Chemistry
posted by Kim .
The decomposition of ehtane, C_2H_6, is a firstorder 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 4.17E4
Aln100/27)/4.17E4
t=52min
B= (ln100/x)=4.17E4*22
lnx=4.0547
e^94.0547)=57.667
10057.667 =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= 4.17E4 *1320seconds ln100
lnx=4.0547
x=E^4.0547
x=57.67
10057.67=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?