A compound decomposed by first order rxn. The concentration of compound decreases from .1180 M to .0950 M in 5.2 min. What fraction of the compound remains after 6.7 minutes?I'm thinking I set up a proportion maybe. Any ideas how to tackle this problem?
To solve this problem, we can use the first order reaction equation:
ln(Ct/C0) = -kt
Where Ct is the final concentration, C0 is the initial concentration, k is the rate constant, and t is the time.
First, we need to determine the rate constant (k). We can use the given information that the concentration decreases from 0.1180 M to 0.0950 M in 5.2 minutes.
ln(0.0950/0.1180) = -k * 5.2 min
Now, solve the equation for k:
k = -(ln(0.0950/0.1180)) / 5.2 min
Next, we can use the determined rate constant to find the fraction of the compound that remains after 6.7 minutes.
ln(Ct/C0) = -kt
ln(Ct/0.1180) = -k * 6.7 min
Now, we can solve the equation for Ct:
Ct = 0.1180 * e^(-k * 6.7 min)
Now you can plug in the value of k we calculated earlier to get the value of Ct:
Ct = 0.1180 * e^(-(ln(0.0950/0.1180))/5.2 * 6.7) M
This will give you the concentration of the compound after 6.7 minutes. To find the fraction of the compound that remains after 6.7 minutes, divide this concentration by the initial concentration (0.1180 M).
The problem states that it is a first order reaction so you use the first order equation of ln(No/N) = akt
Plug in 0.1180 for No and 0.0950 for N, a = 1, t is 5.2 min, solve for k.
Then ln(No/N) = kt. You know k from above, you know t is 6.7 min, solve for No/N, then take the reciprocal (to get N/No) which will be the fraction remaining. Some students get confused with this and there is another way of doing it.
You could substitute No = 0.1180, k from above and 6.7 for t and solve for N which is the concn remaining. Then solve for N/0.1180 to find the fraction remaining. Same answer either way.