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Lena772

  • one year ago

If the rate constant of a reaction is 0.33 min-1 and the initial reactant concentration is 0.088 M, how many minutes will it take for the reactant concentration to equal 0.013 M?

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  1. Lena772
    • one year ago
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    @timo86m

  2. anonymous
    • one year ago
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    R U sure the units of your rate constant is accurate. It usually includes M

  3. anonymous
    • one year ago
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    or M-1 i mean

  4. Lena772
    • one year ago
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    That's what the question says :/

  5. Lena772
    • one year ago
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    Can you help me with this one? @timo86m

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  6. anonymous
    • one year ago
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    Need more info then this will do (reactant side) -> (products side) In that case dont worry about how many reactants. Just know they produce 2 products CS2 and Cl2. Those reactants produce 3 times as much Cl2 per given time So 3/1

  7. Lena772
    • one year ago
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    May it it be -3/1 ? Cause 3/1 is incorrect. @timo86m

  8. anonymous
    • one year ago
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    its not d?

  9. Lena772
    • one year ago
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    It's not D nor C. I only have one more try so I'm not sure whether to chose A or B.

  10. anonymous
    • one year ago
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    i wouldn't chose negatives. The reason could be the equation is not balanced and you have to balance it first :(

  11. Lena772
    • one year ago
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    @Zale101 can you please help me with the original question

  12. anonymous
    • one year ago
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    assuming rate = k * [A] .33 min^-1 * .088 M = 0.02904 M/min Now u can use the rate

  13. anonymous
    • one year ago
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    You can use a simple y = m x + b .013=-.02904 x + .088 to get 2.58 y is your wanted M m is the rate x is the minutes and b is original http://www.wolframalpha.com/input/?i=.013%3D-.02904+x+%2B+.088 graph

  14. Abhisar
    • one year ago
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    Hi @lena772 ! Your given rate constant is 0.33 \(\sf min^{-1}\), check the unit, it indicates that the reaction is of 1st order!! Consider the reaction to be R ----> P We can write that Rate = k[R} or \(\sf \Large \frac{d[R]}{dt}\)\(\sf =-k[r]^1\\\Large \frac{d[R]}{[R]}=-kdt\) Integrating both sides, \(\sf \large\int_{R_0}^{R}\frac{1}{[R]}.d[R]=\large\int_{0}^{t}-k.dt\) That gives us, \(\sf ln[R]-ln[R_0]=-kt\), where [R] is final concentration and \(\sf [R_o]\) is initial concentration of the reactants and t is your time and k is rate constant. Now substituting the values you have, ln[0.013]-ln[0.088]= -0.33 \(\times\) t => \(\sf ln\frac{0.013}{0.088}=-0.33t\) Solve the equation and you can easily find the value of time. Remember you don't need to do all that calculus stuff each time. You just need to identify the order of reaction by seeing the unit of rate constant and use the integral rate formulas accordingly.

  15. Abhisar
    • one year ago
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    This is a hint for your second question. Suppose there is a reaction 2A + 1/3B ----> 6C + 3D Then we can write, Rate = \(\sf \frac{1}{2}\frac{-d[A]}{dt}=3\frac{-d[B]}{dt}=\frac{1}{6}\frac{d[C]}{dt}=\frac{1}{3}\frac{d[D]}{dt}\)

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