Chapter 6: Chemical Kinetics
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: The number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction, is called the molecularity of a reaction.
Answer: The unit of the rate constant for a first-order reaction is $\text{time}^{-1}$ (e.g., $\text{s}^{-1}$, $\text{min}^{-1}$, or $\text{hour}^{-1}$).
Answer: A reaction which has higher true molecularity (or order) but behaves as a first-order reaction under certain experimental conditions (usually when one of the reactants is present in large excess) is called a pseudo-first-order reaction.
Answer: The half-life period ($t_{1/2}$) of a reaction is defined as the time required for the initial concentration of the reactant to reduce to exactly half of its original value.
2 Mark Questions (Short Answer-I)
Answer:
| Order of Reaction | Molecularity of Reaction |
|---|---|
| It is the sum of the powers of concentration terms in the experimentally determined rate law. | It is the number of reacting species participating in an elementary reaction. |
| It is an experimentally determined value. | It is a theoretical concept derived from the reaction mechanism. |
| Order can be zero, fraction, or integer. | Molecularity is always a whole number (1, 2, 3...) and can never be zero or a fraction. |
| Applicable to both elementary and complex reactions. | Applicable only to elementary (single-step) reactions. Complex reactions have no overall molecularity. |
Answer:
1. The integrated rate law for a first-order reaction is:
$$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$$
2. At half-life, $t = t_{1/2}$, the concentration of reactant remaining is half of the initial concentration, i.e., $[A]_t = \frac{[A]_0}{2}$.
3. Substitute these values into the rate law:
$$k = \frac{2.303}{t_{1/2}} \log_{10} \frac{[A]_0}{[A]_0 / 2}$$
$$k = \frac{2.303}{t_{1/2}} \log_{10} 2$$
4. Since $\log_{10} 2 = 0.3010$, we get:
$$k = \frac{2.303 \times 0.3010}{t_{1/2}}$$
$$k = \frac{0.693}{t_{1/2}} \implies t_{1/2} = \frac{0.693}{k}$$
Conclusion: The final expression for $t_{1/2}$ contains only the rate constant $k$ and the constant 0.693. It does not contain the initial concentration term $[A]_0$. Hence, proved.
3 Mark Questions (Short Answer-II / Derivations)
Answer:
1. Consider a general first-order reaction: $A \rightarrow \text{Products}$.
2. The differential rate law is given by:
$$\text{Rate} = -\frac{d[A]}{dt} = k[A]^1$$
Where $[A]$ is the concentration of reactant at time $t$, and $k$ is the first-order rate constant.
3. Rearranging the variables, we get:
$$\frac{d[A]}{[A]} = -k \cdot dt$$
4. Let the initial concentration of $A$ at $t = 0$ be $[A]_0$, and the concentration at time $t = t$ be $[A]_t$. Integrating both sides within these limits:
$$\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt$$
5. We know that $\int \frac{dx}{x} = \ln x$. Thus:
$$[\ln [A]]_{[A]_0}^{[A]_t} = -k [t]_0^t$$
$$\ln [A]_t - \ln [A]_0 = -k(t - 0)$$
$$\ln \left( \frac{[A]_t}{[A]_0} \right) = -kt$$
$$kt = \ln \left( \frac{[A]_0}{[A]_t} \right)$$
6. Converting natural logarithm ($\ln$) to base-10 logarithm ($\log_{10}$) by multiplying with 2.303:
$$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$$
This is the required integrated rate law equation for a first-order reaction.
Answer:
Given Data:
Time ($t$) = 40 minutes
Decomposition = 30%
Let the initial concentration $[A]_0 = 100$.
Since 30% is decomposed, the amount remaining at time $t$, $[A]_t = 100 - 30 = 70$.
Step 1: Calculate Rate Constant ($k$)
Using the integrated rate law for first-order reaction:
$$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$$
$$k = \frac{2.303}{40} \log_{10} \left( \frac{100}{70} \right)$$
$$k = \frac{2.303}{40} \log_{10} (1.428)$$
From log tables, $\log_{10} (1.428) = 0.1548$
$$k = \frac{2.303 \times 0.1548}{40}$$
$$k = \frac{0.3565}{40} = 0.00891 \text{ min}^{-1}$$
Step 2: Calculate Half-Life ($t_{1/2}$)
$$t_{1/2} = \frac{0.693}{k}$$
$$t_{1/2} = \frac{0.693}{0.00891}$$
$$t_{1/2} = 77.77 \text{ minutes}$$
The half-life period of the reaction is 77.77 minutes.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 3-mark numerical combined with a 1-mark theory question).
(b) Define: Activation Energy. [1 Mark] Oct 2014, March 2019
Answer (a): Numerical on Arrhenius Equation
Given Data:
$T_1 = 30^\circ\text{C} = 30 + 273 = 303 \text{ K}$, $\quad k_1 = 1.2 \times 10^{-3} \text{ s}^{-1}$
$T_2 = 40^\circ\text{C} = 40 + 273 = 313 \text{ K}$, $\quad k_2 = 2.1 \times 10^{-3} \text{ s}^{-1}$
$R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}$
Formula:
$$\log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$$
Calculation:
$$\log_{10} \left( \frac{2.1 \times 10^{-3}}{1.2 \times 10^{-3}} \right) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{313 - 303}{303 \times 313} \right]$$
$$\log_{10} \left( \frac{2.1}{1.2} \right) = \frac{E_a}{19.147} \left[ \frac{10}{94839} \right]$$
$$\log_{10} (1.75) = \frac{10 \cdot E_a}{1815882.3}$$
From log tables, $\log_{10}(1.75) = 0.2430$.
$$0.2430 = \frac{10 \cdot E_a}{1815882.3}$$
$$10 \cdot E_a = 0.2430 \times 1815882.3$$
$$10 \cdot E_a = 441259.4$$
$$E_a = 44125.94 \text{ J mol}^{-1}$$
Converting to kiloJoules:
$$E_a = 44.13 \text{ kJ mol}^{-1}$$
Answer (b): Theory
Activation Energy ($E_a$): The minimum extra amount of energy that must be supplied to the reactant molecules so that their energy becomes equal to the threshold energy required to form the activated complex and undergo a chemical reaction.
(b) Give one example of a zero-order reaction. [1 Mark] March 2021, March 2023
Answer (a): Derivation/Proof
For a first-order reaction, the time $t$ is given by:
$$t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$$
Let initial concentration $[A]_0 = 100$.
Case 1: For 99.9% completion
Amount decomposed = 99.9
Amount remaining $[A]_t = 100 - 99.9 = 0.1$
$$t_{99.9\%} = \frac{2.303}{k} \log_{10} \left( \frac{100}{0.1} \right) = \frac{2.303}{k} \log_{10} (1000)$$
$$t_{99.9\%} = \frac{2.303}{k} \log_{10} (10^3) = \frac{2.303}{k} \times 3 \quad \text{--- (Eq 1)}$$
Case 2: For 90% completion
Amount decomposed = 90
Amount remaining $[A]_t = 100 - 90 = 10$
$$t_{90\%} = \frac{2.303}{k} \log_{10} \left( \frac{100}{10} \right) = \frac{2.303}{k} \log_{10} (10)$$
$$t_{90\%} = \frac{2.303}{k} \times 1 \quad \text{--- (Eq 2)}$$
Comparing Eq 1 and Eq 2:
$$t_{99.9\%} = 3 \times \left( \frac{2.303}{k} \times 1 \right)$$
$$t_{99.9\%} = 3 \times t_{90\%}$$
Hence proved.
Answer (b): Theory
Example of Zero-Order Reaction: The decomposition of gaseous ammonia on a hot platinum or tungsten surface at high pressure.
$$2NH_3(g) \xrightarrow{Pt, 1130K} N_2(g) + 3H_2(g)$$
Rate = $k[NH_3]^0 = k$.
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