NTA Abhyas JEE Main2020ChemistryChemical KineticsPractice
The rate constant for the first order decomposition of a certain reaction is described by the equation: log k s -1 = 14.34 - 1.25 × 10 4 K T At what temperature will its half-life period be 256 min ?
Options
- A239.33 kJ ; 669 K
- B259.33 kJ
- C329.33 kJ
- D539.33 kJ
Correct answer
A. 239.33 kJ ; 669 K
Step-by-step solution
When half-life = 256 min, k = ln 2 t 1 / 2 = 0.693 256 × 6 0 s -1 = 4.5 × 1 0 - 5 s -1 ⇒         1.25 × 10 4 T   =   14.34 −   [ l o g   4.5 × 10 − 5 ] ⇒    14 .34 - [ -5+0 .65 ] ⇒ 1 9 · 3 4 - 0 · 6 5 = 1 8 · 6 9 T = 12500 18.69 = 669K