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Chemistry LibreTexts

Solutions 2

  • Page ID
    47370
  • Q1.1

    Calculate the mean free path and the binary number of collisions per liter per second between Ar atoms at 298 K and 1.00 atm. Use 3.62 Å as the collision diameter of the Ar molecules. Assume ideal gas behavior.

    S1.1

    Calculate the mean free path

    Step 1: Calculate \(\frac{N}{V}\) 

       \[PV=NRT=\frac{N}{N_A}RT\]

    $$\dfrac{N}{V}=\dfrac{PN_A}{RT}=\dfrac{(1atm)(6.022\times 10^{23}mol^{-1})}{(0.08206L\,atm\,K^{-1}mol^{-1})(298K)}=2.46\times 10^{22}L^{-1}=2.46\times10^{25}m^{-3}$$

    Step 2: Calculate the mean free path

    $$\lambda=\dfrac{1}{\sqrt{2}\pi d^{2}(\dfrac{N}{V})}=\dfrac{1}{\sqrt{2}\pi (3.62\times 10^{-10}m)^{2}(2.46\times 10^{25}m^{-3})}=6.982\times 10^{-8}m$$

    Calculate the binary number of collisions

    Step 3: Calculate the average molecular speed

    $$\bar{C}=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8(8.3145JK^{-1}mol^{-1}(298K)}{\pi (39.948\times 10^{-3}\dfrac{kg}{mol})}}=397.419 m/s$$

    Step 4: Calculate the binary # of collisions and convert the unit to $$L^{-1}s^{-1}

    $$z_{11}=\dfrac{\sqrt{2}}{2}\pi d^2\bar{C}(\dfrac{N}{V})^{2}= \dfrac{\sqrt{2}}{2}\pi (3.62\times 10^{-10}m)^{2}(397.419ms^{-1})(2.46\times10^{25}m^{-3})^{2}(\dfrac{1m^{3}}{1000L})=7.0012\times10^{31} L^{-1}s^{-1} $$

     

    Q1.2 

    For molecular oxygen at \(56^{o}C\), calculate the number of collisions a single molecule makes in 1 second and the total number of binary collisions at P=1.0 atm and P=0.25 atm. How does pressure relate to these two quantities? The collision diameter of oxygen is 3.02 A.

    S1.2

    Step 1: Average molecular speed

    $$\bar{C}=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8(8.3145JK^{-1}mol^{-1}(273.15+56)K)}{\pi (32\times 10^{-3}\dfrac{kg}{mol})}}=466.67 m/s$$

    Step 2: Calculate \(\frac{N}{V}\) at P= 1 atm

     \[PV=NRT=\frac{N}{N_A}RT\]

    $$\dfrac{N}{V}=\dfrac{PN_A}{RT}=\dfrac{(1atm)(6.022\times 10^{23}mol^{-1})}{(0.08206L\,atm\,K^{-1}mol^{-1})(329.15K)}=2.2295\times 10^{22}L^{-1}=2.2295\times10^{25}m^{-3}$$

    Step 3:Calculate the number of collisions a single molecule makes in 1 second at P=1 atm

    $$z_{1}=\sqrt{2}\pi d^2\bar{C}\dfrac{N}{V}= \sqrt{2}\pi (3.02\times 10^{-10}m)^{2}(466.67ms^{-1})(2.2295\times10^{25}m^{-3})^{2}=4.216\times10^{9}s^{-1} $$

    Step 4: Calculate the binary number of collisions at P=1 atm

    $$z_{11}=\dfrac{\sqrt{2}}{2}\pi d^2\bar{C}(\dfrac{N}{V})^{2}= \dfrac{\sqrt{2}}{2}\pi (3.02\times 10^{-10}m)^{2}(466.67ms^{-1})(2.2295\times10^{25}m^{-3})^{2}=4.6697\times10^{34} L^{-1}s^{-1} $$

    Step 5: Calculate the number of collisions a single molecule makes in a second at P=0.25 atm

    $$\dfrac{N}{V}=\dfrac{PN_A}{RT}=\dfrac{(0.25atm)(6.022\times 10^{23}mol^{-1})}{(0.08206L\,atm\,K^{-1}mol^{-1})(329.15K)}=5.574\times10^{24}m^{-3}$$

    $$z_{1}=\sqrt{2}\pi d^2\bar{C}\dfrac{N}{V}=1.054\times10^{9}s^{-1} $$

    Step 6: Calculate the binary number of collisions at P=0.25 atm

    $$z_{11}=\dfrac{\sqrt{2}}{2}\pi d^2\bar{C}(\dfrac{N}{V})^{2}=2.9186\times10^{33} m^{-3}s^{-1} $$

    Step 7: relationship between pressure and these two quantities

    $$z_{1}=\sqrt{2}\pi d^2\bar{C}\dfrac{N}{V}=\sqrt{2}\pi d^2\bar{C}\dfrac{PN_A}{RT}$$

    $$z_{11}=\dfrac{\sqrt{2}}{2}\pi d^2\bar{C}(\dfrac{N}{V})^{2}(\dfrac{PN_A}{RT})^2$$

    Therefore, \(z_{1}\) is proportional to pressure, \(z_{11}\) is proportional to pressure squared.

    Q1.3 

    If gas 1 has a molar mass of 72 g/mol and gas 2 has a molar mass of 2 g/mol. How much faster or slower does gas 2 effuse from a small opening than gas 1 at the same temperature?

    S1.3

     

    $$\dfrac{r_{1}}{r_{2}}=\sqrt{\dfrac{M_{1}}{M_{2}}}=\sqrt{\dfrac{2g/mol}{72g/mol}}=\dfrac{1}{6}$$

    Therefore, gas 2 effuses at 6 times the rate of gas 1.

     

    Q1.4

    Write the rate of reaction in terms of the rate of disappearance of reactant and the rate of formation of products:

    1. \(NO_{(g)} + O_{3 (g)} \rightarrow NO_{2(g)} + O_{2(g)}\)
    2. \(2C_2H_{6 (g)}  + 7O_{2(g)} \rightarrow 4 CO_{2(g)}  + 6 H_2O_{(aq)}\)
    3. \(H_{2 (g)}  + I_{2 (g)} \rightarrow 2HI_{(g)}  \)
    4. \(4OH_{(g)}  + H_2S_{(g)} \rightarrow SO_{2(g)} + 2H_2O_{(aq)} + H_{2(g)}\)

    S1.4

    a. Ans:

    In terms of the rate of disappearance of reactants:

        $$rate=-\dfrac{d[NO]}{dt}=-\dfrac{d[O_{3}]}{dt}$$

    In terms of the rate of formation of products:

        $$rate=\dfrac{d[NO_{2}]}{dt}=\dfrac{d[O_{2}]}{dt}$$

    b. Ans:

    In terms of the rate of disappearance of reactants:

        $$rate=-\dfrac{1}{2}\dfrac{d[C_{2}H_{6}]}{dt}=-\dfrac{1}{7}\dfrac{d[O_{2}]}{dt}$$

    In terms of the rate of formation of products:

        $$rate=\dfrac{1}{4}\dfrac{d[CO_{2}]}{dt}$$

    c. Ans: 

    In terms of the rate of disappearance of reactants:

        $$rate=-\dfrac{d[H_{2}]}{dt}=-\dfrac{d[I_{2}]}{dt}$$

    In terms of the rate of formation of products:

        $$rate=\dfrac{1}{2}\dfrac{d[HI]}{dt}$$

    d. Ans:

    In terms of the rate of disappearance of reactants:

        $$rate=-\dfrac{1}{4}\dfrac{d[OH]}{dt}=-\dfrac{d[H_{2}S]}{dt}$$

    In terms of the rate of formation of products:

        $$rate=\dfrac{d[SO_{2}]}{dt}=\dfrac{[H_{2}]}{dt}$$

    Q1.5

    Determine the value of the rate constant for the (elementary) reaction:

    \[I_{2(g)} + H_{2 (g)} \rightarrow 2HI_{(aq)}\]

    If the [I2] = 0.15 M, [H2] = 0.2M, and the rate of reaction  is 0.005 M s-1 at 298 K.

    S1.5

    Assuming this is elementary reaction and the reaction rate can be expressed as:

        $$rate=k[Br_{2}][H_{2}]$$

    Then the rate constant k is:

        $$k=\dfrac{rate}{[Br_{2}][H_{2}]}=\dfrac{0.005M s^{-1}}{0.15M  0.2M}=0.1667 M^{-1} s^{-1}$$

    Q1.6

    What are the units of the rate constant for a second-order reaction?

    S1.6

    For a second order reaction, 

        $$rate=k[A]^{2}$$

    So $$k=\dfrac{rate}{[A]^{2}}$$

    The unit of the reaction rate is: \(M s^{-1}\)

    The unit of the concentration of reactant A is: M

    Therefore, the unit of the rate constant k is:

    $$\dfrac{M s^{-1}}{M^{2}}=M^{-1} s^{-1}$$

    Q1.7

    Derive the half-life formula for the 0th, 1st, and 2nd order kinetics.

    S1.7

    For 0th order reaction:

    The rate of reaction is:

      $$-\dfrac{d[A]}{dt}=k[A]^{0}=k$$

    So,

      $$d[A]=-k dt$$

    Integrate on both sides,

    \[\int_{[A]_{0}}^{[A]} d[A] = - \int_{0}^{t} kdt \]

    Then, solve for \([A]\). This provides the integrated form of the rate law.

    \[[A] = [A]_0 -kt \]

    The integrated form of the rate law allows us to find the population of reactant at any time after the start of the reaction.

    To find \(t_{1/2}\),  we can plug in \([A]=[A]_{0}\) in the equation and solve for \(t_{1/2}\).

                                                                                   \[ \dfrac{1}{2}[A]_{0}=[A]_{0}-kt_{1/2} \]

    We get

    \[ t_{1/2}=\dfrac{[A]_{0}}{2k}\]

     

    For 1st order reaction:

    The rate of reaction is: 

    $$-\dfrac{d[A]}{dt}k[A]$$

    Rearrange it, we get:

    $$\dfrac{d[A]}{[A]}=-k dt$$

    \[\int_{[A]_{0}}^{[A]} \dfrac{d[A]}{[A]}=-\int_{0}^{t} kdt\]

    $$ln\dfrac{[A]}{[A]_{0}}=-kt$$

    $$[A]=[A]_{0} e^{-kt}$$

    To find \(t_{1/2}\),  we can plug in \([A]=[A]_{0}\) in the equation and solve for \(t_{1/2}\).

    $$\dfrac{1}{2}[A]_{0}=[A]_{0} e^{-kt_{1/2}}$$

    Take the natural log on both sides,

    $$ln(\dfrac{1}{2})=ln(e^{-kt_{1/2}})=-kt_{1/2}$$

    We get

    $$ t_{1/2}=\dfrac{ln2}{k}$$

     

    For 2nd order reaction:

    The rate of reaction is: 

    $$-\dfrac{d[A]}{dt}=k[A]^{2}$$

    Rearrange and integrate on both sides,

    $$ \int_{[A]}^{[A]_{0}}\dfrac{d[A]}{[A]^{2}}=-\int_{0}^{t} kdt$$

    $$-(\dfrac{1}{[A]}-\dfrac{1}{[A]_{0}})=\dfrac{1}{[A]_{0}}-\dfrac{1}{[A]}=-kt$$

    So

    $$\dfrac{1}{[A]}=\dfrac{1}{[A]_{0}}+kt$$

    To find \(t_{1/2}\),  we can plug in \([A]=[A]_{0}\) in the equation and solve for \(t_{1/2}\).

    $$\dfrac{2}{[A]_{0}}=\dfrac{1}{[A]_{0}}+kt_{1/2}$$

    We get

    $$t_{1/2}=\dfrac{1}{k[A]_{0}}$$