# Solutions 2

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## 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:

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

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

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