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March 26, 2017

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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?

  • Chemistry - ,

    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.

  • Chemistry - ,

    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

  • Chemistry - ,

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

  • Chemistry - ,

    The K comes from dividing it by 212seconds

  • Chemistry - ,

    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.

  • Chemistry - ,

    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%

  • Chemistry - ,

    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.

  • Chemistry - ,

    Thank you for you help

  • Chemistry - ,

    for the same question:

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

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