NTA Abhyas JEE Main2020ChemistryChemical KineticsPractice
For the reaction, A + B → P , - d A d t = - d B d t = k A B and R t = 1 A 0 - B 0 ln A B 0 B A 0 when A 0 ≠ B 0 If A 0 = B 0 then the integrated rate law will be
Options
- Ak t = ln A B
- B1 B = 1 A 0 + k t
- C1 A = 1 B 0 + k t
- D1 A = 1 A 0 + k t or 1 B = 1 B 0 + k t
Correct answer
D. 1 A = 1 A 0 + k t or 1 B = 1 B 0 + k t
Step-by-step solution
For a second order reaction, A + B → P , if [ A ] 0 ≠ [ B ] 0 then the integrated rate law equation will be k t = 1 [ A ] 0 - [ B ] 0 ln [ A ] [ B ] 0 [ B ] [ A ] 0 where, k = rate constant, t = time If [ A ] 0 = [ B ] 0 then for second order reaction, A + A → P , integrated rate law equation will be 1 [ A ] = 1 [ A ] 0 + k t or 1 [ B ] = 1 [ B ] 0 + k t