#IT
1223 Views

For a first order reaction it's frequency factor is 2×10 to the power 12/s and threshold energy is 110.8 kJ/mole.At what temperature the half life period will be 38.5 mins, if the average energy of reactants is 20.3 kJ/mol


Dibbendu Ganguly 17th Jan, 2020
Answer (1)
Uttara 24th Nov, 2020

Hello,

We know that,

t(1/2)=(ln*ln)/k

k=(ln*ln)/38.5*60

k = 3 * 10^-4  s^-1

k = A *e^-(E/RT)

3*10^-4 = 2*10^2 * e^-(E/RT)

ln(1.5)-16ln(10)= -E/RT

T=E/(R*36.43)

and E =110.8-20.3 = 90.5 *10^3 J

therefore T = (90.5*10^3) / (8.314*36.43) = 298 K

Hope this helps!




Related Questions

UPES Integrated LLB Admission...
Apply
Ranked #28 amongst Institutions in India by NIRF | Ranked #1 in India for Academic Reputation by QS University Rankings | 16.6 LPA Highest CTC
Jindal Global Law School Admi...
Apply
Ranked #1 Law School in India & South Asia by QS- World University Rankings | Merit cum means scholarships | Application Deadline: 30th Nov'24
Nirma University Law Admissio...
Apply
Grade 'A+' accredited by NAAC
Great Lakes PGPM & PGDM 2025
Apply
Admissions Open | Globally Recognized by AACSB (US) & AMBA (UK) | 17.3 LPA Avg. CTC for PGPM 2024 | Application Deadline: 1st Dec 2024
ICFAI Business School-IBSAT 2024
Apply
9 IBS Campuses | Scholarships Worth Rs 10 CR
UPES B.Tech Admissions 2025
Apply
Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books