Answer
$k_1 = 0.08949 min^{-1}$
Work Step by Step
A first order reaction is 35.5% completed in (t=) 4.9 minutes.
Let initial concentration of reactant, $a_o$ = 100
Concentration of reaction left after 4.9 minutes, $a_t$= 100 - 35.5 = 64.5
Using the integrated rate law for first order reaction,
$$ k_1 = \frac{1}{t} \times ln (\frac{a_o}{a_t})$$
Substituting the values in the above equation,
$$ k_1 = \frac{1}{4.9 min} \times ln (\frac{100}{64.5})$$
On solving we get,
$$k_1 = 0.08949 min^{-1}$$
