Question

He first-order reaction, so2cl2 → so2 + cl2, has a half-life of 8.75 hours at 593 k. how long will it take for the concentration of so2cl2 to fall to 16.5% of its initial value? the first-order reaction, so2cl2 → so2 + cl2, has a half-life of 8.75 hours at 593 k. how long will it take for the concentration of so2cl2 to fall to 16.5% of its initial value? 22.7 hr 0.143 hr 2.28 hr 6.99 hr

Answer 1

0,693/k = t 1/2
0,693/k = 8,75h
0,693 = 8,75h×k
k = 0,0792 1/h
------------------
C - 1
Ct - 16,5%× 1 = 0,165

lnC/Ct = k*t
ln1/0,165 = 0,0792×t
ln6,06 = 0,0792×t
1,802/0,0792 = t
t = 22,752 h ≈ 22,7 h

:)

Answer 2

Answer:

$t=22.7h$

Explanation:

Hello,

In this case, the rate law for this chemical reaction turns out:

$\frac{dC_{SO_2Cl_2}}{dt}=-kC_{SO_2Cl_2}$

As it is a first-order kinetic, therefore, after integrating, one obtains:

$\int\limits^{C_{SO_2Cl_2}^f}_{C_{SO_2Cl_2}^0} {\frac{1}{C_{SO_2Cl_2}}} \, dC_{SO_2Cl_2} =-k\int\limits^t_0 {} \, dt \\ln(\frac{C_{SO_2Cl_2}^f}{C_{SO_2Cl_2}^0})=-kt$

Now, as it is indicated that the concentration of the reactant falls to 16.5% of its initial value, one does:

$ln(\frac{0.165C_{SO_2Cl_2}^0}{C_{SO_2Cl_2}^0})=-kt$

Nonetheless, the rate constant is obtained via the half life as shown below:

$t_{1/2}=\frac{ln(2)}{k}\\k=\frac{ln(2)}{8.75h}=0.0792h^{-1}$

Finally, the required time results:

$t=-\frac{ln(0.165)}{0.0792h^{-1}} \\t=22.7h$

Best regards.