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 $0.013412s^{-1}$

Explanation:

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

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

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:

$k=\frac{2.303}{120}\log\frac{100}{20}\\\\k=0.013412s^{-1}$

Hence, the rate constant for the reaction is $0.013412s^{-1}$