The reaction C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol. At 600.0 K, the rate constant, 𝑘, is 6.1×10−8 s−1. What is the value of the rate constant at 770.0 K?

Answer 1

Use the Arrhenius equation:

ln (k2/k1) = -Ea/R (1/T2 - 1/T1)

k1 = 6.1x10^-8s^-1

T1 = 600K

k2 = ?

T2 = 770K

Ea = 262 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol to keep units consistent with those for Ea

ln (k2 / 6.1x10^-8) = -262/0.008314 (1/770 - 1/600)

ln k2 - ln 6.1x10^-8 = -31,513 (0.001299 - 0.001667)

ln k2 - (-16.6) = -31513 (-0.000368)

ln k2 + 16.6 = 11.60

ln k2 = -5.00

k2 = 6.7x10^-3 s^-1

(be sure to check the math)