What is the half-life of a compound if 69 percent of a given sample of the compound decomposes in 59 min? Assume first-order kinetics.
The boiling point of diethyl ether CH3CH2OCH2CH3 is 34.500C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is cholesterol .
How many grams of cholesterol, C27H46O (386.6 g/mol), must be dissolved in 263.0 grams of diethyl ether to raise the boiling point by 0.4500C ?
ln(No/N) = kt
Assume No = 100, then N = 31 after 69% has decomposed. Plug in 59 min for t and calculate k.
Then go to k=(0.693/t1/2) and calculate t1/2. The half-life will be in minutes.
To determine the half-life of a compound using first-order kinetics, we can use the following formula:
t1/2 = ln(2) / k
Where:
t1/2 = half-life of the compound
ln = natural logarithm
k = rate constant
In this case, we are given that 69% of the compound decomposes in 59 minutes. To use this information, we need to find the value of the rate constant, k.
Let's assume the initial concentration of the compound is 'A0'. After 59 minutes, 69% of the compound decomposes, leaving us with 31% remaining. So the final concentration, 'A', is 0.31A0.
The rate constant (k) can be determined using the following equation:
k = ln(A0 / A) / t
Where:
A0 = initial concentration
A = final concentration
t = time
In this case, A0 is the initial concentration (100%) and A is the final concentration (31%). t would be 59 minutes.
k = ln(1 / 0.31) / 59
Now that we have the value of k, we can use it to calculate the half-life using the formula mentioned earlier:
t1/2 = ln(2) / k
By substituting the value of k, we can find the answer:
t1/2 = ln(2) / (ln(1 / 0.31) / 59)
Evaluating this expression will give us the half-life of the compound based on the given information.