posted by Kim .
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.
I don't obtain k = your number
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
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?