The Mistake Bank
Class 12 - Chapter 3: Chemical Kinetics
Time waits for no one, especially during a rate of reaction problem. Keep your coefficients aligned, your logarithms straight, and your order of reactions strictly experimental.
1. Order vs Stoichiometry
Rate LawScenario: For the general balanced reaction $2A + B \rightarrow C$, what is the Order of the reaction?
Student looks at the stoichiometric coefficients and blindly sums them up:
$$ \text{Order} = 2 + 1 = 3 $$
(You cannot determine the Order just by looking at a balanced equation!)
Order is purely Experimental!
The order could be 0, 1, 2, or even fractional. It must be determined through experimental data (like initial rate method tables).
2. Rate of Reaction vs Disappearance
Stoichiometric DivisionScenario: For the reaction $N_2 + 3H_2 \rightarrow 2NH_3$, the rate of disappearance of $H_2$ is $6 \times 10^{-3} \text{ M s}^{-1}$. Calculate the overall Rate of Reaction.
Student assumes the rate of disappearance of a reactant is exactly equal to the rate of the reaction.
Answer given: $\mathbf{6 \times 10^{-3} \text{ M s}^{-1}}$.
(This ignores the stoichiometry! Hydrogen is disappearing 3 times faster than the overall reaction is progressing.)
Divide by the Stoichiometric Coefficient!
We are given that the rate of disappearance ($-\frac{\Delta [H_2]}{\Delta t}$) is $6 \times 10^{-3}$.
$$ \text{Rate of Reaction} = \frac{1}{3} \times (6 \times 10^{-3}) = \mathbf{2 \times 10^{-3} \text{ M s}^{-1}} $$
3. Units of Rate Constant (k)
Formulas & UnitsScenario: Determine the unit of the rate constant ($k$) for a Second Order reaction.
Student memorized the first-order unit because it's the most common, and blindly applies it to everything:
$$ \text{Unit} = s^{-1} \text{ or } min^{-1} $$
(This only works for Order = 1. A wrong unit in the final answer costs ½ a mark!)
Use the General Unit Formula!
Where $n$ is the order of the reaction. For a Second Order reaction ($n=2$):
$$ (\text{mol L}^{-1})^{1-2} \cdot s^{-1} = (\text{mol L}^{-1})^{-1} \cdot s^{-1} $$
Final Unit: $\mathbf{L \text{ mol}^{-1} \text{ s}^{-1}}$ (or $M^{-1} s^{-1}$)
4. Half-Life of Zero Order
Half-LifeScenario: For a Zero Order reaction, the initial concentration is doubled. What happens to its half-life ($t_{1/2}$)?
Student recalls the famous first-order rule: "Half-life is independent of initial concentration."
They apply it here and answer: "It remains constant."
(They confused Zero Order with First Order!)
Zero Order Half-Life depends on Initial Concentration!
$$ t_{1/2} = \frac{[R]_0}{2k} $$
Notice that $t_{1/2}$ is directly proportional to the initial concentration $[R]_0$.
If you double the concentration, the half-life also doubles.
5. Catalyst Thermodynamic Illusion
Activation EnergyScenario: A positive catalyst is added to a reaction mixture. How does it affect the standard Gibbs Free Energy ($\Delta G^\circ$) of the reaction?
Student thinks: "A catalyst makes the reaction easier and faster, so it must become more spontaneous."
Answer given: "$\Delta G^\circ$ becomes more negative."
Catalysts affect the Path, NOT the Destination!
It does NOT alter thermodynamic state functions like Gibbs Energy ($\Delta G^\circ$), Enthalpy ($\Delta H$), or the Equilibrium Constant ($K_c$).
Answer: $\Delta G^\circ$ remains absolutely unchanged.
6. Molecularity vs Order Reality
Complex ReactionsScenario: Experimental data shows a reaction has an Order of 1.5. What is the Molecularity of this reaction?
Student assumes Molecularity and Order are roughly the same concept, just derived differently.
Answer given: "1.5"
(Impossible! You cannot have half a molecule colliding in space.)
Molecularity MUST be a Positive Integer!
If a reaction has a fractional order, it is definitively a Complex Reaction (multi-step). Molecularity has no meaning for the overall complex reaction.
7. Arrhenius Plot Slope Sign
Arrhenius EquationScenario: You plot a graph of $\ln k$ versus $1/T$ for a chemical reaction. What does the slope of this graph equal?
Student memorizes the variables but forgets the algebraic structure.
Answer given: "Slope = $E_a / R$"
(They forgot the crucial negative sign that indicates the downward slope!)
It is a Negative Slope!
Take the natural log: $\ln k = \ln A - \frac{E_a}{RT}$
Compare this to a straight line $y = mx + c$, where $y = \ln k$ and $x = 1/T$.
The slope ($m$) is exactly: $\mathbf{-\frac{E_a}{R}}$
(If the plot is $\log_{10} k$ vs $1/T$, the slope is $-\frac{E_a}{2.303R}$)
8. Pseudo-First Order Misidentification
Reaction OrderScenario: The acid-catalyzed hydrolysis of ethyl acetate is: $CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$. What is its order?
Student sees two reactants (the ester and water).
They assume the rate depends on both: $\text{Rate} = k[Ester][H_2O]$.
Answer given: Second Order.
Water is in Massive Excess!
The rate law simplifies to $\text{Rate} = k'[Ester]$.
Because it behaves like a first-order reaction despite having a molecularity of 2, it is a Pseudo-First Order Reaction.
9. The 2.303 Conversion Factor
Integrated Rate LawScenario: A student writes the integrated rate law for a first-order reaction to solve for time ($t$) using standard base-10 logarithms.
Student writes: $t = \frac{1}{k} \log_{10} \frac{[A]_0}{[A]}$
They plug in the numbers and get a time that is significantly smaller than the correct answer.
(They forgot the mathematical conversion factor between natural log $\ln$ and $\log_{10}$!)
$\ln(x) = 2.303 \log_{10}(x)$
$t = \frac{1}{k} \ln \frac{[A]_0}{[A]}$
To use base-10 logs (which is standard for calculators and tables), you MUST multiply by 2.303.
Correct Formula: $\mathbf{t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}}$
10. Temperature Coefficient Exponentiation
Effect of TemperatureScenario: The rate of a reaction doubles for every $10^\circ\text{C}$ rise in temperature. If the temperature is raised from $20^\circ\text{C}$ to $50^\circ\text{C}$, by what factor does the rate increase?
Student thinks linearly: "$20^\circ$ to $50^\circ$ is three $10^\circ$ jumps. It doubles, so $2 \times 3 = 6$."
Answer given: 6 times.
(Kinetics is exponential, not linear!)
Use the Exponent Formula: Rate $\propto 2^{(\Delta T / 10)}$
- $20^\circ \to 30^\circ$: Rate $\times 2$
- $30^\circ \to 40^\circ$: Rate $\times 2 \times 2 = 4$
- $40^\circ \to 50^\circ$: Rate $\times 4 \times 2 = 8$
Using formula: $2^{((50-20)/10)} = 2^3 = \mathbf{8 \text{ times faster}}$.
11. Activation Energy for Reverse Reaction
Energy ProfilesScenario: For an exothermic reaction, the activation energy of the forward reaction is $50 \text{ kJ/mol}$. Is the activation energy for the reverse reaction also $50 \text{ kJ/mol}$?
Student assumes the energy hill is the same height from both sides.
Answer given: Yes, $E_{a(rev)} = 50 \text{ kJ/mol}$.
Check the Enthalpy Change ($\Delta H$)!
To go backwards, the products must climb back up that extra depth PLUS the original activation energy hill.
Formula: $\Delta H = E_{a(fwd)} - E_{a(rev)}$
Therefore, $E_{a(rev)}$ will be greater than $50 \text{ kJ/mol}$.
12. The Solid Surface Area Trap
Factors Affecting RateScenario: How does crushing a solid reactant into a fine powder affect the rate of reaction?
Student thinks: "Since the mass of the solid hasn't changed, and active mass of solids is taken as unity, the rate remains unchanged."
Rate Increases due to Surface Area!
Reactions involving solids only happen at the boundary/surface. Crushing the solid exponentially increases the exposed Surface Area, providing more active sites for collisions.
Result: The rate of reaction increases significantly.
13. Zero-Order vs First-Order Graphical Slopes
GraphsScenario: You plot a graph of Reactant Concentration $[A]$ versus Time $t$ and get a straight downward sloping line. What is the order of the reaction?
Student memorized that kinetics graphs usually involve natural logs for first-order.
They get confused and guess First Order.
$[A]$ vs $t$ is strictly Zero Order!
- First Order: The equation is $\ln[A] = -kt + \ln[A]_0$. Plotting $[A]$ vs $t$ yields an exponential decay curve. You must plot $\ln[A]$ vs $t$ to get a straight line.
Answer: Zero Order.
14. Initial Rate vs Instantaneous Rate
DefinitionsScenario: How is the Initial Rate of a reaction determined from a concentration-time graph?
Student calculates the average rate over the first few seconds (e.g., from $t=0$ to $t=5$ seconds): $\frac{\Delta [C]}{\Delta t}$.
Draw a Tangent at $t=0$!
It cannot be calculated accurately using a wide time interval ($\Delta t$). You must draw a tangent to the concentration-time curve precisely at $t=0$ and calculate its slope.
15. Order and Molecularity at Zero
Theoretical LimitsScenario: Can a reaction have a Molecularity of Zero? Can it have an Order of Zero?
Student thinks: "If there are zero molecules, there is no reaction. So neither can be zero."
Answer given: "No to both."
Order CAN be zero; Molecularity CANNOT!
Order: It is purely mathematical. A Zero Order reaction simply means the rate of the reaction is entirely independent of the reactant's concentration (e.g., decomposition of gases on a saturated metal surface).
Confess Your Sins!
"Kinetics determines how fast you get the answer, but Accuracy determines if you get marks."
Did one of these catch you? Or do you have a different horror story from your last exam?
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