C2H6 --> 2CH3
at a temperature where the half-life is 21 minutes.
If the initial concentration is 8.0 mM, what would the concentration be after 46.0 minutes?
How long would it take to convert 95% of C2H6 at this temperature?
Fist, solve for k.
k = 0.693/t1/2.
The b part of the question is solved with the following.
Set No = 8.0 mM
Solve for N. Units will be in mM.
Substitute k from the first part.
ln(No/N) = kt
For c part.
Use the same equation as part b but us No = 100
N = 5
k form part a
Solve for t.
Post your work if you get stuck.
To answer these questions, we need to understand how the concentration of C2H6 changes over time using the half-life.
The half-life is the time it takes for half of the initial concentration to react. In this case, the half-life of C2H6 is given as 21 minutes.
Question 1: What would the concentration be after 46.0 minutes?
To find the concentration after a certain time, we need to determine the number of half-lives that have passed.
Let's divide the given time (46.0 minutes) by the half-life (21 minutes):
46.0 minutes / 21 minutes = 2.19 half-lives
Since we can't have fractions of half-lives, we can round down to 2 half-lives.
During each half-life, the concentration is reduced by half. After the first half-life, the concentration becomes:
8.0 mM / 2 = 4.0 mM
After the second half-life, the concentration becomes:
4.0 mM / 2 = 2.0 mM
Therefore, the concentration after 46.0 minutes is 2.0 mM.
Question 2: How long would it take to convert 95% of C2H6 at this temperature?
To convert 95% of C2H6, we need to find the time it takes for the concentration to decrease to 5% of the initial concentration.
Since the half-life is 21 minutes, each half-life reduces the concentration by a factor of 2. We can calculate the number of half-lives needed to reach 5% of the initial concentration:
Initial concentration = 100%
Final concentration (5%) = 5%
Reduction factor = 5% / 100% = 0.05
Let's calculate how many times the concentration needs to be halved to reach 0.05:
Reduction factor = 0.5^n (Where n is the number of half-lives)
0.05 = 0.5^n
Taking the logarithm base 0.5 of both sides:
log(0.05) = log(0.5^n)
log(0.05) = n * log(0.5)
Let's calculate n:
n = log(0.05) / log(0.5) ≈ 6.86
The number of half-lives needed to reach 5% of the initial concentration is approximately 6.86, but since we can't have fractions of half-lives, we round up to 7.
Therefore, it would take approximately 7 half-lives to convert 95% of C2H6 at this temperature. Using the given half-life of 21 minutes, the time required would be:
7 half-lives * 21 minutes per half-life = 147 minutes.