For the reaction cyclobutane(g)$\rightarrow$ 2ethylene(g)at 800 K,the half-life is 0.43 s.Calculate the time needed for the concentration of cyclobutane to fall to 1/64 of its initial value.

Question

For the reaction cyclobutane(g)$\rightarrow$  2ethylene(g)at 800 K,the half-life is 0.43 s.Calculate the time needed for the concentration of cyclobutane to fall to 1/64 of its initial value.

Quizplus · Accepted Answer

Since the half-life is 0.43 s, we can use the half-life formula:

t1/2 = ln 2 / k 
where k is the rate constant

Solving for k:
k = ln 2 / t1/2
k = ln 2 / 0.43 s
k = 1.61 s^-1

Now, we can use the first order reaction equation:
ln [cyclobutane] / [cyclobutane]0 = -kt

We want to find the time needed for [cyclobutane] / [cyclobutane]0 = 1/64, so we can plug in those values and solve for t:

ln (1/64) = -1.61 s^-1 t
t = ln (1/64) / (-1.61 s^-1)
t = 2.6 s

Therefore, the correct answer is B) 2.6 s.

For the Reaction Cyclobutane(g) →\rightarrow→ 2ethylene(g)at 800 K,the Half-Life Is 0

For the Reaction Cyclobutane(g) $\rightarrow$ 2ethylene(g)at 800 K,the Half-Life Is 0