The first-order reaction, SO2Cl2 �¨ SO2 + Cl2, has a rate constant equal to 2.20 �~ 10-5 s-1 at 593 K. What percentage of the initial amount of SO2Cl2 will remain after 6.00 hours?
To determine the percentage of the initial amount of SO2Cl2 that remains after 6.00 hours, we can use the first-order reaction kinetics equation:
ln([A]/[A]0) = -kt
where [A] is the concentration of SO2Cl2 at a given time, [A]0 is the initial concentration of SO2Cl2, k is the rate constant, t is the time, and ln is the natural logarithm.
Since we know the rate constant, k, we can rearrange the equation to solve for [A]/[A]0:
[A]/[A]0 = e^(-kt)
or
[A]/[A]0 = e^-(2.20 × 10^-5 s^-1 × 6.00 hours)
Now let's solve for [A]/[A]0.