Reaction A(g) → 2B(g) + C(g) is a first order reaction. It was started with pure A t/min Pressure of system at time t/mm Hg 10 160 ∞ 240 Which of the following option is incorrect?

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q56 – JEE-MAIN 2025

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Home Chemistry Chemical KineticsQuestion #285

Reaction A(g)

\to

2B(g) + C(g) is a first order reaction. It was started with pure A

\begin{array}{cc} t/min & Pressure of system at time t/mm Hg \\ 10 & 160 \\ \infty & 240 \end{array}

Which of the following option is incorrect?

Clue 1 of 3: Stoichiometry and Total Pressure

Relate the total pressure of the system to the initial pressure of reactant A and the extent of reaction using the given stoichiometry.

🥷

Ninja Strategy•Fundamental Reaction Properties

Recognize that first-order reactions never truly go to completion, making that option likely correct and narrowing down the search for the incorrect statement.

🧠 Full Solution Path

Step 1: Determine initial pressure of A (

P_{0}

) and relate total pressure to extent of reaction.✦ Active

The reaction is A(g) $\to$ 2B(g) + C(g). Let the initial pressure of A be $P_{0}$ . At time $t$ , if $x$ pressure of A has reacted, the partial pressures are $P_{A} = P_{0} - x$ , $P_{B} = 2 x$ , $P_{C} = x$ .

The total pressure at time $t$ is $P_{t} = P_{A} + P_{B} + P_{C} = (P_{0} - x) + 2 x + x = P_{0} + 2 x$ .

At $t = \infty$ , the reaction goes to completion, so $x = P_{0}$ . The total pressure $P_{\infty} = P_{0} + 2 P_{0} = 3 P_{0}$ .

Given $P_{\infty} = 240$ mm Hg, so $3 P_{0} = 240$ mm Hg, which gives $P_{0} = 80$ mm Hg.

Option 1 states initial pressure of A is 80 mm Hg, which is correct.

Step 2: Calculate partial pressure of A at

t = 10

min and evaluate Option 2.○ Expand

At $t = 10$ min, the total pressure $P_{10} = 160$ mm Hg. Using $P_{t} = P_{0} + 2 x$ , we have $160 = 80 + 2 x$ .

Solving for $x$ , we get $2 x = 80$ , so $x = 40$ mm Hg.

The partial pressure of A at $t = 10$ min is $P_{A} = P_{0} - x = 80 - 40 = 40$ mm Hg.

Option 2 states partial pressure of A after 10 minute is 40 mm Hg, which is correct.

Step 3: Calculate the rate constant

k

and evaluate Option 3 and 4.○ Expand

For a first-order reaction, the rate constant $k$ is given by $k = \frac{1}{t} \ln (\frac{P_{0}}{P_{A}})$ .

Substituting values at $t = 10$ min: $P_{0} = 80$ mm Hg and $P_{A} = 40$ mm Hg.

k = \frac{1}{10} \ln (\frac{80}{40}) = \frac{1}{10} \ln (2) = \frac{0.693}{10} = 0.0693 {min}^{- 1} .

Option 3 states the rate constant is $1.693 {min}^{- 1}$ , which is incorrect.

Option 4 states the reaction never goes to completion. First-order reactions theoretically never reach 100% completion in finite time, so this statement is correct.

Since the question asks for the incorrect option, option 3 is the answer.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 56 – JEE-MAIN 2025

Clue 1 of 3: Stoichiometry and Total Pressure

🧠 Full Solution Path