Surface Chemistry

First, calculate the initial moles of acetic acid:

$\text{Moles of CH}_3\text{COOH} = \frac{0.45}{60} = 0.0075$

Now, find its initial molarity. The solution volume is 50 mL = 0.05 L:

$\text{Molarity} = \frac{0.0075}{0.05} = 0.15\ \mathrm{M}$

After adsorption, the given pH is 3. Therefore, the concentration of hydrogen ions is:

$[\text{H}^+] = 10^{-3}\ \mathrm{M}$

The dissociation of acetic acid is represented as:

$\mathrm{CH_3COOH(aq)} \rightleftharpoons \mathrm{CH_3COO^-(aq)} + \mathrm{H^+(aq)}$

According to the expression of the acid dissociation constant,

$K_a = \frac{[\text{CH}_3\text{COO}^-] [\text{H}^+]}{[\text{CH}_3\text{COOH}]}$

Substitute the known values:

$1.0 \times 10^{-5} = \frac{10^{-3} \times 10^{-3}}{C - 10^{-3}}$

Solving for $ C $:

$C = 0.1\ \mathrm{M}$

Now, the amount of acetic acid adsorbed on 1 g of charcoal is the difference between initial and final concentrations:

$x = \frac{(0.15 - 0.1) \times 50 \times 60}{1000} = 0.15\ \mathrm{g}$

From the Freundlich isotherm relation,

$\log \frac{x}{m} = \log K + \frac{1}{n} \log C$

Since the slope is 1, $ \frac{1}{n} = 1 $. Hence,

$\log 0.15 = \log K + \log 0.1$

Combining the logarithmic terms,

$\log 0.15 - \log 0.1 = \log K$

$\log 1.5 = \log K$

Therefore,

$K = 1.5\ \mathrm{L\ mol^{-1}}$

$\begin{aligned} & \frac{x}{m}=K(C)^{\frac{1}{n}} \\\\ & 4=K(10)^{\frac{1}{n}}........(i) \\\\ & 10=K(16)^{\frac{1}{n}}.......(ii)\end{aligned}$

From (ii) and (i)

$ \begin{aligned} & \log 4-\log 10=\frac{1}{n}(\log 10-\log 16) \\ & 2 \log 2-1=\frac{1}{n}(1-4 \log 2) \\ & 0.6-1=\frac{1}{x}(1-1.2) \\ & -0.4=\frac{-0.2}{n} \\ & n=\frac{2}{4}=\left(\frac{1}{2}\right) \end{aligned} $

$\begin{aligned} \text { Now } 10 & =K(16)^2 \\ K & =\frac{10}{256} \\ \text { So, } \frac{x}{m} & =K(20)^2 \\ \frac{x}{m} & =\frac{10}{256} \times 400 \\ \frac{x}{m} & =15.625\end{aligned}$

or

$ \begin{aligned} & 4=K(10)^2 \\ & K=\frac{4}{100} \\ & \frac{x}{m}=K(20)^2 \\ & \frac{x}{m}=\frac{4}{100} \times 400 \\ & \frac{x}{m}=16 \end{aligned} $

Let's first understand and break down the given information. We are provided with the following details to solve for the value of $\mathbf{P}$:

1. Volume of acetic acid solution used: $100 \mathrm{~mL}$ of $0.5 \mathrm{M}$ acetic acid.

2. Volume of $\mathrm{NaOH}$ solution used for neutralization: $40 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{NaOH}$.

3. Surface area of the charcoal: $1.5 \times 10^2 \mathrm{~m}^2 \mathrm{~g}^{-1}$.

4. Avogadro's number: $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$.

Let's start with the concentration and volume of acetic acid used:

$ \text{Moles of acetic acid initially} = 0.5 \text{ M} \times \frac{100 \text{ mL}}{1000} = 0.05 \text{ moles} $

Next, let's determine the moles of $\mathrm{NaOH}$ used for neutralization:

$ \text{Moles of } \mathrm{NaOH} = 1 \text{ M} \times \frac{40 \text{ mL}}{1000} = 0.04 \text{ moles} $

Since $\mathrm{NaOH}$ completely neutralizes the unadsorbed acetic acid:

$ \text{Moles of unadsorbed acetic acid} = 0.04 \text{ moles} $

Therefore, the moles of acetic acid adsorbed on the charcoal is given by:

$ \text{Moles of adsorbed acetic acid} = 0.05 - 0.04 = 0.01 \text{ moles} $

Now, we need to calculate the total number of molecules of acetic acid adsorbed:

$ \text{Number of molecules} = 0.01 \text{ moles} \times 6.0 \times 10^{23} \text{ molecules/mole} = 6.0 \times 10^{21} \text{ molecules} $

Given the surface area of charcoal, we can find the total surface area:

$ \text{Total surface area} = 1.5 \times 10^2 \text{ m}^2 / \text{g} \times 1 \text{ g} = 1.5 \times 10^2 \text{ m}^2 $

Finally, we need to determine the area occupied by each molecule of acetic acid:

$ \text{Area per molecule of acetic acid} = \frac{\text{Total surface area}}{\text{Number of molecules}} = \frac{1.5 \times 10^2 \text{ m}^2}{6.0 \times 10^{21} \text{ molecules}} $

Let's perform the calculation:

$ \text{Area per molecule of acetic acid} = \frac{1.5 \times 10^2}{6.0 \times 10^{21}} = 0.25 \times 10^{-19} \text{ m}^2 = 2.5 \times 10^{-20} \text{ m}^2 $

Since the problem states that the area per molecule is $\mathbf{P} \times 10^{-23} \mathrm{~m}^2$, we equate:

$ 2.5 \times 10^{-20} = \mathbf{P} \times 10^{-23} $

Solving for $\mathbf{P}$:

$ \mathbf{P} = \frac{2.5 \times 10^{-20}}{10^{-23}} = 2.5 \times 10^3 = 2500 $

Therefore, the value of $\mathbf{P}$ is 2500.

The coagulating value of an electrolyte is defined as the minimum amount of the electrolyte required to coagulate a unit volume of the colloidal solution in a given time. In this problem, we are given that $20\ \mathrm{mL}$ of $0.5\ \mathrm{M}$ $\mathrm{NaCl}$ solution is required to coagulate $200\ \mathrm{mL}$ of $\mathrm{As}_2\mathrm{S}_3$ solution in 2 hours.

To find the coagulating value of $\mathrm{NaCl}$, we need to determine the amount of $\mathrm{NaCl}$ required to coagulate 1 liter (1000 mL) of the $\mathrm{As}_2\mathrm{S}_3$ solution in 2 hours. This can be calculated using the following formula:

$\text{Coagulating value} = \frac{\text{amount of electrolyte required to coagulate 1 L of sol in 2 hours}}{1\ \mathrm{L}}$

To find the amount of $\mathrm{NaCl}$ required to coagulate 200 mL of the $\mathrm{As}_2\mathrm{S}_3$ solution in 2 hours, we can use the formula for the number of moles of $\mathrm{NaCl}$ used:

$n_{\mathrm{NaCl}} = C_{\mathrm{NaCl}} \times V_{\mathrm{NaCl}}$

where $C_{\mathrm{NaCl}}$ is the concentration of $\mathrm{NaCl}$ in moles per liter, and $V_{\mathrm{NaCl}}$ is the volume of $\mathrm{NaCl}$ solution used in liters. Substituting the given values, we get:

$n_{\mathrm{NaCl}} = 0.5\ \mathrm{M} \times 0.02\ \mathrm{L} = 0.01\ \mathrm{mol}$

To find the amount of $\mathrm{NaCl}$ required to coagulate 1 L of the $\mathrm{As}_2\mathrm{S}_3$ solution, we need to scale up the amount of $\mathrm{NaCl}$ used by a factor of 5 (since 5 times the given volume of $\mathrm{NaCl}$ solution is required to make 1 L of the $\mathrm{As}_2\mathrm{S}_3$ solution). This gives:

$n_{\mathrm{NaCl}} = 0.01\ \mathrm{mol} \times 5 = 0.05\ \mathrm{mol}$

The coagulating value of $\mathrm{NaCl}$ can now be calculated as:

$\text{Coagulating value} = \frac{0.05\ \mathrm{mol}}{1\ \mathrm{L}} \times 1000 = 50\ \mathrm{mM}$

Therefore, the coagulating value of $\mathrm{NaCl}$ is $\boxed{50}$ (in millimoles per liter).

Let's analyze each statement about modern adsorption theory of heterogeneous catalysis:

A. The catalyst is diffused over the surface of reactants.

This statement is incorrect. In heterogeneous catalysis, the catalyst is usually a solid, and the reactants are in a different phase (gas or liquid). The reactants are adsorbed on the catalyst's surface, not the other way around.

B. Reactants are adsorbed on the surface of the catalyst.

This statement is correct. In the adsorption theory of heterogeneous catalysis, reactants are adsorbed onto the surface of the catalyst, where they undergo a chemical reaction.

C. Occurrence of chemical reaction on the catalyst's surface through formation of an intermediate.

This statement is correct. In the adsorption theory, the reaction takes place on the catalyst's surface through the formation of an intermediate species.

D. It is a combination of intermediate compound formation theory and the old adsorption theory.

This statement is correct. The modern adsorption theory of heterogeneous catalysis combines the concepts of intermediate compound formation and the old adsorption theory.

E. It explains the action of the catalyst as well as those of catalytic promoters and poisons.

The modern adsorption theory of heterogeneous catalysis does explain the action of the catalyst. However, the roles of catalytic promoters and poisons are not entirely explained by this theory. The NCERT textbook states that the modern adsorption theory explains the action of the catalyst but does not elaborate on the roles of promoters and poisons.

Therefore, considering the information provided by the NCERT textbook, point E is not entirely correct. There are three correct statements (B, C, and D) about the modern adsorption theory of heterogeneous catalysis.

The coagulating power of an electrolyte is inversely proportional to its coagulating value. That is,

$ \text{Coagulating power} = \frac{1}{\text{Coagulating value}} $

Therefore, the coagulating power of AlCl₃ is $ \frac{1}{0.09} $

and the coagulating power of NaCl is $ \frac{1}{50.04} $

We are asked to find the factor by which the coagulating power of AlCl₃ is greater than the coagulating power of NaCl. This factor is given by the ratio

$ x = \frac{\text{Coagulating power of AlCl₃}}{\text{Coagulating power of NaCl}} = \frac{\frac{1}{0.09}}{\frac{1}{50.04}} = \frac{50.04}{0.09} $

$ x = \frac{50.04}{0.09} \approx 556.22 $

So, the coagulating power of AlCl₃ is approximately 556 times the coagulating power of NaCl.

However, since the problem asks for the nearest integer, the answer would be 556.

Paints, milk, froth, soap lather and hair cream have liquid as dispersion medium.

Step 1: Identify Activation Energies

• Activation energy for platinum: $E_{a1} = 30 \, \mathrm{kJ/mol} = 30 \times 10^3 \, \mathrm{J/mol}$
• Activation energy for nickel: $E_{a2} = 41.4 \, \mathrm{kJ/mol} = 41.4 \times 10^3 \, \mathrm{J/mol}$

Step 2: Given Constants

• Gas constant: $R = 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \mathrm{mol}^{-1}$
• Temperature: $T = 300 \, \mathrm{K}$
• $\ln 10 = 2.3$

Step 3: Calculate Ratio of Rate Constants

Using the Arrhenius equation, the ratio of the rate constants can be calculated as:

$ \frac{K_2}{K_1} = \exp\left(\frac{E_{a1} - E_{a2}}{RT}\right) $

Step 4: Find Logarithm of the Ratio

Taking the natural logarithm and using $\ln 10 = 2.3$ to convert to base-10 logarithm:

$ \log \frac{K_2}{K_1} = \frac{\ln\left(\frac{K_2}{K_1}\right)}{\ln 10} = \frac{E_{a1} - E_{a2}}{2.3 \cdot RT} $

Step 5: Substitute Values

$ \log \frac{K_2}{K_1} = \frac{(30 \times 10^3 - 41.4 \times 10^3) \, \mathrm{J/mol}}{2.3 \times 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 300 \, \mathrm{K}} \approx 1.99 $

Step 6: Final Answer

The nearest integer to the calculated value is 2.

So, the logarithm of the ratio of the rates of chemisorption on equal areas of platinum and nickel at 300 K is 2.

$\mathrm{{1 \over n} = \tan (45^\circ )}$

$\therefore$ $\mathrm{n = 1}$

$\mathrm{\log k = 0.602}$

$\therefore$ $\mathrm{k = 4}$

$\mathrm{{x \over m} = k.{p^{1/n}}}$

$ = 4 \times 0.4 = 1.6$ g

The correct statements are

(B) Surface tension is due to uneven forces acting on the molecules present on the surface

(D) The molecules on the surface are responsible for vapor pressure if the system is a closed system

Option A :

$\to$ When moisture is added in anhydrus CaCl₂ then absorption occurs.

$\to$ When silica gel is kept in presence of moisture then it get's wet from outside only so it is a example of adsorption.

Option C :

$\Delta $H is always negative for adsorption. As the adsorption proceeds $\Delta $H becomes less and less negative, ultimately becomes equal to T$\Delta $S and $\Delta $G becomes zero.

So, statements (A) and (C) are incorrect.

A. It is non-specific

B. It is low

C. Extent of adsorption decreases with increase of temperature

D. It results in multimolecular layer

E. No activation energy is needed

No. of correct statements $=2$

The following curves are not in accordance with Freundlich adsorption isotherm.

$ \begin{aligned} &\text { Initial moles of acetic acid }=M_{1} \times V_{1} \\\\ &=100 \times 0.3=30 \mathrm{~m} \mathrm{~mol} . \end{aligned} $

$ \begin{aligned} &\text { moles after adsorption }=100 \times 0.125=12.5 \mathrm{~m} \mathrm{~mol} \text {. } \\\\ &\text { moles adsorbed }=30-12.5=17.5 \mathrm{~m} \mathrm{~mol} \end{aligned} $

for acetic acid; weight adsorbed.

$ \begin{aligned} &\text { moles }=\frac{\text { Weight }}{\text { Molar Mass }} \\\\ &\text { weight }=17.5 \times 10^{-3} \times 60=1050 \times 10^{-3} \mathrm{~g} \\\\ &\text { for } 0.8 \mathrm{~g} \text { charcoal mass of acetic acid adsorbed } \\\\ &=1050 \times 10^{-3} \mathrm{~g} \\\\ &1 \mathrm{~g} \text { charcoal }=\frac{1050 \times 10^{-3}}{0.8}=1312.5 \times 10^{-3} \mathrm{~g} \\\\ &=13125 \times 10^{-4} \mathrm{~g} \end{aligned} $

Given that $\log \mathrm{K}=$ intercept $=0.602=\log 4$

$\therefore \mathrm{K}=4$

Slope $=\frac{1}{\mathrm{n}}=1$

and initial pressure $=0.03 \mathrm{~atm}$

$\frac{x}{m}=K(p)^{1 / n}=4 \times 0.03=0.12=12 \times 10^{-2}$

mass of gas absorbed per gm of adsorbent

$ =12 \times 10^{-2} \mathrm{~g} $

Mass of wood charcoal $=0.6 \mathrm{~g}$

Initial moles of acetic acid $=0.2 \times 0.2=0.04$

Final moles of acetic acid $\quad=0.1 \times 0.2=0.02$

Moles of acetic acid adsorbed $=0.04-0.02 = 0.02$

Mass of acetic acid adsorbed per gm of charcoal

$ =\frac{0.02 \times 60}{0.6}=2.0 \mathrm{~g} $

We know

${x \over m} = K{P^{1/n}}$; using (x $\propto$ V)

$ \Rightarrow {{10} \over 1} = K \times {(100)^{1/n}}$ ..... (1)

${{15} \over 1} = K \times {(200)^{1/n}}$ ..... (2)

${V \over 1} = K \times {(300)^{1/n}}$ ..... (3)

Divide

(2) / (1)

${{15} \over {10}} = {2^{1/n}}$

$\log \left( {{3 \over 2}} \right) = {1 \over n}\log 2$

${1 \over n} = {{\log 3 - log2} \over {\log 2}} = {{0.4771 - 0.3010} \over {0.3010}}$

${1 \over n} = 0.585$

Divide

(3) / (1)

${V \over {10}} = {3^{1/n}}$

$\log \left( {{V \over {10}}} \right) = {1 \over n}\log 3$

$\log \left( {{V \over {10}}} \right) = 0.585 \times 0.4771 = 0.2791$

${V \over {10}} = {10^{0.279}} \Rightarrow V = 10 \times {10^{0.279}}$

$ \Rightarrow V = {10^{1.279}} = {10^x}$

$\Rightarrow$ x = 1.279

$\Rightarrow$ x = 128 $\times$ 10^$-$2 (Nearest integer)

Freundlich isotherm;

${x \over m} = k.{p^{{1 \over n}}}$

Substituting values ;

$\left( {{{64} \over 1}} \right) = {\left( 2 \right)^{{1 \over n}}} \Rightarrow n = {1 \over 6} = 0.166$

$ \cong 17 \times {10^{ - 2}}$

Coagulation value The minimum concentration of electrolyte in millimoles required to cause coagulation of 1 L of colloidal solution.

Given, 0.0018 g Cl^$-$ present in 100 mL solution

Coagulation value of Cl^$-$ is $ = {{{{0.0018} \over {35.5}} \times {{10}^3}} \over {0.1}} = 0.5070 \simeq 1$

According to Freundlich adsorption isotherm, in the median range of pressure

       ${x \over m} \propto {P^{{1 \over n}}}$

$ \Rightarrow $  ${x \over m}$ = kP$^{{1 \over n}}$

taking log both sides, we get,

log${x \over m}$ = logk + ${1 \over n}$ logP

Here in graph between log${x \over m}$ and logP, slope is ${1 \over n}$ and intercepts = log k.

$ \therefore $ ${1 \over n}$ = 2 $ \Rightarrow $ n = ${1 \over 2}$

and log k = 0.4771 = log 3

$ \Rightarrow $ k = 3

$ \therefore $ ${x \over m}$ = 3P$^{{2}}$

So, mass of gas adsorbed per gram of adsorbent

= 3 $ \times $ (0.04)²

= 48 $ \times $ 10^–4

According to Freundlich adsorption isotherm, in the median range of pressure

       ${x \over m} \propto {P^{{1 \over n}}}$

$ \Rightarrow $  ${x \over m}$ = kP$^{{1 \over n}}$ ....(1)

taking log both sides, we get,

log${x \over m}$ = logk + ${1 \over n}$ logP

Here in graph between log${x \over m}$ and logP, slope is ${1 \over n}$ and intercepts = log k.

Given, log k = 0.4771 $ \Rightarrow $ k = 3

Slope ${1 \over n}$ = 2

put in eq. (1), ${x \over m}$ = 3 $ \times $ (4)² = 48

Given Data:

P = 0.01 atm

T = 298 K

V = 2.46 cm$^3$

Density of surface sites $=6.023\times10^{14}/\mathrm{cm^3}$

Surface area (A) = 1000 cm$^2$

Goal:

We want to find out how many surface sites are covered by each nitrogen (N$_2$) molecule.

Step 1: Find the Total Number of Surface Sites

We are given the number of sites per cm$^3$ and the total surface area. Multiply these to get the total number of sites:

Total number of surface sites = Density $\times$ Total surface area

$=6.023\times10^{14}\mathrm{~cm}^{-3}\times1000\mathrm{~cm^2}$

$=6.023\times10^{17}$ sites

Step 2: Calculate the Number of Sites Occupied by N$_2$

Nitrogen covers just 20% of the sites. So, multiply the total sites by 20% (or 0.20):

Sites occupied by N$_2$ = $\frac{20}{100}\times6.023\times10^{17}$

$=1.2046\times10^{17}$ sites

Step 3: Find Number of N$_2$ Molecules Released (Using Gas Laws)

To get the number of N$_2$ molecules, first calculate how many moles of N$_2$ gas are collected using the ideal gas law:

$n = \frac{PV}{RT}$

Where R = 0.0821 L-atm/K·mol; V = 2.46 cm$^3$ = 2.46 × 10$^{-3}$ L

Now, plug in the values:

$n = \frac{0.01\,\text{atm} \times 2.46 \times 10^{-3}\,\text{L}}{0.0821\, \text{L-atm/(K·mol)} \times 298\,\text{K}}$

$n = 1.01 \times 10^{-7}$ moles

Now, multiply the number of moles by Avogadro's number (6.023 × 10$^{23}$) to get molecules:

Total N$_2$ molecules released = $n \times N_A$

$= 1.01\times 10^{-7}\times 6.023\times10^{23} = 6.08\times10^{16}$ molecules

Step 4: Number of Surface Sites Occupied by Each N$_2$ Molecule

Now divide the number of sites occupied by the number of N$_2$ molecules:

$\text{Number of sites per N}_2 = \frac{1.2046\times10^{17}}{6.08\times10^{16}} = 1.98 \approx 2$

Final Answer

Each nitrogen molecule occupies about 2 surface sites.

The number of moles of acetic acid in 100 mL (before adding charcoal) = 0.05 mol

The number of moles of acetic acid in 100 mL (after adding charcoal) = 0.049 mol

The number of moles of acetic acid adsorbed on the surface of charcoal = 0.05 - 0.049 = 0.001 mol

The number of molecules of acetic acid adsorbed on the surface of charcoal = 0.001 $\times$ 6.02 $\times$ 10²³ = 6.02 $\times$ 10²⁰

Given that the surface area of charcoal = 3.01 $\times$ 10² m² , so the area occupied by single acetic acid molecule on the surface of charcoal is

= ${{3.01 \times {{10}^2}} \over {6.02 \times {{10}^{20}}}}$ = 5 $\times$ 10^-19 m²