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Solutions 4

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    Name: ______________________________

    Section: _____________________________

    Student ID#:__________________________


    Compounds A and B both decay by first-order kinetics. The half-life of A is 20 minutes and the half-life of B is 48 minutes. If a container initially contains equal concentrations of compounds A and B, after how long will the concentration of B be twice that of A?

    1. Write, in mathematical terms, the information given in the problem and what the problem is asking for.

    \[ t_{1/2, A} = 20.0 min \]

    \[ t_{1/2, B} = 48.0 min \]

    \[ [A]_0 = [B]_0 \]

    A and B decay by first-order, so

    \[ -\dfrac{d[A]}{dt} = k_A [A] \]

    \[ -\dfrac{d[B]}{dt} = k_B [B] \]

    (Note: the rate constants for A and B are not equal, so indicate which is which with subscripts.)

    Want find t at which the following is true:

    \[ [B] = 2[A] \]

    2. Substitute the integrated rate equations for [A] and [B]

    \[ [B]_0 e^{-k_B t} = 2 [A]_0 e^{-k_A t} \]

    3. Write expressions for the rate constants in terms of half-lives, and substitute into the equation.

    \[ t_{1/2, A} = \dfrac{ln2}{k_A} \Rightarrow k_A = \dfrac{ln2}{t_{1/2, A}} \]

    \[ t_{1/2, B} = \dfrac{ln2}{k_B} \Rightarrow k_B = \dfrac{ln2}{t_{1/2, B}} \]

    \[ [B]_0 e^{-\dfrac{ln2}{t_{1/2, B}} t} = 2 [A]_0 e^{-\dfrac{ln2}{t_{1/2, B}} t} \]

    4. Solve for t

    Since initial concentrations of A and B are equal:

    \[ e^{-\dfrac{ln2}{t_{1/2, B}} t} = 2 e^{-\dfrac{ln2}{t_{1/2, B}} t} \]

    Take natural log of both sides:

    \[ -\dfrac{t}{t_{1/2, B}} ln2 = \left ( 1 - \dfrac{t}{t_{1/2, A}} \right ) ln2 \]

    \[ t = \dfrac{1}{\dfrac{1}{t_{1/2, A}} - \dfrac{1}{t_{1/2, B}}} \]

    5. Plug in values for half-lives

    \[t = 34.2 min\]

    Answer: 34.2 minutes


    If a reaction has come to thermodynamic equilibrium, can we say anything in particular about the system's kinetics?

    \[ A\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}B \]

    When the system is at equilibrium,

    \[ \dfrac{d[A]}{dt} = \dfrac{d[B]}{dt} = 0 \]

    \[ k_{-1} [A] = k_1 [B] \]

    \[ \dfrac{k_{-1}}{k_1} = \dfrac{[B]}{[A]} \]

    \[ K_{eq} = \dfrac{k_{-1}}{k_1} \]


    Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?

    If the reactants in the collision do not have enough energy to overcome the activation barrier (1). Or if energetic enough reactants collide a reaction may not occur because of steric effects. (2)


    How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.

    Generally an increase in temperature increases the rate of reaction. From the collision theory point of view as you increase the temperature you increase the speed of the reactants leading to more collisions and to the collisions having more energy.


    An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?

    \[k = Ae^{\frac{-E_a}{RT_1}} (1)\]

    \[1.47k = Ae^{\frac{-E_a}{RT_2}} (2)\]

    Dividing equation 1 by equation 2 and taking the natural log yields:

    \[-.385 = \frac{-E_a}{RT_1} + \frac{E_a}{RT_2}\]

    \[E_a = 43 kJ/mol\]


    The hydrolysis of the sugar sucrose to the sugars glucose and fructose,

    \[\ce{C12H22O11 + H2O ⟶ C6H12O6 + C6H12O6}\]

    follows a first-order rate equation for the disappearance of sucrose: Rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)

    1. In neutral solution, k = 2.1 × 10−11 s−1 at 27 °C and 8.5 × 10−11 s−1 at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
    2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 × 10−7 M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
    3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?


    \[2.1x10^{-11} = Ae^{\frac{-E_a}{300R}} (4.6.1)\]

    \[8.5x10^{-11}= Ae^{\frac{-E_a}{310R}} (4.6.2)\]

    Dividing equation 4.6.1 by equation 4.6.2 and taking the natural log yields:

    \[-1.398 = \frac{-E_a}{2494.2} + \frac{E_a}{2521.54}\]

    \[-1.398 = 4.35x10^{-6}E_a\]

    \[E_a = 322 kJ/mol\]

    Now knowing the activation energy we can solve for the frequency factor A:

    \[2.1x10^{-11} = A*e^{-125}\]

    \[A = 2.45x10^{45}\]

    \[k = 2.45x10^{45}*e^{-121}\]

    \[k = 6.7x10^{-8}\]

    (b) 1.81 × 108 h or 7.6 × 106 day.

    (c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state.

    Solutions 4 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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