Lena772
  • Lena772
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?
Chemistry
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SOLVED
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katieb
  • katieb
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Lena772
  • Lena772
@timo86m
anonymous
  • anonymous
R U sure the units of your rate constant is accurate. It usually includes M
anonymous
  • anonymous
or M-1 i mean

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Lena772
  • Lena772
That's what the question says :/
Lena772
  • Lena772
Can you help me with this one? @timo86m
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anonymous
  • anonymous
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
Lena772
  • Lena772
May it it be -3/1 ? Cause 3/1 is incorrect. @timo86m
anonymous
  • anonymous
its not d?
Lena772
  • Lena772
It's not D nor C. I only have one more try so I'm not sure whether to chose A or B.
anonymous
  • anonymous
i wouldn't chose negatives. The reason could be the equation is not balanced and you have to balance it first :(
Lena772
  • Lena772
@Zale101 can you please help me with the original question
anonymous
  • anonymous
assuming rate = k * [A] .33 min^-1 * .088 M = 0.02904 M/min Now u can use the rate
anonymous
  • anonymous
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
Abhisar
  • Abhisar
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.
Abhisar
  • Abhisar
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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