Theophylline is a pharmaceutical drug that is sometimes used to help with lung f

Question

Theophylline is a pharmaceutical drug that is sometimes used to help with lung function. You observe a case where the lab results indicate that a sample of theophylline (initial concentration of 2.0 times 10^-3 M) has been metabolized to a concentration of 1.0 times 10^-3 M in 24 hours. In another 24 hours the drug concentration was 5.0 times 10^-4 M. What is the rate of decomposition four (4) hours after the initial sample was taken? 4.17 times 10^-5 M/h 2.98 times 10^-3 M/h 2.89 times 10^-2 M/h 1.03 times10^-3 M/h 5.15 times 10^-5 M/h

Explanation / Answer

First find the rate constant:

Initial concentration = 2.0*10-3 M

After 24hrs, concentration =1.0*10-3 M

lnA = -kt + lnA0

where, A0 is initial concentration, t is time   

ln[1.0*10-3M] = -k(24hrs) + ln[2.0*10-3 M]

k= 2.8881132*10-2 hr-1

Now, we will calculate the concentration after 4hrs

lnA = -(2.8881132*10-2 hr-1*4) + ln2.0*10-3 M

A = 1.781797*10-3 M

concentration = 1.781797*10-3 M

Thus, Rate of decomposition = kC

where, C is concentration = 1.781797*10-3 M

Rate of decomposition = 2.8881132*10-2hr-1 * 1.781797*10-3 M = 5.15*10-5 M/hr

So, the answer is e)

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