mu\text{g}$

For first order reaction:
$$t = \frac{2.303}{k}\log\frac{[R]_0}{[R]}$$

Time for 90% completion ($t_{90}$):

If 90% is completed, 10% remains: $[R] = 0.10\,[R]_0$
$$t_{90} = \frac{2.303}{k}\log\frac{[R]_0}{0.10[R]_0} = \frac{2.303}{k}\log 10 = \frac{2.303}{k} \times 1 = \frac{2.303}{k}$$

Time for 99% completion ($t_{99}$):

If 99% is completed, 1% remains: $[R] = 0.01\,[R]_0$
$$t_{99} = \frac{2.303}{k}\log\frac{[R]_0}{0.01[R]_0} = \frac{2.303}{k}\log 100 = \frac{2.303}{k} \times 2 = \frac{4.606}{k}$$

Ratio:
$$\frac{t_{99}}{t_{90}} = \frac{4.606/k}{2.303/k} = 2$$

$$\boxed{t_{99} = 2\,t_{90}}$$

Hence proved that time required for 99% completion is twice the time required for 90% completion.

Given:
- $t = 40\,\text{min}$
- 30% decomposed, so 70% remains: $[R] = 0.70\,[R]_0$

Step 1: Find k:
$$k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} = \frac{2.303}{40}\log\frac{[R]_0}{0.70[R]_0}$$
$$= \frac{2.303}{40}\log\left(\frac{1}{0.70}\right) = \frac{2.303}{40} \times \log(1.4286)$$
$$= \frac{2.303}{40} \times 0.1549$$
$$= \frac{0.3567}{40} = 8.918 \times 10^{-3}\,\text{min}^{-1}$$

Step 2: Calculate $t_{1/2}$:
$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.918 \times 10^{-3}} = 77.7\,\text{min}$$

Answer: $t_{1/2} \approx 77.7\,\text{min}$

Reaction: $(\text{CH}_3)_2\text{CHN}=\text{NCH}(\text{CH}_3)_2 \rightarrow \text{N}_2 + \text{C}_6\text{H}_{14}$

For every mole of azoisopropane decomposed, 2 moles of products are formed.

Let initial pressure $= P_0 = 35.0\,\text{mm Hg}$, and pressure decrease of azoisopropane at time $t = p$.

$$P_{\text{total}} = (P_0 - p) + p + p = P_0 + p$$
$$p = P_{\text{total}} - P_0$$
$$P_{\text{azoisopropane}} = P_0 - p = P_0 - (P_{\text{total}} - P_0) = 2P_0 - P_{\text{total}}$$

At $t = 360\,\text{s}$:
$$P_{\text{azoisopropane}} = 2(35.0) - 54.0 = 70.0 - 54.0 = 16.0\,\text{mm Hg}$$

At $t = 720\,\text{s}$:
$$P_{\text{azoisopropane}} = 2(35.0) - 63.0 = 70.0 - 63.0 = 7.0\,\text{mm Hg}$$

Using first order rate equation:
$$k = \frac{2.303}{t}\log\frac{P_0}{P_t}$$

At $t = 360\,\text{s}$:
$$k = \frac{2.303}{360}\log\frac{35.0}{16.0} = \frac{2.303}{360} \times \log(2.1875)$$
$$= \frac{2.303}{360} \times 0.3398 = 2.175 \times 10^{-3}\,\text{s}^{-1}$$

At $t = 720\,\text{s}$:
$$k = \frac{2.303}{720}\log\frac{35.0}{7.0} = \frac{2.303}{720} \times \log 5$$
$$= \frac{2.303}{720} \times 0.6990 = 2.235 \times 10^{-3}\,\text{s}^{-1}$$

Average $k$:
$$k \approx \frac{(2.175 + 2.235) \times 10^{-3}}{2} \approx 2.21 \times 10^{-3}\,\text{s}^{-1}$$

Answer: $k \approx 2.21 \times 10^{-3}\,\text{s}^{-1}$

Reaction: $\text{SO}_2\text{Cl}_2(\text{g}) \rightarrow \text{SO}_2(\text{g}) + \text{Cl}_2(\text{g})$

Let initial pressure of $\text{SO}_2\text{Cl}_2 = P_0 = 0.5\,\text{atm}$, and decrease in pressure $= p$ at time $t$.

$$P_{\text{total}} = (P_0 - p) + p + p = P_0 + p$$
$$p = P_{\text{total}} - P_0$$
$$P_{\text{SO}_2\text{Cl}_2} = P_0 - p = 2P_0 - P_{\text{total}}$$

At $t = 100\,\text{s}$, $P_{\text{total}} = 0.6\,\text{atm}$:
$$P_{\text{SO}_2\text{Cl}_2} = 2(0.5) - 0.6 = 1.0 - 0.6 = 0.4\,\text{atm}$$

Finding rate constant $k$:
$$k = \frac{2.303}{t}\log\frac{P_0}{P_t} = \frac{2.303}{100}\log\frac{0.5}{0.4}$$
$$= \frac{2.303}{100} \times \log(1.25) = \frac{2.303}{100} \times 0.09691$$
$$= 2.231 \times 10^{-3}\,\text{s}^{-1}$$

When total pressure $= 0.65\,\text{atm}$:
$$P_{\text{SO}_2\text{Cl}_2} = 2(0.5) - 0.65 = 1.0 - 0.65 = 0.35\,\text{atm}$$

Rate of reaction:
$$\text{Rate} = k \times P_{\text{SO}_2\text{Cl}_2} = 2.231 \times 10^{-3} \times 0.35$$
$$= 7.81 \times 10^{-4}\,\text{atm s}^{-1}$$

Answer: Rate of reaction $= 7.81 \times 10^{-4}\,\text{atm s}^{-1}$

Converting temperatures to Kelvin and calculating ln k:

| T/°C | T/K | $1/T$ (K⁻¹) | $k \times 10^5$/s⁻¹ | $\ln k$ |
|---|---|---|---|---|
| 0 | 273 | $3.663 \times 10^{-3}$ | 0.0787 | $-11.75$ |
| 20 | 293 | $3.413 \times 10^{-3}$ | 1.70 | $-10.98$ |
| 40 | 313 | $3.195 \times 10^{-3}$ | 25.7 | $-10.57$ |
| 60 | 333 | $3.003 \times 10^{-3}$ | 178 | $-8.63$ |
| 80 | 353 | $2.833 \times 10^{-3}$ | 2140 | $-6.15$ |

Graph: Plot $\ln k$ (y-axis) vs $1/T$ (x-axis) — a straight line with negative slope.

From Arrhenius equation:
$$\ln k = \ln A - \frac{E_a}{RT}$$

Slope $= -E_a/R$

Calculating slope (using extreme points):
$$\text{slope} = \frac{-6.15 - (-11.75)}{(2.833 - 3.663) \times 10^{-3}} = \frac{5.60}{-0.830 \times 10^{-3}} = -6.747 \times 10^3\,\text{K}$$

$$E_a = -\text{slope} \times R = 6.747 \times 10^3 \times 8.314 = 56,092\,\text{J mol}^{-1} \approx 56.1\,\text{kJ mol}^{-1}$$

Finding A (using $T = 273\,\text{K}$, $k = 0.0787 \times 10^{-5}\,\text{s}^{-1}$):
$$\ln A = \ln k + \frac{E_a}{RT} = -11.75 + \frac{56092}{8.314 \times 273}$$
$$= -11.75 + \frac{56092}{2269.7} = -11.75 + 24.72 = 12.97$$
$$A = e^{12.97} \approx 4.3 \times 10^5\,\text{s}^{-1}$$

Rate constant at 30°C (303 K):
$$\log k = \log A - \frac{E_a}{2.303RT} = \log(4.3 \times 10^5) - \frac{56092}{2.303 \times 8.314 \times 303}$$
$$= 5.633 - \frac{56092}{5800.6} = 5.633 - 9.67 = -4.037$$
$$k_{30} = 10^{-4.037} \approx 9.16 \times 10^{-5}\,\text{s}^{-1}$$

Rate constant at 50°C (323 K):
$$\log k = 5.633 - \frac{56092}{2.303 \times 8.314 \times 323} = 5.633 - \frac{56092}{6183.5} = 5.633 - 9.07 = -3.437$$
$$k_{50} = 10^{-3.437} \approx 3.65 \times 10^{-4}\,\text{s}^{-1}$$

Answer:
- $E_a \approx 56.1\,\text{kJ mol}^{-1}$
- $A \approx 4.3 \times 10^5\,\text{s}^{-1}$
- $k_{30°C} \approx 9.16 \times 10^{-5}\,\text{s}^{-1}$
- $k_{50°C} \approx 3.65 \times 10^{-4}\,\text{s}^{-1}$