Chapter 6: Chemical Kinetics
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. Define Chemical Kinetics.
Chemical Kinetics: It is the branch of physical chemistry that deals with the study of the rate (speed) of chemical reactions, the factors affecting the rates, and the mechanism by which the reactions proceed.
2. Define Rate of Reaction. State its SI unit.
Rate of Reaction: The change in the concentration of any one of the reactants or products per unit time is called the rate of the reaction.
Rate = Decrease in concentration of reactant / Time taken = Increase in concentration of product / Time taken
SI Unit: mol L$^{-1}$ s$^{-1}$ (moles per liter per second) or M s$^{-1}$. For gaseous reactions, it is atm s$^{-1}$.
3. Distinguish between Average Rate and Instantaneous Rate of a reaction.
- Average Rate: It is the change in the concentration of reactants or products divided by a finite, macroscopic time interval ($\Delta t$).
Average Rate = $\frac{\Delta [C]}{\Delta t}$ - Instantaneous Rate: It is the rate of reaction at a specific, exact instant of time. It is obtained when the time interval $\Delta t$ approaches zero.
Instantaneous Rate = $\lim_{\Delta t \to 0} \frac{\Delta [C]}{\Delta t} = \frac{d[C]}{dt}$
4. Write the expression for the rate of reaction for: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$.
The rate of reaction is expressed by dividing the rate of change of concentration of each substance by its stoichiometric coefficient in the balanced chemical equation.
$$\text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}$$
(Negative sign indicates consumption of reactants, positive indicates formation of products).
5. State the factors influencing the rate of a chemical reaction.
- Concentration of reactants: Rate generally increases with an increase in concentration.
- Temperature: Rate almost universally increases with a rise in temperature.
- Presence of a Catalyst: A positive catalyst increases the rate of reaction by providing an alternate path with lower activation energy.
- Surface Area: For solid reactants, greater surface area (e.g., powdered form) increases the rate.
- Nature of reactants: Physical state and chemical nature (bond strength) of reactants affect the rate.
6. State the Rate Law.
Rate Law: It is an experimentally determined mathematical expression that relates the rate of a reaction to the molar concentrations of the reactants, with each concentration term raised to some power (which may or may not be equal to the stoichiometric coefficients).
For $aA + bB \rightarrow \text{Products}$, Rate Law: $\text{Rate} = k[A]^x[B]^y$
Where $x$ and $y$ are experimentally determined.
7. Define Rate Constant (Specific Reaction Rate, $k$).
Rate Constant ($k$): In the rate law expression, if the concentration of all reactants is taken as unity (1 mol L$^{-1}$), then the rate of the reaction is equal to the rate constant.
If $\text{Rate} = k[A]^x[B]^y$, and $[A] = [B] = 1 \text{ M}$, then $\text{Rate} = k$.
It is independent of concentration but depends on temperature.
8. How does the rate constant ($k$) depend on temperature?
The rate constant ($k$) increases exponentially with an increase in temperature according to the Arrhenius equation:
$$k = A e^{-E_a/RT}$$
Generally, for every 10°C rise in temperature, the rate constant (and hence the rate of reaction) nearly doubles.
9. Define the Order of a reaction. Can it be fractional?
Order of Reaction: It is defined as the sum of the powers (exponents) of the concentration terms of reactants present in the experimentally determined rate law equation.
If $\text{Rate} = k[A]^x[B]^y$, then Overall Order = $x + y$.
Yes, the order of a reaction can be zero, a whole number, or a fraction. It is a purely experimental quantity.
10. Define Molecularity of a reaction. Can it be zero?
Molecularity: It is defined as the number of reactant molecules (or atoms or ions) taking part in an elementary reaction, which must collide simultaneously to bring about the chemical reaction.
No, molecularity can never be zero, negative, or fractional. It must be an integer ($1, 2, 3$). A molecularity of zero would mean no reactant molecules are colliding, which is impossible for a reaction to occur.
11. Distinguish between Order and Molecularity of a reaction. (Any 4 points)
| Order of Reaction | Molecularity of Reaction |
|---|---|
| Sum of powers of concentration terms in rate law. | Number of colliding molecules in an elementary step. |
| It is an experimentally determined value. | It is a theoretical concept based on reaction mechanism. |
| Can be zero, fractional, or an integer. | Can only be a whole number (1, 2, 3). Cannot be zero or fractional. |
| Applicable to both elementary and complex reactions. | Applicable only to elementary reactions (steps). |
12. What are elementary and complex reactions?
- Elementary Reactions: Reactions that occur in a single step and have no reactive intermediates. For these, order generally equals molecularity.
- Complex Reactions: Reactions that take place in a sequence of two or more elementary steps. The overall mechanism involves intermediates.
13. What is the Rate Determining Step (RDS) in a complex reaction?
In a complex reaction taking place in several sequential elementary steps, the slowest step governs the overall rate of the reaction. This slowest step is called the Rate Determining Step (RDS). The overall rate law is derived from this step.
14. Derive the units of rate constant ($k$) for a zero-order reaction.
Rate Law for zero order: $\text{Rate} = k[A]^0$
Since $[A]^0 = 1$, we get: $\text{Rate} = k$
Therefore, the unit of $k$ is the same as the unit of Rate.
Unit of $k$ = mol L$^{-1}$ s$^{-1}$ (or M s$^{-1}$).
15. Derive the units of rate constant ($k$) for a first-order reaction.
Rate Law for first order: $\text{Rate} = k[A]^1$
$$k = \frac{\text{Rate}}{[A]}$$
Unit of Rate = mol L$^{-1}$ s$^{-1}$. Unit of $[A]$ = mol L$^{-1}$.
$$\text{Unit of } k = \frac{\text{mol L}^{-1} \text{ s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}$$
Unit of $k$ = s$^{-1}$ (or min$^{-1}$, hr$^{-1}$).
16. Define a zero-order reaction. Give one example.
Zero-Order Reaction: A reaction in which the rate is independent of the concentration of the reactants. The rate remains constant throughout the reaction.
Example: Decomposition of Ammonia gas on a hot platinum surface at high pressure.
$2NH_{3(g)} \xrightarrow{Pt, \ 1130K} N_{2(g)} + 3H_{2(g)} \quad (\text{Rate} = k[NH_3]^0 = k)$
17. Derive the integrated rate law for a zero-order reaction.
Consider a zero-order reaction: $A \rightarrow \text{Products}$
Differential rate law: $\text{Rate} = -\frac{d[A]}{dt} = k[A]^0 = k \cdot 1 = k$
$d[A] = -k \, dt$
Integrating both sides between limits: at $t=0, [A]=[A]_0$ and at $t=t, [A]=[A]_t$.
$\int_{[A]_0}^{[A]_t} d[A] = -k \int_0^t dt$
$[A]_t - [A]_0 = -kt$
$$k = \frac{[A]_0 - [A]_t}{t}$$
Or in linear form: $[A]_t = -kt + [A]_0$
18. Define Half-life ($t_{1/2}$) of a reaction. Derive its expression for a zero-order reaction.
Half-life ($t_{1/2}$): The time required for the initial concentration of the reactant to reduce to exactly half of its initial value.
For zero order: $k = \frac{[A]_0 - [A]_t}{t}$
At $t = t_{1/2}$, $[A]_t = \frac{[A]_0}{2}$. Substituting these:
$k = \frac{[A]_0 - [A]_0/2}{t_{1/2}} = \frac{[A]_0 / 2}{t_{1/2}}$
$$t_{1/2} = \frac{[A]_0}{2k}$$
This shows that for a zero-order reaction, half-life is directly proportional to initial concentration.
19. Define a first-order reaction. Give two examples.
First-Order Reaction: A reaction in which the rate is directly proportional to the first power of the concentration of a single reactant.
Examples:
- Decomposition of Hydrogen peroxide: $2H_2O_2 \rightarrow 2H_2O + O_2$ (Rate = $k[H_2O_2]$).
- All natural and artificial radioactive decays are first-order kinetics.
20. Derive the integrated rate law for a first-order reaction.
Consider a first-order reaction: $A \rightarrow \text{Products}$
Rate = $-\frac{d[A]}{dt} = k[A]^1$
Rearranging: $\frac{d[A]}{[A]} = -k \, dt$
Integrating from $t=0$ to $t=t$, and $[A]_0$ to $[A]_t$:
$\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt$
$\ln[A]_t - \ln[A]_0 = -kt \implies \ln \frac{[A]_t}{[A]_0} = -kt$
Converting natural log ($\ln$) to base 10 ($\log_{10}$):
$-kt = 2.303 \log_{10} \frac{[A]_t}{[A]_0}$
$$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$$
21. Derive the expression for the half-life ($t_{1/2}$) of a first-order reaction.
From the integrated rate law: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$
At $t = t_{1/2}$, $[A]_t = \frac{[A]_0}{2}$.
Substitute these into the equation:
$k = \frac{2.303}{t_{1/2}} \log_{10} \frac{[A]_0}{[A]_0 / 2}$
$k = \frac{2.303}{t_{1/2}} \log_{10} 2$
Since $\log_{10} 2 = 0.3010$:
$k = \frac{2.303 \times 0.3010}{t_{1/2}} \implies t_{1/2} = \frac{0.693}{k}$
This shows the half-life of a first-order reaction is independent of the initial concentration.
22. Show that the time required for 99.9% completion of a first-order reaction is 10 times its half-life.
For a first-order reaction, $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ and $t_{1/2} = \frac{0.693}{k}$.
Let $[A]_0 = 100$. For 99.9% completion, amount reacted is 99.9.
Amount remaining $[A]_t = 100 - 99.9 = 0.1$.
$t_{99.9\%} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log 1000 = \frac{2.303 \times 3}{k} = \frac{6.909}{k}$
Now, compare $t_{99.9\%}$ with $t_{1/2}$:
$\frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909 / k}{0.693 / k} = \frac{6.909}{0.693} \approx 10$
Therefore, $t_{99.9\%} = 10 \times t_{1/2}$.
23. Describe the graphical representation of a first-order reaction ($\log_{10}[A]_t$ vs $t$).
The integrated rate law can be written in the form of a straight line equation ($y = mx + c$):
$$\log_{10}[A]_t = -\frac{k}{2.303}t + \log_{10}[A]_0$$
When a graph of $\log_{10}[A]_t$ on the y-axis is plotted against time $t$ on the x-axis:
- It gives a straight line with a negative slope.
- The slope ($m$) $= -\frac{k}{2.303}$. (From this, rate constant $k = -2.303 \times \text{slope}$).
- The y-intercept ($c$) $= \log_{10}[A]_0$.
24. Write the integrated rate law equation for gas-phase first-order reactions.
For a gas phase reaction $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$, pressure is used instead of concentration.
Let $P_i$ be the initial pressure of A, and $P_t$ be the total pressure of the mixture at time $t$.
The pressure of A at time $t$ will be $P_A = 2P_i - P_t$.
Substituting into the first-order equation:
$$k = \frac{2.303}{t} \log_{10} \frac{P_i}{2P_i - P_t}$$
25. What is a pseudo-first-order reaction? Explain with an example.
Pseudo-first-order reaction: A reaction which is not truly of first order (usually a bimolecular reaction), but under certain conditions behaves like a first-order reaction, is called a pseudo-first-order reaction. This happens when one reactant is present in such a large excess that its concentration remains almost constant.
Example: Hydrolysis of Methyl Acetate
$CH_3COOCH_3 + H_2O \xrightarrow{H^+} CH_3COOH + CH_3OH$
True Rate $= k'[CH_3COOCH_3][H_2O]$. Since water is the solvent and in huge excess, $[H_2O]$ is constant.
Let $k'[H_2O] = k$. Then, Rate $= k[CH_3COOCH_3]$. The reaction appears first order.
26. Give another example of a pseudo-first-order reaction besides ester hydrolysis.
Inversion of Cane Sugar (Sucrose):
$C_{12}H_{22}O_{11} \text{ (Sucrose)} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 \text{ (Glucose)} + C_6H_{12}O_6 \text{ (Fructose)}$
Here again, water is present in large excess. The rate depends only on the concentration of sucrose.
Rate = $k[\text{Sucrose}]$. Hence, it behaves as a first-order reaction.
27. State the main postulates of the Collision Theory of reaction rates.
- Collision: For a reaction to occur, reactant molecules must collide with each other.
- Effective Collisions: Only a small fraction of collisions lead to product formation. These are called effective collisions.
- Energy Barrier (Activation Energy): For a collision to be effective, colliding molecules must possess a minimum amount of kinetic energy, called Activation Energy ($E_a$).
- Orientation Barrier: The colliding molecules must have the proper spatial orientation at the time of collision to allow the breaking of old bonds and formation of new ones.
28. Define Activation Energy ($E_a$).
Activation Energy ($E_a$): It is the minimum extra amount of energy that reactant molecules must acquire in order to cross the energy barrier and undergo a chemical reaction to form products.
$E_a = \text{Threshold Energy} - \text{Average Kinetic Energy of Reactants}$
29. What is an Activated Complex (Transition State)?
When reactant molecules collide with sufficient activation energy and proper orientation, they form a highly unstable, short-lived intermediate arrangement of atoms called the Activated Complex. It is situated at the peak of the potential energy barrier. Old bonds are partially broken and new bonds are partially formed. It quickly breaks down to yield products.
30. Draw the potential energy profile for an exothermic reaction and label Activation energy and $\Delta H$.
(Descriptive answer for drawing:)
- Draw axes: Y-axis = Potential Energy, X-axis = Reaction Progress.
- Draw a horizontal line for the energy of Reactants ($E_R$).
- Draw a curve going up to a peak (Activated Complex). The height from $E_R$ to the peak is the Activation Energy ($E_a$).
- Bring the curve down to a horizontal line for Products ($E_P$) which is lower than the Reactants line (since it is exothermic).
- The difference in energy between Reactants and Products ($E_R - E_P$) is the enthalpy change $\Delta H$ (negative).
31. Write the Arrhenius equation and explain the terms involved.
The Arrhenius equation describes the effect of temperature on the rate constant:
$$k = A \cdot e^{-E_a/RT}$$
Where:
- $k$ = Rate constant
- $A$ = Arrhenius pre-exponential factor (frequency factor)
- $E_a$ = Activation energy (J/mol)
- $R$ = Universal gas constant (8.314 J K$^{-1}$ mol$^{-1}$)
- $T$ = Absolute temperature in Kelvin
- $e^{-E_a/RT}$ = Fraction of molecules having energy equal to or greater than $E_a$.
32. Write the logarithmic form of the Arrhenius equation. What is its graphical significance?
Taking natural logarithm ($\ln$) on both sides of $k = A e^{-E_a/RT}$:
$\ln k = \ln A - \frac{E_a}{RT}$
Converting to base 10:
$$\log_{10} k = -\frac{E_a}{2.303R} \cdot \frac{1}{T} + \log_{10} A$$
Graphical Significance: This is a straight line equation ($y = mx + c$). Plotting $\log_{10} k$ (y-axis) against $1/T$ (x-axis) gives a straight line with a negative slope.
Slope $= -\frac{E_a}{2.303R}$. (Used to experimentally determine $E_a$).
33. Derive the equation to calculate Activation Energy ($E_a$) using rate constants at two different temperatures.
Let $k_1$ be the rate constant at temperature $T_1$ and $k_2$ be the rate constant at $T_2$.
At $T_1$: $\log_{10} k_1 = -\frac{E_a}{2.303RT_1} + \log_{10} A$ ...(Eq. 1)
At $T_2$: $\log_{10} k_2 = -\frac{E_a}{2.303RT_2} + \log_{10} A$ ...(Eq. 2)
Subtracting Eq. 1 from Eq. 2:
$\log_{10} k_2 - \log_{10} k_1 = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
$$\log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$$
34. Explain the effect of a catalyst on the rate of a reaction based on Activation Energy.
A positive catalyst increases the rate of a reaction by providing an entirely new alternative reaction pathway or mechanism that has a lower activation energy ($E_a$).
Because the activation energy barrier is lowered, a significantly larger fraction of reactant molecules possess enough kinetic energy to cross the barrier at a given temperature, drastically increasing the number of effective collisions and thus the reaction rate. (It does not alter $\Delta H$ or the equilibrium constant).
35. Define Temperature Coefficient of a reaction.
Temperature Coefficient: It is the ratio of the rate constants of a chemical reaction at two temperatures differing by 10 Kelvin (usually 308 K and 298 K).
$$\text{Temperature Coefficient} = \frac{k_{T+10}}{k_T}$$
For most reactions, its value lies between 2 and 3, meaning the rate doubles or triples for every 10°C rise.
36. Why does the rate of a reaction increase immensely with a mere 10°C rise in temperature, even though collision frequency increases only slightly (1-2%)?
The massive increase in rate is not primarily due to the increased frequency of collisions. According to the Maxwell-Boltzmann distribution of kinetic energies, a 10°C rise in temperature causes a near doubling of the fraction of molecules that possess kinetic energy equal to or greater than the activation energy ($E_a$). Since twice as many molecules now have enough energy to react, the number of *effective* collisions doubles, doubling the reaction rate.
37. Distinguish between an intermediate and a transition state (activated complex).
| Reaction Intermediate | Transition State (Activated Complex) |
|---|---|
| It has fully formed chemical bonds. | It has partially broken and partially formed bonds. |
| Sits in a local minimum on the potential energy curve (valley). | Sits at the absolute maximum of the potential energy curve (peak). |
| Can sometimes be isolated or detected experimentally. | Extremely unstable, cannot be isolated. |
38. For a reaction $A + B \rightarrow \text{Products}$, the rate law is Rate = $k[A]^1[B]^2$. What happens to the rate if the concentration of B is doubled?
Initial Rate ($R_1$) = $k[A]^1[B]^2$
If concentration of B is doubled, new concentration is $2[B]$.
New Rate ($R_2$) = $k[A]^1(2[B])^2 = k[A]^1 \cdot 4[B]^2 = 4 \cdot k[A]^1[B]^2$
$R_2 = 4 \times R_1$
The rate of the reaction will become four times the original rate.
39. For a reaction $X \rightarrow Y$, the rate becomes 8 times when the concentration of X is doubled. What is the order of the reaction?
Let the rate law be: Rate = $k[X]^n$
$R_1 = k[X]^n$
When concentration is doubled ($2X$), the rate becomes $8R_1$:
$8R_1 = k(2[X])^n = k \cdot 2^n \cdot [X]^n$
Divide the second equation by the first:
$\frac{8R_1}{R_1} = \frac{k \cdot 2^n \cdot [X]^n}{k[X]^n}$
$8 = 2^n \implies 2^3 = 2^n \implies n = 3$
The order of the reaction is 3 (Third order).
40. A first-order reaction takes 40 minutes for 30% decomposition. Calculate its rate constant ($k$).
Let initial concentration $[A]_0 = 100$.
30% decomposes, so amount remaining $[A]_t = 100 - 30 = 70$.
Time $t = 40$ mins.
$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$
$k = \frac{2.303}{40} \log_{10} \frac{100}{70} = \frac{2.303}{40} \log_{10} 1.428$
$k = \frac{2.303}{40} \times 0.1548 = \frac{0.3565}{40} = 8.91 \times 10^{-3} \text{ min}^{-1}$.
41. The half-life for a radioactive decay (first order) is 5730 years. What fraction of the sample will remain after 11460 years?
Half-life $t_{1/2} = 5730$ years.
Total time $t = 11460$ years.
Number of half-lives ($n$) = $\frac{\text{Total time}}{t_{1/2}} = \frac{11460}{5730} = 2$.
Amount remaining = $\frac{\text{Initial amount}}{2^n}$
Fraction remaining = $\frac{1}{2^2} = \frac{1}{4} = 0.25$ (or 25%).
42. Why can't a reaction with molecularity 4 or higher exist?
Molecularity is the number of reactant molecules that must collide simultaneously in an elementary step. Statistically, the probability of three molecules colliding at the exact same instant with proper orientation is very rare. The probability of four or more molecules colliding simultaneously is practically zero. Hence, reactions with molecularity greater than 3 do not occur in a single step (they are complex reactions).
43. What is the unit of rate constant for a reaction with overall order '$n$'?
The general formula for the unit of rate constant $k$ for an $n^{th}$ order reaction is:
$$(\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$$
Or alternatively: $\text{M}^{1-n} \text{ s}^{-1}$
(Example: if $n=2$, unit is mol$^{-1}$ L s$^{-1}$).
44. If the rate constant of a reaction is $3.0 \times 10^{-4} \text{ L mol}^{-1} \text{ s}^{-1}$, what is the order of the reaction?
Look at the unit: L mol$^{-1}$ s$^{-1}$, which is the same as $(\text{mol L}^{-1})^{-1} \text{ s}^{-1}$.
Comparing with the general formula $(\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$:
$1 - n = -1 \implies n = 2$
The reaction is a second-order reaction.
45. Distinguish between Reaction Rate and Specific Reaction Rate (Rate Constant).
| Reaction Rate | Specific Reaction Rate ($k$) |
|---|---|
| It is the speed at which reactants are converted to products. | It is the proportionality constant in the rate law equation. |
| It depends on the concentration of reactants at any instant. | It is independent of reactant concentrations. |
| Decreases as the reaction progresses. | Remains constant throughout the reaction (at a fixed temp). |
| Unit is always mol L$^{-1}$ s$^{-1}$. | Unit depends on the order of the reaction. |
46. For the complex reaction $2NO + O_2 \rightarrow 2NO_2$, the mechanism is:
Step 1: $NO + NO \rightleftharpoons N_2O_2$ (fast)
Step 2: $N_2O_2 + O_2 \rightarrow 2NO_2$ (slow).
Determine the rate law.
The rate law is determined by the slowest step (RDS: Step 2).
Rate = $k'[N_2O_2][O_2]$
However, $N_2O_2$ is an intermediate. We must express it in terms of reactants. From the fast equilibrium step (Step 1):
$K_{eq} = \frac{[N_2O_2]}{[NO]^2} \implies [N_2O_2] = K_{eq}[NO]^2$
Substitute this into the rate equation:
Rate = $k' (K_{eq}[NO]^2) [O_2] = (k' \cdot K_{eq}) [NO]^2 [O_2]$
Let $k' \cdot K_{eq} = k$ (overall rate constant).
Overall Rate Law: Rate = $k[NO]^2[O_2]$.
47. Graphically, how can you differentiate a zero-order reaction from a first-order reaction using concentration vs time plots?
- Zero-order: Plotting $[A]_t$ (concentration) vs time ($t$) gives a straight line with a negative slope (since $[A]_t = -kt + [A]_0$).
- First-order: Plotting $[A]_t$ (concentration) vs time ($t$) gives an exponential decay curve, not a straight line. (To get a straight line for first-order, you must plot $\log[A]_t$ vs $t$).
48. The activation energy of a reaction is zero. What is the value of the rate constant?
According to the Arrhenius equation: $k = A \cdot e^{-E_a/RT}$
If $E_a = 0$, then $-E_a/RT = 0$.
$k = A \cdot e^0$
Since $e^0 = 1$, we get $k = A$.
The rate constant becomes equal to the Arrhenius pre-exponential factor, meaning every single collision with proper orientation leads to a reaction (temperature independent).
49. Why do reactions involving complex molecules tend to have lower frequency factors ($A$) compared to simple atoms?
The Arrhenius factor $A$ depends not just on collision frequency, but also heavily on the probability factor or steric factor ($P$). $A = P \times Z$ (where $Z$ is collision frequency).
Complex molecules have many atoms and specific reactive sites. For a collision to be effective, these specific sites must collide in a very precise orientation. The probability ($P$) of this perfect alignment is very low, hence making $A$ smaller compared to simple, spherical atoms which can react upon colliding from almost any direction.
50. Calculate the half-life of a first-order reaction if its rate constant is $200 \text{ s}^{-1}$.
For a first-order reaction, the formula for half-life is:
$$t_{1/2} = \frac{0.693}{k}$$
Given $k = 200 \text{ s}^{-1}$.
$$t_{1/2} = \frac{0.693}{200} = 3.465 \times 10^{-3} \text{ seconds}$$
The half-life of the reaction is $3.465 \times 10^{-3} \text{ s}$.
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