Question
Half life of a first order reaction is 900 second. If initial concentration of reactant is 0.08\, mol dm ^ -3 find concentration that remains after 35 minute ?
MHT CET
Chemistry
Chemical Kinetics
2026
MHT CET 2026 (19 April Shift 2)
MHT CET Chemistry Question (2026) — Solution
Options
- A. 0.159\, mol dm ^ -3
- B. 0.0159\, mol dm ^ -3
- C. 1.05\, mol dm ^ -3
- D. 0.759\, mol dm ^ -3
Answer
B. 0.0159\, mol dm ^ -3
Step-by-step solution
Given t_ 1/2 = 900 s and t = 35 min = 35 60 = 2100 s. Number of half-lives n = t t_ 1/2 = 2100 900 = 7 3 . For a first order reaction, the concentration remaining after n half-lives is given by: [A] = [A]_0 2^n Substituting the given values: [A] = 0.08 2^ 7/3 [A] = 0.08 4 2^ 1/3 Since 2^ 1/3 1.26: [A] 0.08 4 1.26 = 0.08 5.04 0.0159 mol dm ^ -3 Answer: 0.0159\, mol dm ^ -3
Related: Chemistry — Chemical Kinetics · All PYQ Banks