Question

please show all work. i need to know process for test

A compound decomposes by a first-order process. If 60% of the compound remains after 60 minutes, the half-life of the compound is minutes. -5 O 44 0 -18 45 81

Answer 1

As we know that first order process:

ln(C/Co) = -kt

here C = The concentration of the remaining at time t;
Co =The initial concentration;
k = THe rate constant

As from the question it is given 60% decomposes therefore 40% remains

=>C/Co * 100% = 40% => C/Co = 0.40

ln(0.40) = -k * 60 =>-0.91 = -k*60

=>rate constant k = 0.0152 min-1

and the Half-life = ln(2)/k

= 0.6931/0.0152

= 45.59 ≈ 45 minutes