Question

a specie with initial concentration 1.4×10^-2 mol L-1 decay by first order kinetics with a rate constant 2.35×10^2L mol-1 s-1. what is the half-life if this specie?​

Answer 1

0.293 / 2.35 × 10^2 seconds.

Step-by-step explanation:

The half-life of a first order reaction can be calculated using the equation:

t1/2 = 0.693 / k

where t1/2 is the half-life, k is the rate constant, and 0.693 is the natural logarithm of 2.

Substituting the values given in the question:

t1/2 = 0.693 / 2.35 × 10^2 L mol-1 s-1 = 0.293 / 2.35 × 10^2 s

So, the half-life of the species is approximately 0.293 / 2.35 × 10^2 seconds.

Answer 2

293 ms

Step-by-step explanation:

The half-life of a first-order reaction can be calculated using the following formula:

t1/2 = 0.693 / k

Where t1/2 is the half-life and k is the rate constant.

For this reaction, the rate constant is 2.35 x 10^2 L/mol/s, so:

t1/2 = 0.693 / (2.35 x 10^2 L/mol/s) = 0.293 / (2.35 x 10^2) s = 0.000293 s = 293 ms