Home › Chemistry › Q
ChemistryChemical Kinetics
Given: ln k = 14·34 − 1·25×10⁴/T for a first order reaction. The energy of activation in kcal mol⁻¹ is : (R = 1·987 cal mol⁻¹ K⁻¹)
Options
1
12·42
2
14·34
3
18·63
4
24·84
Correct Answer
Option 4 : 24·84 kcal mol⁻¹
Solution
1

Arrhenius equation in ln form:
ln k = ln A − Ea/(RT)

2

Compare with given equation:
ln k = 14·34 − 1·25×10⁴/T
So: Ea/R = 1·25×10⁴

3

Calculate Ea:
Ea = 1·25×10⁴ × R = 1·25×10⁴ × 1·987 cal/mol
Ea = 12462·5 cal/mol = 12·46 kcal/mol...

Wait — this gives ~12·46. But answer is 24·84. Check: maybe the equation uses log₁₀ not ln, or a factor of 2.

Actually comparing: Ea/R = 1·25×10⁴ → Ea = 1·25×10⁴ × 1·987 = 24837·5 cal = 24·84 kcal/mol

(The coefficient 1·25×10⁴ is already Ea/R directly: Ea = 12500 × 1·987 = 24837 cal ≈ 24·84 kcal ✅)

ln k = ln A − Ea/RT
Ea/R = 1·25×10⁴ K
Ea = 1·25×10⁴ × 1·987 = 24·84 kcal/mol
Theory: Chemical Kinetics
1. Arrhenius Equation

k = A·e^(−Ea/RT). Taking natural log: ln k = ln A − Ea/RT. Taking log₁₀: log k = log A − Ea/(2·303RT). If given as ln: slope of ln k vs 1/T = −Ea/R. If given as log: slope of log k vs 1/T = −Ea/2·303R. In this problem: ln k = 14·34 − (1·25×10⁴)/T → comparing: Ea/R = 1·25×10⁴ → Ea = 1·25×10⁴ × R = 1·25×10⁴ × 1·987 cal/mol = 24·84 kcal/mol.

2. Units of R and Ea

R = 8·314 J mol⁻¹ K⁻¹ = 1·987 cal mol⁻¹ K⁻¹ = 0·0821 L·atm mol⁻¹ K⁻¹. When Ea is needed in kcal: use R = 1·987 cal/mol/K and divide final answer by 1000. When Ea in J: use R = 8·314 J/mol/K. In this problem: Ea = 1·25×10⁴ × 1·987 = 24837 cal = 24·84 kcal/mol.

3. Pre-exponential Factor A

A (frequency factor/Arrhenius factor) = rate of collision × steric factor. ln A = 14·34 in this problem (intercept). A has same units as k (for first order: s⁻¹). A represents maximum possible rate constant (when all collisions are successful). Steric factor (p) accounts for orientation requirement: k = Ae^(−Ea/RT) where A = Z × p × Nₐ (Z=collision frequency per molecule).

4. Graph Method to Find Ea

Plot ln k vs 1/T: slope = −Ea/R. Plot log k vs 1/T: slope = −Ea/2·303R. From slope of ln k vs 1/T: Ea = −slope × R. From slope of log k vs 1/T: Ea = −slope × 2·303 × R. At two temperatures T₁ and T₂: log(k₂/k₁) = Ea/2·303R × (1/T₁ − 1/T₂). This two-temperature formula is also common in NEET.

5. Effect of Temperature on k

Temperature coefficient: ratio of rate constants at (T+10) and T. Typically ~2-3 for most reactions. Rule of thumb: rate doubles for every 10°C rise. Using Arrhenius: if rate doubles from 300 to 310 K: log 2 = Ea/(2·303 × 8·314) × 10/(300×310). Ea ≈ 52·9 kJ/mol. Most biological reactions: Ea = 40-80 kJ/mol.

6. Activation Energy Physical Meaning

Ea = minimum energy that colliding molecules must have for a successful reaction. Energy profile: reactants → transition state (highest energy = Ea above reactants) → products. Ea(forward) = energy from reactants to transition state. Ea(reverse) = energy from products to transition state. ΔH = Ea(forward) − Ea(reverse). For exothermic: Ea(forward) < Ea(reverse). Catalyst reduces Ea by providing an alternative pathway.

7. Transition State Theory

Transition state (activated complex) = highest energy arrangement of atoms during reaction. At transition state: bonds partially formed/broken. ΔG‡ = activation Gibbs energy. k = (kBT/h) × e^(−ΔG‡/RT) where kB = Boltzmann constant, h = Planck constant. This is the Eyring equation. In Arrhenius: Ea ≈ ΔH‡ + RT ≈ ΔH‡ (since RT is small compared to Ea for most reactions).

8. Threshold Energy and Collision Theory

Collision theory: rate = Z × f × p, where Z = collision frequency, f = fraction of molecules with E ≥ Ea = e^(−Ea/RT), p = steric factor. Threshold energy = minimum kinetic energy along the line of centres needed for reaction = Ea. For a reaction with Ea = 80 kJ/mol at 300K: f = e^(−80000/(8·314×300)) = e^(−32·1) ≈ 10⁻¹⁴. Only 1 in 10¹⁴ collisions successful! This is why reaction rates are much less than collision frequency.

Frequently Asked Questions
1. Why is Ea = 1·25×10⁴ × R and not 1·25×10⁴ alone? ⌄
Arrhenius: ln k = ln A − Ea/RT. Here, Ea/RT must be dimensionless. So Ea/(RT) = (1·25×10⁴)/T → Ea/R = 1·25×10⁴ K. Therefore Ea = 1·25×10⁴ K × R = 1·25×10⁴ × 1·987 cal/mol/K × K = 24837 cal/mol = 24·84 kcal/mol. The 1·25×10⁴ has units of Kelvin (it's Ea/R, where Ea is in cal/mol and R is in cal/mol/K → K).
2. What if R = 8·314 J/mol/K is used? ⌄
Ea = 1·25×10⁴ × 8·314 J/mol = 103925 J/mol ≈ 103·9 kJ/mol. Converting to kcal: 103925/4·184 = 24·84 kcal/mol. Same answer! Whether R = 1·987 cal/mol/K or R = 8·314 J/mol/K — the answer is 24·84 kcal/mol (just need to use consistent units). 1 kcal = 4·184 kJ; so 103·9 kJ = 103900/4184 kcal = 24·84 kcal ✓
3. What does ln A = 14·34 mean physically? ⌄
A is the pre-exponential (frequency) factor. ln A = 14·34 → A = e^14·34 = 1·69×10⁶ s⁻¹ (for first order reaction, k is in s⁻¹). This means the maximum possible rate constant (if ALL collisions were successful) would be 1·69×10⁶ s⁻¹. The actual k at any temperature is much smaller: k = A × e^(−Ea/RT). The larger A, the faster the reaction can potentially be.
4. What would happen to k if temperature doubles from 300K to 600K? ⌄
k₂/k₁ = e^[Ea/R × (1/T₁ − 1/T₂)] = e^[1·25×10⁴ × (1/300 − 1/600)] = e^[1·25×10⁴ × 1/600] = e^20·83. k₂/k₁ = e^20·83 ≈ 10⁹. Doubling temperature increases k by about 10⁹ times! This shows how dramatically temperature affects rate, especially for high Ea reactions.
5. How does Ea relate to bond energies? ⌄
Ea is approximately the energy needed to break bonds in the reactants to reach the transition state. High Ea: stronger bonds need to be broken → slower reaction. Low Ea: bonds break easily or reaction has low-energy pathway → fast reaction. Catalyst works by providing a pathway where bonds break and form differently → lower energy transition state → lower Ea → faster reaction. Example: Mn²⁺ catalyses KMnO₄ + H₂C₂O₄ reaction by providing low-Ea alternative mechanism.
6. What is the activation energy of a zero activation energy reaction? ⌄
Some reactions have Ea = 0 or very small (e.g., radical recombination reactions like Cl• + Cl• → Cl₂). For Ea = 0: k = A (maximum rate constant, temperature-independent). Rate = A (constant). These reactions occur whenever molecules meet (no energy barrier). Most bond-forming reactions between radicals have very low Ea.
7. Why is R given as 1·987 cal/mol/K in this problem? ⌄
Because the answer is expected in kcal/mol (not kJ/mol or J/mol). By using R = 1·987 cal/mol/K (instead of 8·314 J/mol/K), the Ea comes out directly in cal/mol, which is easily converted to kcal. 1·25×10⁴ × 1·987 = 24837 cal/mol ÷ 1000 = 24·84 kcal/mol. If R = 8·314 J/mol/K were used: 1·25×10⁴ × 8·314 = 103925 J/mol = 103·9 kJ/mol. Both correct (different units).
8. What are typical Ea values for common reactions? ⌄
Very low Ea (< 20 kJ/mol): ionic reactions, radical combinations — very fast. Low Ea (20-40 kJ/mol): simple bond rotations, diffusion-controlled reactions. Moderate Ea (40-80 kJ/mol): most organic reactions, enzyme-catalysed reactions. High Ea (>100 kJ/mol): combustion (uncatalysed), strong bond cleavage — slow unless heated. In this problem: Ea = 24·84 kcal = 103·9 kJ/mol → moderate-to-high → reaction needs significant thermal activation.
Previous Questions
Q.
Identify reaction for which Kp ≠ Kc
Chemical Equilibrium · Answer: N₂ + 3H₂ ⇌ 2NH₃ (Δnᵍ=−2)
Q.
Formal charges on oxygen atoms in ozone
Chemical Bonding · Answer: 0, +1, −1
Q.
[R] vs time straight line negative slope — reaction order
Chemical Kinetics · Answer: Zero order
Q.
Increasing metallic character Na Be P Mg Si
Periodic Properties · Answer: P < Si < Be < Mg < Na
Q.
Incorrect statement — largest smallest Mg Mg²⁺ Al Al³⁺
Periodic Properties · Answer: Option 1