Question

A first order reaction is 40% complete in 80 minutes. calculate the value of rate constant k in what time will the reaction be 90%

Answer 1

40 % is completed = 80 mintues

1 % = 80/40

K = 2

90% completed = 2 x 90

180 mintues

Answer 2

Answer:

Concept:

The rate and orientation of a chemical reaction are quantified in chemical kinetics by the reaction rate constant, or reaction rate coefficient, k. Simulations of molecular dynamics can be used to determine the rate constant for simple processes. Calculating the molecule's mean residence time in the reactant state is one strategy that might be used. This is possible for small systems with brief residence durations, but it is not universally applicable because reactions are frequently uncommon events at the molecular level. Divided Saddle Theory is a strategy that is straightforward to solve this issue. For calculating rate constants, various techniques as the Bennett Chandler process and mile stoning are also developed.

Given:

In 80 minutes, a first order reaction is 40% finished.

Find:

calculate the value of rate constant k in what time will the reaction be 90%

Answer:

• Find out how many atoms are involved in the reaction's fundamental step.
• For each atom taking part in the process, determine its sequence of reaction.
• Increase each reactant's initial concentration until it is in the sequence in which it will react, then multiply all of them together.
• Subtract the rate from the outcome of the preceding step.
• The units of your rate constant will be determined by the overall order of the reaction.

For the initial reaction, $k$

$=(\frac{2.303}{t})log\frac{a}{a-x}$

Assume,

$a=100$%,

$x=40$%,

$t=80min$

$a-x=100-40$

         $=60$

$=k(\frac{2.303}{80} )log(\frac{100}{60} )$

$k=0.00638min^{-1}$

Consequently, the rate constant's value is $0.00638min^{-1}$

$2.t=?$

when

$x=90%$%

$a-x=100-90$

         $=10$

from above,

$k = 0.00638 min^{-1}$

$t=(\frac{2.303}{0.00638} )log(\frac{100}{10} )$

$t=360.97$

$90$% of the reaction will have occurred after $360.97$ minutes.

#SPJ2