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Homework Help: [Chem] Reaction Rate (Differential Equations)

  1. Dec 31, 2012 #1
    I'm only in a high school class. I just made up this problem to see if I'm getting the right idea in the long run. That, and then there's the fact that differential equations are SO MUCH FUN! :biggrin:
    1. The problem statement, all variables and given/known data
    Molecular bromine (initial concentration: 0.49 M) and formic acid (initial concentration: 0.81 M) react in an aqueous solution in a sealed container, following the equation [itex]Br_2 (aq) + HCOOH (aq) \rightarrow 2Br^- (aq) + 2H^+ (aq) + CO_2 (g)[/itex] at a constant temperature of 25°C. Given that the rate constant (k) for 25°C is 3.50 x 10-3 M-1s-1, find...

    (a)...the molar concentration of molecular bromine as a function of time.
    (b)...the instantaneous rate of appearance of bromine ions at 100 seconds, given [HCOOH] = 0.64 M.
    (c)...the pressure as a function of molar concentration of CO2 (assuming CO2 is the only cause of pressure increase in the container) if the initial pressure is at 1 atm.

    3. The attempt at a solution
    (a) Finding, simply, the rate of the reaction in terms of [itex]Br_2[/itex] concentration. Using the rate equation, we can say that [itex]\left|\frac{d[Br_2]}{dt}\right| = k[Br_2][HCOOH][/itex]. Using the formula [itex]\frac{[HCOOH]}{[Br_2]} = \frac{[HCOOH]_0}{[Br_2]_0} e^{([HCOOH]_0 - [Br_2]_0) kt}[/itex], we can write [HCOOH] as a function of [Br2] over time. Thus, rearranging the equations, we get [itex]\left|\frac{d[Br_2]}{dt}\right| = k[Br_2]([Br_2] \frac{[HCOOH]_0}{[Br_2]_0} e^{([HCOOH]_0 - [Br_2]_0) kt}) = k[Br_2]^2 \frac{[HCOOH]_0}{[Br_2]_0} e^{([HCOOH]_0 - [Br_2]_0) kt}[/itex]. Rearranging, we get [itex]\frac{d[Br_2]}{[Br_2]^2} = k \frac{[HCOOH]_0}{[Br_2]_0} e^{([HCOOH]_0 - [Br_2]_0) kt} dt[/itex]. Integrating, we get [itex]\frac{1}{[Br_2]} = \frac{[HCOOH]_0}{[Br_2]_{0}^{2} - [Br_2]_0[HCOOH]_0} e^{([HCOOH]_0 - [Br_2]_0)kt} + C[/itex], where C is the constant of integration. Taking the reciprocal of both sides, we get [itex]\displaystyle [Br_2] = \frac{[Br_2]_0([Br_2]_0 - [HCOOH]_0)}{C[Br_2]_0([Br_2]_0 - [HCOOH]_0) + [HCOOH]_0 e^{([HCOOH]_0 - [Br_2]_0)kt}}[/itex]. Thus, our function for [Br2] over time is given by [itex][Br_2] = \frac{0.19358}{0.1568C + e^{0.00112t}}[/itex] (I pulled out the negatives here because, in reality, we've been working with the absolute value of the derivative of bromine molarity with respect to time. The answer here has to be positive). Knowing that the reaction starts with an initial concentration of 0.49 M, it becomes clear that C = 3.85803, so our final equation is [itex][Br_2] = \frac{0.19358}{0.604939 + e^{0.00112t}}[/itex].

    (b) I took the liberty of using the equation I found in (a) to get the concentration of [Br2]100 and then using a completely made up number for [HCOOH]100. [itex]\frac{1}{2} \frac{d[Br^-]}{dt} = k[HCOOH]_{100}[Br_2]_{100} \Rightarrow \frac{d[Br^-]}{dt} = 2k[HCOOH]_{100}[Br_2]_{100} = 0.00169 \ Ms^{-1}[/itex].

    (c) This was a little more fun. I decided that, to avoid complexity, I would assume that CO2 would act as an ideal gas. [itex]\frac{d[CO_2]}{dt} = \frac{1}{RT} \frac{dP}{dt}[/itex]. Time differentials cancel, leaving [itex]d[CO_2] = \frac{1}{RT} dP[/itex]. Integrating, we get [itex][CO_2] = \frac{P}{RT} + C[/itex], where C is the constant of integration. Rearanging, we get [itex]P = [CO_2]RT + C[/itex] (I chose to ignore that C is now a different value, though still a constant, because it's easier to solve for just one symbol at the end). Knowing that our initial pressure is 1 atm and our initial concentration of carbon dioxide is 0, we can say that C = 1. Thus, our final equation is [itex]P = [CO_2]RT + 1[/itex].

    If you notice anything wrong, tell me. I want to feel secure in my knowledge of this before I do the stuff my class is doing (which is rather simple, math-wise).
  2. jcsd
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