Question

Part A

A certain first-order reaction has a rate constant of 3.00×10⁻² s−1 at 25 ∘C . What is the value of k at 69 ∘C if Ea = 77.0 kJ/mol ?

Express your answer using two significant figures.

Part B

Another first-order reaction also has a rate constant of 3.00×10⁻² s−1 at 25 ∘C . What is the value of k at 69 ∘C if Ea = 115 kJ/mol ?

Express your answer using two significant figures.

Answer 1

Part A)

We know the formula

ln(k₂/k₁)=(-Ea/RT₂)+(Ea/RT₁)

So by putting given values in above formula , we get

ln(k₂/ 3.00 x 10⁻² s⁻¹ ) = (-77000 J/mol / (8.31 x 342 K ) + ( 77000 J/mol / (8.31 x 298 K )

ln(k₂/ 3.00 x 10⁻² s⁻¹ ) =   4.00

So

k₂ = e⁴ x 3.00 x 10⁻² s⁻¹

= 1.63 s⁻¹

Part B)

We know the formula

ln(k₂/k₁)=(-Ea/RT₂)+(Ea/RT₁)

So by putting given values in above formula , we get

ln(k₂/ 3.00 x 10⁻² s⁻¹ ) = (-115000 J/mol / (8.31 x 342 K ) + ( 115000 J/mol / (8.31 x 298 K )

ln(k₂/ 3.00 x 10⁻² s⁻¹ ) = 5.97

So

k₂ = e^5.97 x 3.00 x 10⁻² s⁻¹

= 11.75 s⁻¹