Question

Some nuclear power plants use plutonium-239 which is produced in breeder reactors. The rate-determining step is the second B-emission. How long does it take to make 1.00 kg of 299Pu if the reaction is complete when the product is 90% 39Pu ?

Answer 1

0.357 days is required for the completion of 90% of 1kg 239Pu at the rate-determining step for the reaction of the second B-emission.

What is the rate-determining step?

In chemical kinetics, the slowest step, sometimes referred to as the rate-determining step, or rate-limiting step frequently determines the overall pace of a process. This approximation of the rate-determining step is frequently used to simplify the prediction of the relevant rate equation for a given reaction mechanism (for comparison with the experimental rate law).

The collection of simultaneous rate equations for the various phases of the mechanism can theoretically be used to estimate the temporal evolution of the reactant and product concentrations. However, it is not always simple to solve these differential equations analytically; in some circumstances, numerical integration may even be necessary.  The single rate-determining step hypothesis can be very beneficially simplified.

Calculation:

We know that,

Half-life of 239Pu is 2.35 days

Then,

K=㏑$\frac{2}{t1/2}$;

⇒K=$\frac{ln2}{2.35days}$;

⇒K=0.294956day∧-1;

Given

N0=1kg;

Nf=0.90kg;

K=$\frac{1}{t}$×㏑$\frac{N0}{Nf}$;

⇒t=$\frac{1}{K}$×㏑$\frac{N0}{Nf}$;

⇒t=$\frac{1}{0.294956day^-1}$×㏑$\frac{1.00}{0.900}$;

⇒t=0.357days

0.357 days is required for the completion of 90% of 1kg 239Pu at the rate-determining step for the reaction of the second B-emission.

To refer more about Rate determining step visit:

brainly.com/question/27908759

#SPJ4