see in description
Asked by Sunny Acharjee | 13th Jan, 2013, 01:31: PM
The Arrhenius equation is
log10 k2/k1 = Ea/(R×2.303) [(T2-T1)/(T1 T2 )]
Given: k2/k1 = 3; R=8.314 JK-1 mol1; T1 = 20 + 273 = 293 K
and T2 = 50 + 273 = 323 K
Subtracting the given values in the Arrhenius equation,
log10 3 = Ea/(8.314×2.303) [(323-293)/(323×293)]
Ea = (2.303 × 8.314×323 × 293×0.477)/30
= 28811.8 J mol-1
= 28.8118 kJ mol-1
The Arrhenius equation is
log10 k2/k1 = Ea/(R×2.303) [(T2-T1)/(T1 T2 )]
Given: k2/k1 = 3; R=8.314 JK-1 mol1; T1 = 20 + 273 = 293 K
and T2 = 50 + 273 = 323 K
Subtracting the given values in the Arrhenius equation,
log10 3 = Ea/(8.314×2.303) [(323-293)/(323×293)]
Ea = (2.303 × 8.314×323 × 293×0.477)/30
= 28811.8 J mol-1
= 28.8118 kJ mol-1
Answered by | 13th Jan, 2013, 05:46: PM
Related Videos
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change