Question

For a first-order reaction, after 2.00 min, 20% of the reactants remain. Calculate the rate constant for the reaction. 1. 74.6 s−1 2. 0.00582 s−1 3. 0.000808 s−1 4. 0.00186 s−1 5. 0.013412 s−1

Answer 1

Answer: The rate constant for the reaction is
`0.013412s^(-1)`

Step-by-step explanation:

Rate law expression for first order kinetics is given by the equation:

`k=(2.303)/(t)\log([A_o])/([A])`

where,

k = rate constant = ?

t = time taken for decay process = 2.00 min = 120 seconds (Conversion factor: 1 min = 60 seconds)

`[A_o]` = initial amount of the sample = 100 grams

[A] = amount left after decay process = 20 grams

Putting values in above equation, we get:

`k=(2.303)/(120)\log(100)/(20)\\\\k=0.013412s^(-1)`

Hence, the rate constant for the reaction is
`0.013412s^(-1)`