Question

The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36 X 10^-4 s ^-1 at 700˚C: C2H6 ---> 2CH3. Calculate the half-life of the reaction in minutes.

Answer 1

The half-life of the reaction is 21.54 minutes.

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, and k is the rate constant. Given that the rate constant for the given reaction is 5.36 x 10^-4 s^-1, we can calculate the half-life as follows:

t1/2 = (0.693 / 5.36 x 10^-4 s^-1)

t1/2 = 1292.16 s = 21.54 minutes (rounded to two decimal places).

Answer 2

The half-life of the reaction is 21.6 minutes.

Step-by-step explanation:

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

t1/2 = ln(2) / k

where t1/2 is the half-life, ln(2) is the natural logarithm of 2, and k is the rate constant. In this case, the rate constant is 5.36 x 10-4 s-1. Substituting this value into the formula:

t1/2 = ln(2) / (5.36 x 10-4)

t1/2 = 1293.9 s

Converting seconds to minutes:

t1/2 = 1293.9 s * (1 min / 60 s)

t1/2 = 21.6 min