Chapter 6: Chemical Kinetics Mock Test
Time: 1 Hour | Maximum Marks: 25
- All questions are compulsory.
- Section A contains Q1 (Multiple Choice) and Q2 (Very Short Answer).
- Section B contains Short Answer Type I questions (2 marks each). Attempt any 4.
- Section C contains Short Answer Type II questions (3 marks each). Attempt any 2.
- Section D contains Long Answer questions (4 marks each). Attempt any 1.
- Use of logarithmic tables is allowed. Calculators are not permitted.
SECTION A
Q1. Select and write the most appropriate answer from the given alternatives: [4 Marks]
-
The unit of the rate constant ($k$) for a zero-order reaction is:
(A) $\text{s}^{-1}$(B) $\text{mol L}^{-1} \text{ s}^{-1}$(C) $\text{L mol}^{-1} \text{ s}^{-1}$(D) $\text{L}^2 \text{ mol}^{-2} \text{ s}^{-1}$
-
Which of the following about the molecularity of a reaction is true?
(A) It can be zero(B) It can be a fraction(C) It is always a whole number(D) It is an experimentally determined quantity
-
In the Arrhenius equation $k = A \cdot e^{-E_a/RT}$, the factor $e^{-E_a/RT}$ represents:
(A) Frequency of collisions(B) Fraction of molecules having energy $\ge E_a$(C) Activation energy(D) Pre-exponential factor
-
For a first-order reaction, the graph of $\log_{10}[A]_t$ against time ($t$) is a straight line with a slope equal to:
(A) $-k / 2.303$(B) $-k$(C) $k / 2.303$(D) $2.303 / k$
Q2. Answer the following questions in one sentence: [3 Marks]
- Define: Half-life period of a reaction.
- Give one example of a pseudo-first-order reaction.
- What is the overall order of radioactive decay processes?
SECTION B
Attempt any FOUR of the following: [8 Marks]
- Distinguish between the Order and Molecularity of a reaction. (Any 2 points).
- Derive the relationship between the half-life ($t_{1/2}$) and rate constant ($k$) for a first-order reaction.
- The rate of a first-order reaction is $1.5 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$ when the concentration of the reactant is 0.5 M. Calculate the rate constant of the reaction.
- Explain the effect of a catalyst on the activation energy and the rate of a chemical reaction.
- What is an elementary reaction? Give one example.
SECTION C
Attempt any TWO of the following: [6 Marks]
- Derive the integrated rate law equation for a first-order reaction.
- A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life period.
- Explain the graphical representation of a zero-order reaction by plotting (i) Concentration $[A]_t$ vs Time ($t$) and (ii) Rate vs Concentration $[A]_t$.
SECTION D
Attempt any ONE of the following: [4 Marks]
- (a) The rate constant of a reaction is $1.2 \times 10^{-3} \text{ s}^{-1}$ at 303 K and $2.1 \times 10^{-3} \text{ s}^{-1}$ at 313 K. Calculate the activation energy ($E_a$) of the reaction. ($R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}$). [3 Marks]
(b) Define: Activation Energy. [1 Mark] - (a) Show that the time required for 99.9% completion of a first-order reaction is three times the time required for 90% completion. [3 Marks]
(b) Write the integrated rate law for a zero-order reaction. [1 Mark]
Solutions & Marking Scheme
SECTION A [7 Marks]
Q1. Multiple Choice Answers:
1. (B) $\text{mol L}^{-1} \text{ s}^{-1}$ [1 Mark for correct option]
2. (C) It is always a whole number [1 Mark for correct option]
3. (B) Fraction of molecules having energy $\ge E_a$ [1 Mark for correct option]
4. (A) $-k / 2.303$ [1 Mark. From equation $\log_{10}[A]_t = \frac{-k}{2.303}t + \log_{10}[A]_0$]
Q2. Very Short Answers:
1. Half-life period:
It is defined as the time required for the initial concentration of the reactant to reduce to exactly half of its original value. [1 Mark for correct definition]
2. Pseudo-first-order reaction example:
Acid-catalyzed hydrolysis of an ester (e.g., Ethyl acetate) OR Inversion of cane sugar. [1 Mark for correct example]
3. Order of radioactive decay:
All natural and artificial radioactive decay processes follow First-order kinetics. [1 Mark for correct answer]
SECTION B [8 Marks]
Q3. Order vs Molecularity:
| Order of Reaction | Molecularity of Reaction |
|---|---|
| It is the sum of powers of concentration terms in the 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 mechanism. |
| It can be zero, fraction, or integer. | It is always a whole number (cannot be zero). |
[1 Mark for each correct point of distinction. Total 2 Marks]
Q4. Derivation of $t_{1/2}$ for first-order:
Integrated rate law: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$ [1/2 Mark]
At half-life, $t = t_{1/2}$ and $[A]_t = \frac{[A]_0}{2}$. [1/2 Mark]
Substitute in formula: $k = \frac{2.303}{t_{1/2}} \log_{10} \frac{[A]_0}{[A]_0/2} = \frac{2.303}{t_{1/2}} \log_{10} 2$ [1/2 Mark]
Since $\log_{10} 2 = 0.3010$, $t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k}$ [1/2 Mark]
Q5. Numerical on Rate Law:
Given: First-order reaction, Rate = $1.5 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$, $[A] = 0.5 \text{ M}$. [1/2 Mark]
Formula: Rate $= k \cdot [A]^1$ (for first order) $\implies k = \frac{\text{Rate}}{[A]}$ [1/2 Mark]
Calculation: $k = \frac{1.5 \times 10^{-2}}{0.5}$ [1/2 Mark]
Answer: $k = 3.0 \times 10^{-2} \text{ s}^{-1}$ [1/2 Mark for correct answer with unit]
Q6. Effect of Catalyst:
A catalyst provides an alternative reaction pathway or mechanism that has a lower activation energy ($E_a$). [1 Mark]
Because the activation energy is lowered, a larger fraction of molecules possess enough energy to cross the energy barrier. Consequently, the rate of the reaction increases. [1 Mark]
Q7. Elementary Reaction:
Definition: A reaction that occurs in a single step with no intermediate products formed is called an elementary reaction. [1 Mark]
Example: $O_3(g) \rightarrow O_2(g) + O(g)$ [1 Mark for any valid elementary reaction]
SECTION C [6 Marks]
Q8. Integrated Rate Law (First-Order):
1. For $A \rightarrow \text{Products}$, Rate $= -\frac{d[A]}{dt} = k[A]$. [1 Mark]
2. Rearranging: $\frac{d[A]}{[A]} = -k \cdot dt$. [1/2 Mark]
3. Integrating within limits $[A]_0$ to $[A]_t$ and $0$ to $t$:
$\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt$ [1/2 Mark]
4. $[\ln[A]]_{[A]_0}^{[A]_t} = -k[t]_0^t \implies \ln[A]_t - \ln[A]_0 = -kt \implies \ln\left(\frac{[A]_0}{[A]_t}\right) = kt$ [1/2 Mark]
5. Converting to log base 10: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$. [1/2 Mark]
Q9. First-Order Numerical ($t_{1/2}$):
Given: $t = 40 \text{ min}$, 30% decomposed. Let $[A]_0 = 100$. Then $[A]_t = 100 - 30 = 70$. [1/2 Mark]
Formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$ [1/2 Mark]
$k = \frac{2.303}{40} \log_{10}\left(\frac{100}{70}\right) = \frac{2.303}{40} \log_{10}(1.428) = \frac{2.303 \times 0.1548}{40} = 0.00891 \text{ min}^{-1}$ [1 Mark]
Half-life: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.00891}$ [1/2 Mark]
Answer: $t_{1/2} = 77.77 \text{ minutes}$. [1/2 Mark]
Q10. Graphical Representation (Zero-Order):
(i) $[A]_t$ vs $t$: Integrated rate law is $[A]_t = -kt + [A]_0$. This is in the form of $y = mx + c$. [1/2 Mark]
Plotting $[A]_t$ on y-axis and $t$ on x-axis gives a straight line with a negative slope (slope = $-k$) and y-intercept = $[A]_0$. [1 Mark for description or graph]
(ii) Rate vs $[A]_t$: For zero order, Rate = $k[A]^0 = k$. [1/2 Mark]
The rate is independent of concentration. Plotting Rate vs $[A]_t$ gives a straight horizontal line parallel to the x-axis (concentration axis). [1 Mark for description or graph]
SECTION D [4 Marks]
Q11. (a) Arrhenius Eq. Numerical [3 Marks] (b) Activation Energy [1 Mark]
(a) Numerical:
Given: $T_1 = 303\text{K}, k_1 = 1.2\times10^{-3}$; $T_2 = 313\text{K}, k_2 = 2.1\times10^{-3}$. [1/2 Mark]
Formula: $\log_{10}\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left[\frac{T_2 - T_1}{T_1 T_2}\right]$ [1/2 Mark]
$\log_{10}\left(\frac{2.1}{1.2}\right) = \frac{E_a}{2.303 \times 8.314}\left[\frac{313 - 303}{303 \times 313}\right]$ [1/2 Mark]
$\log_{10}(1.75) = \frac{E_a}{19.147}\left[\frac{10}{94839}\right] \implies 0.2430 = \frac{10 E_a}{1815882.3}$ [1/2 Mark]
$E_a = \frac{0.2430 \times 1815882.3}{10} = 44125.9 \text{ J/mol} = 44.13 \text{ kJ/mol}$. [1 Mark]
(b) Activation Energy ($E_a$): The minimum extra amount of energy required by reactant molecules to form the activated complex and undergo a chemical reaction. [1 Mark]
Q12. (a) 99.9% vs 90% Proof [3 Marks] (b) Zero Order Eq [1 Mark]
(a) Proof:
Formula: $t = \frac{2.303}{k}\log_{10}\frac{a}{a-x}$. Let $a = 100$.
For 99.9%: $x = 99.9$, so $a-x = 0.1$.
$t_{99.9\%} = \frac{2.303}{k}\log_{10}\left(\frac{100}{0.1}\right) = \frac{2.303}{k}\log_{10}(1000) = \frac{2.303}{k} \times 3 \quad \text{--- (1)}$ [1 Mark]
For 90%: $x = 90$, so $a-x = 10$.
$t_{90\%} = \frac{2.303}{k}\log_{10}\left(\frac{100}{10}\right) = \frac{2.303}{k}\log_{10}(10) = \frac{2.303}{k} \times 1 \quad \text{--- (2)}$ [1 Mark]
Comparing (1) and (2): $t_{99.9\%} = 3 \times t_{90\%}$. (Hence Proved). [1 Mark]
(b) Zero Order Integrated Rate Law: $k = \frac{[A]_0 - [A]_t}{t}$ OR $[A]_t = -kt + [A]_0$. [1 Mark]
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