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.

Question

A. How long in minutes will it take to decompose ethane so that 27% remain
B. What percentage of ethane is decomposed after 22minutes?

Answer 1

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.

Answer 2

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

Answer 3

I don't obtain k = your number
ln(0.00839/0.00768)= 21k
k = ??

Answer 4

The K comes from dividing it by 212seconds

Answer 5

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.

Answer 6

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%

Answer 7

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.

Answer 8

Thank you for you help

Answer 9

for the same question:

what is the rate of decomposition (in mol/L*hr) when C2H6= 0.00422 M?