Question

A second order reaction is 41.5% complete in 500 sec.?

a) Calculate the rate constant? 

b) What is the value of the half-life? 

c) How long will it take for the reaction to go to 25%, 80% completion?

Answer 1

second order kinetics

$\frac{1}{[A]} = \frac{1}{[A_0]} + kt$

since it's 41.5% complete

so remaining amount is 58.5 %

[A] = 0.585 [A₀]

$\frac{1}{[A]} - \frac{1}{[A_0]} = k×500$

$\frac{1}{0.585[A_0]} - \frac{1}{[A_0]} = k×500$

$\frac{1.70-1}{[A_0]} = k×500$

$k = \frac{0.70}{[A_0]×500}$

$k = \frac{0.0014}{[A_0]}$

$t_{1/2} = \frac{1}{k[A_0]}$

$t_{1/2} = \frac{1}{\frac{0.0014}{[A_0]}×[A_0]}$

$t_{1/2} = 714.2 sec$

for 25%

$\frac{1}{[A]} - \frac{1}{[A_0]} = kt$

$\frac{1}{0.75[A_0]} - \frac{1}{[A_0]} = \frac{0.0014}{[A_0]}×t$

$t = \frac{0.33}{0.0014} = 235 sec$

for 80%

$\frac{1}{[A]} - \frac{1}{[A_0]} = kt$

$\frac{1}{0.20[A_0]} - \frac{1}{[A_0]} = \frac{0.0014}{[A_0]}×t$

$t = \frac{4}{0.0014} = 2857.14 sec$