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smorlaz Group Title

This reaction was monitored as a function of time: AB --> A + B A plot of 1/[AB] versus time yields a straight line with slope of -0.055/M*s. a) What is the value of the rate constant (k) for this reaction t this temperature? b) Write the rate law for the reaction. c) What is the half-life when the initial concentration is 0.55M? d) If the initial concentration of AB is 0.250M, and the reaction mixture initially contains no products, what are the concentrations of A and B after 75s?

  • 2 years ago
  • 2 years ago

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  1. Xishem Group Title
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    Well, the integrated rate law for a second-order reaction written in slope-intercept form when concentration is being plotted versus time is this... \[\frac{1}{[AB]}=kt+\frac{1}{[AB]_0}\]The slope here would be k, the rate constant. That should give you the answer for number one.

    • 2 years ago
  2. smorlaz Group Title
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    Thanks, I was wondering about that but wasn't sure. I knew it was second order but forgot that the slope would be k. That's what I get from insomnia.

    • 2 years ago
  3. Xishem Group Title
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    The rate law will take a general form of this...\[Rate=k[AB]^x\]We know that the exponent must be 2, and we already know the value of the rate constant, k...\[Rate=-0.055[AB]^2\]I'm not sure about the rate constant being negative, though. It seems strange and counter-intuitive. It would basically mean that if there is any positive amount of reactant that the reaction would produce reactant. Maybe someone else who knows more about it could input, but I don't think the slope of the graph should be negative, ever. It can be negative for a first-order reaction, but that's because the equation is...\[\ln([A])=-kt+\ln([A]_0)\]Someone else may have to help you with it in this case, because a negative k doesn't make sense to me.

    • 2 years ago
  4. smorlaz Group Title
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    Thanks for your help though. I did get that far but the negative answer and slope doesn't make sense to me so far. I'm trying to figure out part d right now but am a it stuck.

    • 2 years ago
  5. Xishem Group Title
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    Well, using the integrated rate law equation, you can plug in initial concentration, k, and t to find the concentration after t seconds. And you know that for every mole of reactant that is used you'll produce 1 mole of each of the reactants...\[1AB \rightarrow 1A+1B\]Does that help?

    • 2 years ago
  6. Xishem Group Title
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    1 mole of each of the products*

    • 2 years ago
  7. smorlaz Group Title
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    Hmm, yes. It does help. I know that much. Thanks for your help. I shouldn't have changed majors since it's been 4 years since I've used Chemistry. I take it that after you find 1/[AB]subscript t, you can use that to find molarity of [AB]t. Then, you use that value to find [A] and [B] after 75 seconds? I could be wrong but that's my intuition speaking.

    • 2 years ago
  8. Xishem Group Title
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    Yes. For example, if the concentration of reactant changes from 0.250M to 0.100M after 75 seconds, then the concentrations of the products will be equal to the difference. In this case, 0.150M.

    • 2 years ago
  9. smorlaz Group Title
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    Hmm, I got a negative answer for 1/[AB]t which throws me off. Units should be in /M*s so that sets me off for how to find [A] and [B] since we don't have a given volume to cancel out units of /M.

    • 2 years ago
  10. Xishem Group Title
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    The negative k value throws this off, yes. You don't need a volume unit here. Given that the volume of the resulting solution is the same (which it will be), the change in concentration of reactant will be the same as the change in concentration of the product. If you have 1 mole of decomposes according to this reaction, you'll have 1 mole of each of the products, yes? And since the volume of the solution stayed the same, the concentrations will therefore be the same. Make sense? You'll need to figure out what the negative rate constant is all about first, though.

    • 2 years ago
  11. smorlaz Group Title
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    Yes, that makes sense. I'm still trying to figure out what the negative rate constant is all about but am still stumped on that part. I do appreciate all your help though.

    • 2 years ago
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