For reaction A P, rate constant k = 1.5 103 s-1 at 27°C If activation energy for the above reaction is 60 kJ mol-1, then the temperature (in °C) at which rate constant, k = 4.5 103 s-1 is . (Nearest integer) Given : log 2 = 0.30, log 3 = 0.48, R = 8.3 J K-1 mol-1, ln 10 = 2.3
Answer
41
Solution
T1 = 27 + 273 = 300 K, k2k1 = 4.5 1031.5 103 = 3
k2k1 = EaR ( 1T1- 1T2 )
⇒ 3 = 2.3 3 = 2.3 0.48 = 1.104
⇒ 1.104 = 600008.3 ( 1300- 1T2 )
⇒ 1300- 1T2 = 1.104 8.360000
⇒ 1300- 1T2 = 1.527 10-4
⇒ 1T2 = 3.333 10-3 - 0.153 10-3
⇒ 1T2 = 3.181 10-3 K-1
⇒ T2 = 314.4 K
∴ Temperature = 314.4 − 273 = 41.4°C ≈ 41°C
Answer: 41