Question
The half life of first order reaction is 850 s. The initial concentration of the reactant is 0.06 mol dm ^ -3 . What concentration would remain after 1200 s ?
MHT CET
Chemistry
Chemical Kinetics
2026
MHT CET 2026 (11 April Shift 2)
MHT CET Chemistry Question (2026) — Solution
Options
- A. 0.023 mol dm ^ -3
- B. 0.035 mol dm ^ -3
- C. 5.25 mol dm ^ -3
- D. 6.25 mol dm ^ -3
Answer
A. 0.023 mol dm ^ -3
Step-by-step solution
For a first order reaction, the rate constant k is given by: k = 0.693 t_ 1/2 Substituting the given half-life t_ 1/2 = 850 s: k = 0.693 850 = 8.15 10^ -4 s ^ -1 The integrated rate law for a first order reaction is: k = 2.303 t ( [A]_0 [A]_t ) Substituting the values t = 1200 s and [A]_0 = 0.06 mol dm ^ -3 : 8.15 10^ -4 = 2.303 1200 ( 0.06 [A]_t ) ( 0.06 [A]_t ) = 8.15 10^ -4 1200 2.303 0.425 Taking the antilog on both sides: 0.06 [A]_t = 10^ 0.425 2.66 [A]_t = 0.06 2.66 0.0226 mol dm ^ -3 Rounding off, the concentration remaining is approximately 0.023 mol dm ^ -3 . Answer: 0.023 mol dm ^ -3
Related: Chemistry — Chemical Kinetics · All PYQ Banks