Question

The decomposition of so2cl2 is first order in so2cl2 and has a rate constant of 1.42 x 10^-4 if the initial concentration of so2cl2 is 1.00 m how long will it take for the concentration to decrease to .78 m

Answer 1

To determine the time for the concentration of SO2Cl2 to decrease from 1.00 M to 0.78 M, one can use the integrated rate law for first-order reactions and solve for time t.

Step-by-step explanation:

The student is dealing with the first-order decomposition of sulfuryl chloride (SO2Cl2). To calculate the time it takes for the concentration to decrease from 1.00 M to 0.78 M, you can use the first-order integrated rate law, which is ln([A]t/[A]0) = -kt. Here, [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.

Solving for t, we get:

ln([A]t/[A]0) = -kt

ln(0.78 M / 1.00 M) = -(1.42 × 10-4 s-1)t

t = ln(0.78) / (-1.42 × 10-4 s-1)

By calculating, you'll find the value of t that represents the time required for the concentration of SO2Cl2 to decrease to 0.78 M.

Answer 2

For a first-order reaction, the rate law would be expressed as:

r = dC/dt = -kC

Integrating it from time zero and the initial concentration, Co, to time, t, and the final concentration, C. We will obtain the second-order integrated law as follows:

ln (C/Co) = -kt

To determine the time needed to change the concentration, we simply substitute the given values from the problem statement as follows:

ln (C/Co) = -kt
ln (.78/1.00) = -1.42x10^-4 (t)
t = 1749.73 s

Therefore, the time it takes to decrease the concentration of SO2Cl2 from 1.00 M to 0.78 M is approximately 1749.72 s assuming that it follows a first - order reaction.