Question

If the activation energy for the reaction is 268 KJ mol^-1 & the rate constant at 660K is 8.1*10^-3 sec^-1 ;what will be the rate constant at 690K?​

Answer 1

Explanation:

By using arrhenius equation,

⟹logk= 2.303 RT - E a + log A

We get

⟹logA=log(1.8×10-5 )+2.303*8.314*313

= 94140

⟹(log1.8)−5+15.7082)

⟹0.2553−5+15.7082=10.9635

∴logA=(10.9634)=9.914×10 > 10