# 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:

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

## 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 [I_{2}] = 0.15 M, [H_{2}] = 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

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