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Homework Help: Properties of Reactions

  1. May 5, 2005 #1
    1.How would temperature effect the rate of a reaction? Would it increase its kinetic energy thereby speeding up the reaction. Is it right in saying that as the temperature increases, the frequency of collisions increases between molecules, and the orientation changes?

    2. If we have [tex] 4Fe + 3O_{2} \rightarrow 2Fe_{2}O_{3} [/tex] and we know that the concentration of [tex] O_{2} [/tex] changes 0.05 mol/L every 2 seconds what is the rate of disappearance of [tex] Fe [/tex] in [tex] mol/L*sec [/tex]? So [tex] -\frac{1}{4} \frac{\Delta[Fe]}{\Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = \frac{1}{2} \frac{\Delta[Fe_{2}O_{3}]}{\Delta t} [/tex]. In one second the rate of disappearance of [tex] O_[2} [/tex] would be 0.025 mol/L. So would it be [tex] \frac{0.025}{3} [/tex] for the rate of disappearance of [tex] Fe [/tex]? For the rate of evolution of [tex] Fe_{2}O_{3} [/tex] would be [tex] 2 \times 0.025 [/tex]?

    3. If we have a reaction between X and Z and the rate law is [tex] k[X]^{2}[Z]^{3} [/tex] what would happen if:
    a. concentration of X doubled while Z remains constant
    b. concentration of Z triples while X remains constant
    c. concentrations of both X and Z are doubled

    For (a) would the rate quadruple because X has a reaction order of 2? For (b) would the rate go up by a factor of 8? For (c) would the rate go up by a factor of 16?
  2. jcsd
  3. May 6, 2005 #2
    1. The orientation does not necessarily have to change, but you are correct in assuming the frequency of collisions increases. As the temperature increases the kinetic energy of the molecules increases which = more collisions. The increased number of collisions increases the likelihood that a collision will occur at the proper orientation with the proper activation energy.

    2. Not quite. Keep in mind that the rates being made equivalent are coming from their ratios in the initial equation.

    Writing the formula like this: [tex]-\frac{1}{3}\frac{\Delta[O_2]}{\Delta T}=\frac{1}{2}\frac{\Delta [Fe_2O_3]}{\Delta T}[/tex]

    Is very similar to writing:

    [tex] \frac{2}{3}\Delta[O_2]=\Delta [Fe_2O_3][/tex]

    While not specifically the same as the equation when solving for the mole ratio, I am just trying to draw on the comparison.
    When comparing the rates, you need to take into account the ratio's of each reactant/product when finding the comparable rates.

    The rate of disappearance of [tex]O_2[/tex] is just the value for [tex]\Delta [O_2][/tex]. When finding the rates for the other two substances you need to factor in the numerical ratios aswell.

    3. a) Yes it would quadruple [tex]k(2X)^2 (Z)^3 = 4k(X^2)(Z^3)[/tex]
    b) Careful, it is the Z concentration that is tripled (3Z)^3 = ...
    c) [tex]k(2X)^2 (2Z)^3 = (4)(8)k(X^2)(Z^3)[/tex]
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