Chemical Kinetics — NCERT Solutions

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