Sulphuryl chloride, SO2Cl2, decomposes according to the following first-order reaction:
SO2Cl2(g) �¨ SO2(g) + Cl2(g)
At a certain temperature, it is observed that it takes 147 minutes for 23.0% of the sulphuryl chloride to react. Calculate the half-life for the reaction in units of min.
ln(No/N) = kt
Take No = 100, then N = 77, t is 147 min and solve for k.
Then k = 0.693/t1/2 and solve for t1/2
Post your work if you get stuck.
how did you get No and N?
I just assumed 100 for No (the number of molecules at the beginning), then if 23% react, that leaves 77. You could take ANY number for No and take 23% of that and subtract from No to obtain N. OR, you could take No = No, then N = 0.77*No but it's easier to stick in a number. Does that help.
To calculate the half-life of a first-order reaction, you can use the formula:
t1/2 = ln(2) / k
Where:
- t1/2 is the half-life of the reaction.
- ln(2) is the natural logarithm of 2, approximately 0.693.
- k is the rate constant of the reaction.
In this case, we need to determine the rate constant, k.
First, we need to rearrange the integrated rate law for a first-order reaction:
ln([SO2Cl2]t / [SO2Cl2]0) = -kt
Where:
- [SO2Cl2]t is the concentration of SO2Cl2 at a given time, t.
- [SO2Cl2]0 is the initial concentration of SO2Cl2.
- k is the rate constant of the reaction.
- t is the time.
We are given that it takes 147 minutes for 23.0% of the sulphuryl chloride to react. Therefore, at this time (t = 147 min), the concentration of SO2Cl2 is 0.230 times the initial concentration.
ln(0.230) = -k * 147
Now we can solve for k:
k = -ln(0.230) / 147
Now that we have the rate constant, we can calculate the half-life:
t1/2 = ln(2) / k
t1/2 = ln(2) / (-ln(0.230) / 147)
Simplifying further:
t1/2 = (147 * ln(2)) / -ln(0.230)
Calculating this expression will give us the half-life of the reaction in units of minutes.