WebAug 19, 2024 · A first order reaction takes 69.3 min for 50% completion. See answers Advertisement Advertisement Brainly User Brainly User To be calculated K. 50 % completion that means half life period is given i.e. 69.3 seconds. and for a first order reaction. K = 0.693/half life time. K = 0.693/69.3. WebNov 18, 2024 · A first order reaction takes 69.3 minute for 50% completion. How much time will be needed for 80% completion - (A) 160.97 minute (B) 170.97 minute (C) 150.97 minute (D) 0 minute chemical kinetics jee jee mains 1 Answer +1 vote answered Nov 18, 2024 by kajalk (78.1k points) selected Nov 19, 2024 by faiz Correct Option (A) 160.97 minute …
A first order reaction takes 69.3 minutes for 50
A first order reaction takes 69.3 minutes for 50% completion. How much time will be needed for 80% completion? Medium Solution Verified by Toppr t 1/2=69.3 min= Kln 2 K= 69.3ln 2min −1 For 80 % conversion, if we assume initial concentration to be a o, concentration left would be 5a o t× 69.3ln 2=ln(a o/5a o) t= ln 269.3 ln 5=161 min −1 WebNov 25, 2024 · A first order reaction takes 69.3 minutes for 50% completion. How much time will be needed for 80% completion? See answer Advertisement sharmeelavasu … how many grams in 1 cup dark brown sugar
A first order reaction takes 69.3 min for 50% completion.
WebSep 18, 2024 · A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% of the reaction to get completed? (Given: log 5 =0.6990, log 8 = 0.9030, log 2 = 0.3010) class-12; Share It On Facebook Twitter Email. 1 Answer. 0 votes . answered Sep 18, 2024 by Haren (305k ... WebSep 18, 2024 · A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% of the reaction to get completed? (Given : log 5 =0.6990, log 8 = 0.9030, log 2 = 0.3010) CBSE Chemistry Sample Paper 2024 Class 12 Sample paper solutions Share … WebThe correct option is A 230.3 minutes. For a first order reaction, Rate constant, k= 0.693 t1/2 = 0.693 69.3 =0.01 min−1. k = 2.303 t log10 a a−x. where, a= initial amount of reactant. … hoverforecolor