Question

At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 800.0 K?

Answer 1

The rate constant from the Arrhenius equation is
$k=Ae^{-E_{a}/(RT)}}$
where
A = frequency factor
Ea = activation energy. It is assumed to be 50 kJ/mol, because it s not given.
R =  gas constant = 8.31 J/(K-mol)
T =  temperature, K

Evaluate Ea/R = 50000/8.31 = 6.0168 x 10³ K.
Therefore
$k=Ae^{-6.0168\times10^{3}/T}$

When T₁ = 600 K, k₁ = 6.1 x 10⁻⁸ s⁻¹
Therefore when T₂ = 800 K, the rate constant is
$\frac{k_{2}}{6.1\times10^{-8}}= \frac{e^{-6016.8/800}}{e^{-6016.8/600}} \\ =e^{-6016.8(1/800-1/600)}\\ =e^{2.507}\\ =12.268$
k₂ = 7.484 x 10⁻⁷ s⁻¹
The rate has increased by a factor of about 12.

Answer: The rate constant is approximately 7.5 x 10⁻⁷ s⁻¹