Question

It has been established that the denaturation of a virus obeys first order kinetics with an activation energy equal to 586 KJ.mol-1. The reaction half-life is 4.5h at 29.6%. Calculate the half-life of the reaction at 37%.

Answer 1

We have a first order kinetics.

Let's call:

$\begin{gathered} \lbrack Ao\rbrack\text{ = Concentration of virus. Initial at to} \\ \lbrack A\rbrack\text{ = Final Conc. }at\text{ t} \end{gathered}$

$\begin{gathered} to\text{ Initial time} \\ t\text{ final time} \end{gathered}$$\lbrack A\rbrack\text{ = }\lbrack Ao\rbrack\text{ x }e^{-k\text{ x t }}$

That is the formula to calculate the concentration of the virus

You can have it from this:

$-\frac{d\lbrack A\rbrack}{dt}\text{ = k }\lbrack A\rbrack$

we don't have the "k" specific velocity constant

Half-life is when (A) = (Ao)/2

We have t=4.5 h ar 29.6%

$\frac{0.296}{2}=\text{ 0.296 }e^{-k\text{ x 4.5}}\text{ }$

Then k= 0.15 1/h

At first order, half-time doesn't depend on Concentration

So, the time it is going to be the same:

$\frac{0.37}{2}=0.37\text{ x }e^{-0.15xt}$

Then the half-time will be 4.5 h.