Question

The rate constant of a reaction is 4.7×10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C?

Answer 1

Answer:

k is 3,18*10⁻² s⁻¹ at 75°C

Explanation:

following Arrhenius equation:

k= k₀*e^(-Ea/RT)

where k= rate constant , k₀= frequency factor , Ea= activation energy , R= universal gas constant T=absolute temperature

then for T₁=25°C =298 K

k₁= k₀*e^(-Ea/RT₁)

and for T₁=75°C = 348 K

k₂= k₀*e^(-Ea/RT₂)

dividing both equations

k₂/k₁= e^(-Ea/RT₂+Ea/RT₁ )

k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )]

replacing values

k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )] = 4,7*10⁻³ s⁻¹ *e^[-33.6*1000 J/mol /8.314 J/molK*(1/ 348 K -1/298 K )] = 3,18*10⁻² s⁻¹

thus k is 3,18*10⁻² s⁻¹ at 75°C