Verified Solution • 2025–26 Session moderate

QJEEAdv05Che6af: JEE Advanced 2005 Chemistry - Class 12 JEEAdv05Che6af Solution: At constant temperature and volume, X decomposes as 2X(g) 3Y(g) …

At constant temperature and volume, X decomposes as 2X(g) →\to 3Y(g) + 2Z(g); Px is the partial pressure of X.
Observation No. Time (in minute) Px (in mm of Hg)
1 0 800
2 100 400
3 200 200

(i) What is the order of the reaction to X?
(ii) Find the rate constant
(iii) Find the time for 75% completion of the reaction.
(iv) Find the total pressure when pressure of X is 700 mm of Hg

Step-by-Step Verified Solution

The given reaction is : 2X(g) →\to 3Y(g) + 2Z(g)

(1) Partial pressure (of X) trend :

800 mm Hg ⟶100min 400 mm Hg ⟶100min 200 mm Hg Half-life is constant, hence, it is a first order reaction.

(2) For a first order reaction, rate constant k=0.693half−lifek = {{0.693} \over {half - life}}

or, k=0.693100=6.93×10−3k = {{0.693} \over {100}} = 6.93 \times {10^{ - 3}} min −- 1

(3) Time required for the completion of 75% of the reaction is equal to 2 half lives. Hence, Time taken = (2 ×\times 100) = 200 min

(4) IIT-JEE 2005 Chemistry - Chemical Kinetics and Nuclear Chemistry Question 28 English Explanation

∴\therefore Total pressure = (700 + 150 + 100) mm Hg = 950 mm Hg

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring