The decomposition of ehtane, C_2H_6, is a first-order reaction. It is found that it takes 212seconds to decompose 0.00839M C_2H_6 to 0.00768M.
A. How long in minutes will it take to decompose ethane so that 27% remain
B. What percentage of ethane is decomposed after 22minutes?
ln(No/N) = kt
Substitute 0.00839 for No
0.00768 for N and calculate k. Then you can use the same formula for A and B after finding k.
A. I suggest using 100 for No and 27 for N.
B. I suggest using 100 for No, substitute for k and t and solve for N. If you start with 100 then N will be in percent.
Post your work if you get stuck.
k- 4.17E-4
A-ln100/27)/4.17E-4
t=52min
B= (ln100/x)=4.17E-4*22
lnx=4.0547
e^94.0547)=57.667
100-57.667 =42%
Thank you
I don't obtain k = your number
ln(0.00839/0.00768)= 21k
k = ??
The K comes from dividing it by 212seconds
I guess it would help if I keyed in the right numbers. The answer to A is OK. I think 27% usually is understood to means 27.0%; however, if it really is 27% then 53% is the right answer; otherwise you would be allowed another s.f. and you could write 52.3 min.
B. Look at B again. If you are using k in seconds, shouldn't you convert k to minutes OR convert 22 min to seconds? You must have done so because the numbers you have don't give 42% but 42% is correct.
I converted 22 min into second
-lnx= 4.17E-4 *1320seconds -ln100
lnx=4.0547
x=E^4.0547
x=57.67
100-57.67=42.33%
I thought that's what you did but you didn't type it that way the first time. 42%, to two s.f., is correct.
Thank you for you help
for the same question:
what is the rate of decomposition (in mol/L*hr) when C2H6= 0.00422 M?
To solve these problems, we can use the first-order reaction kinetics equation:
ln([A] / [A]0) = -kt
Where:
[A] is the concentration of the reactant at a given time,
[A]0 is the initial concentration of the reactant,
k is the rate constant, and
t is the time.
Given that the decomposition reaction of ethane is first-order, we can use this equation to find the answers to both questions.
A. How long in minutes will it take to decompose ethane so that 27% remains?
To solve this, we can rearrange the equation:
ln([A] / [A]0) = -kt
Rearranging gives us:
kt = -ln([A] / [A]0)
Substituting the given values:
k * 212 seconds = -ln(0.00768M / 0.00839M)
Solving for k:
k = -ln(0.00768M / 0.00839M) / 212 seconds
Now, we can find the time it takes to decompose 27% of ethane. Let's call it t1:
ln(0.27M / 0.00839M) = -k * t1
Substituting the value of k, we get:
ln(0.27M / 0.00839M) = (-ln(0.00768M / 0.00839M) / 212 seconds) * t1
Simplifying further, we have:
t1 = (ln(0.27M / 0.00839M) * 212 seconds) / ln(0.00768M / 0.00839M)
To convert the time to minutes, we divide by 60:
t1 in minutes = [(ln(0.27M / 0.00839M) * 212 seconds) / ln(0.00768M / 0.00839M)] / 60
B. What percentage of ethane is decomposed after 22 minutes?
Using the same equation:
ln([A] / [A]0) = -kt
We can rearrange it to find the percentage of ethane decomposed:
[A] = [A]0 * e^(-kt)
Substituting the given values:
[A] = 0.00839M * e^(k * 22 minutes)
Now, we can calculate the percentage of ethane decomposed:
% Decomposed = ([A]0 - [A]) / [A]0 * 100
Substituting the values, we get:
% Decomposed = (0.00839M - 0.00839M * e^(k * 22 minutes)) / 0.00839M * 100
So, the percentage of ethane decomposed after 22 minutes can be found by substituting these values and calculating the expression.