Question

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E = 56.0 kJ/mol. If the rate constant of this 3 - 1 - reaction is 2.7 x 10 M isat 252.0 °C, what will the rate constant be at 274.0 °C? Round your answer to 2 significant digits. -1 k = ï m-. 1 x s ?

Answer 1

Activation energy = Ea = 56.0 KJ /mole

Rate constant = K1 = 2.7 x10^3 M-1s-1

T1 = 252.0C = 252.0 + 273 = 525.0 K

T2 = 274.0C = 274.0 + 273 = 547.0 K

Rate constant = K2 = ?

R = 8.314 x10^-3 KJ

Clausius-Clapeyron equation

log(K2/k1) = Ea/2.303R [ 1/T1 - 1/T2]

log(K2/2.7 x10^3) = 56.0 / 2.303 x 8.314 x10^-3 [ 1/525.0 - 1/ 547.0]

log( K2/2.7 x10^3 ) = 56.0 / 2.303 x 8.314 x10^-3 [ 547.0 - 525.0 / 525.0 x 547.0]

log( K2/2.7 x10^3) = 56.0 / 2.303 x 8.314 x10^-3 ( 22/287175)
log( K2/2.7 x10^3) = 56.0 x 22 / 2.303 x 8.314 x10^-3 x 287175

log( K2/2.7 x10^3) = 1232/ 5498.58

log( K2/2.7 x10^3) = 0.224

K2 / 2.7 x10^3 = 10^0.224

K2 / 2.7 x10^3 = 1.675

K2 = 2.7 x 10^3 x 1.675

K2 = 4.5225 x10^3

K2 = 4.5 x10^3 M-1s-1

Rate constant at 274.0C = K2 = 4.5 x10^3 M-1 s-1