Question 223 marksShort AnswerSection C

For the first order thermal decomposition reaction, following data was obtained:

C₂H₅Cl(g) → C₂H₄(g) + HCl(g)

S. No.Time(s)Total Pressure (atm)
100.30
2300.50

Calculate rate constant. [Given: log⁡3=0.48\log 3 = 0.48]

Chemical Kinetics
First-order rate constant from pressure data

Official answer

Setting up the pressures


At t=0t = 0 only C₂H₅Cl is present: p0=0.30 atmp_0 = 0.30 \text{ atm}. Let p be the pressure of C₂H₅Cl decomposed at t=30 st = 30 \text{ s}. Each mole gives one mole each of C₂H₄ and HCl:


  • Total pressure Pt=(p0−p)+p+p=p0+pP_t = (p_0 - p) + p + p = p_0 + p
  • 0.50=0.30+p→0.50 = 0.30 + p \to p=0.20 atmp = 0.20 \text{ atm}
  • Pressure of C₂H₅Cl left = p0−p=0.30−0.20=p_0 - p = 0.30 - 0.20 = 0.10 atm0.10 \text{ atm}

Applying the first-order equation


k=2.303t⋅log⁡[p0p0−p]k = \frac{2.303}{t} \cdot \log\left[\frac{p_0}{p_0 - p}\right]


  • k=2.30330⋅log⁡(0.300.10)=2.30330⋅log⁡3k = \frac{2.303}{30} \cdot \log\left(\frac{0.30}{0.10}\right) = \frac{2.303}{30} \cdot \log 3
  • k=2.30330×0.48k = \frac{2.303}{30} \times 0.48

Result


k = 0.0368 s−1≈3.68×10−2 s−10.0368 \text{ s}^{-1} \approx 3.68 \times 10^{-2} \text{ s}^{-1}

Answer keywords:first order reactiontotal pressure p₀ + premaining reactant pressure 0.10 atmintegrated rate lawk = (2.303/t) log(p₀/p)log 3 = 0.48rate constant 3.68 × 10⁻² s⁻¹gas phase decomposition

Marking scheme

  1. 1.1 mark: relating total pressure to reactant pressure (Pt=p0+pP_t = p_0 + p) and finding remaining reactant pressure = 0.10 atm.
  2. 2.1 mark: writing the correct first-order integrated equation k=2.303tlog⁡[p0p0−p]k = \frac{2.303}{t} \log\left[\frac{p_0}{p_0 - p}\right].
  3. 3.1 mark: substituting log⁡3=0.48\log 3 = 0.48 to get k=0.0368 s−1k = 0.0368 \text{ s}^{-1} (accept 3.68×10−2 s−13.68 \times 10^{-2} \text{ s}^{-1}) with correct unit.

Hint

Total pressure=p0+p\text{Total pressure} = p_0 + p (decomposed); find p from 0.50=0.30+p0.50 = 0.30 + p, so reactant left = 0.10 atm; then k=2.30330log⁡(0.300.10)k = \frac{2.303}{30} \log\left(\frac{0.30}{0.10}\right).

Quick oral answer

Since total pressure equals initial pressure plus the pressure decomposed, the reactant left is 0.30−0.20=0.10 atm0.30 - 0.20 = 0.10 \text{ atm}; putting this into k=2.30330log⁡(0.300.10)=2.30330(0.48)k = \frac{2.303}{30} \log\left(\frac{0.30}{0.10}\right) = \frac{2.303}{30}(0.48) gives k≈3.68×10−2 s−1k \approx 3.68 \times 10^{-2} \text{ s}^{-1}.

Common mistakes

  1. 1.Using total pressure 0.50 atm inside the log instead of the remaining reactant pressure 0.10 atm.
  2. 2.Forgetting that one reactant molecule forms two product molecules, so misreading how total pressure relates to decomposition.
  3. 3.Giving the wrong unit — a first-order rate constant must be in s⁻¹, not atm s⁻¹ or mol L⁻¹ s⁻¹.

Analysis and explanation

Concept — pressure as a concentration measure


For a gaseous reaction at constant T and V, partial pressure is proportional to concentration, so the integrated first-order law can be written in pressures. The trick is to relate the measured total pressure to the partial pressure of the reactant that remains.


Reading the stoichiometry


  • One mole of reactant produces two moles of gaseous product, so total moles (and total pressure) rise as the reaction proceeds.
  • The increase in total pressure (0.50−0.30=0.20 atm0.50 - 0.30 = 0.20 \text{ atm}) exactly equals the pressure of reactant consumed.

Exam trap


  • Do not put the total pressure 0.50 into the log term — you need the reactant's remaining pressure, 0.10 atm.
  • Keep units of k as s⁻¹ (first order), and use the given log 3 = 0.48 rather than a calculator value.

Real-world link


Monitoring total pressure is the standard laboratory technique for following gas-phase decompositions (like N₂O₅ or ethyl chloride) because pressure is far easier to measure continuously than concentration.

Interesting facts

The gas-phase pyrolysis of ethyl chloride to ethene and HCl is a classic first-order unimolecular reaction studied since the 1930s.

Because 1 mole of reactant makes 2 moles of gas, the total pressure of such a reaction eventually rises to twice the initial value when decomposition is complete.

Rate constants with units of s−1\text{s}^{-1} (inverse time only) are a fingerprint of first-order kinetics — the units immediately reveal the order.

Question details

Q22 of 37 questions

Board
CBSE
Class
Class 12
Subject
Chemistry
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Chemical Kinetics
Topic
First-order rate constant from pressure data
Bloom's level
apply
Syllabus topic
Unit: Chemical Kinetics — Integrated rate equation for a first-order gaseous reaction using total pressure
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why does the total pressure increase during this reaction?

Each mole of C₂H₅Cl decomposes into one mole of C₂H₄ and one mole of HCl — two gas molecules from one. Since more gas molecules occupy the fixed volume, the total pressure rises as the reaction proceeds, from 0.30 atm toward a final 0.60 atm at completion.

How do I find the pressure of reactant remaining from total pressure?

If p is the pressure of reactant decomposed, the total pressure is p0+pp_0 + p. From 0.50=0.30+p0.50 = 0.30 + p you get p=0.20 atmp = 0.20 \text{ atm}, so the reactant remaining is p0−p=0.30−0.20=0.10 atmp_0 - p = 0.30 - 0.20 = 0.10 \text{ atm}. This 0.10 atm is what goes into the first-order log term.