Sample Solution

BCHCT-133 Solved Assignment

PART A: CHEMICAL ENERGETICS AND EQUILIBRIA

1. (a) (i) Define chemical thermodynamics and outline its significance.
   (ii) Explain the concept of thermodynamic reversibility with the help of a suitable example.
   (b) (i) 223 J of heat was supplied to a thermodynamic system and its internal energy was found to decrease by 92 J . Calculate the amount of associated work and state whether the work was done by the system or done on the system?
   (ii) Derive the relationship between the temperature and pressure for a reversible adiabatic process.
2. (a) (i) 0.5 mole of an ideal gas is taken in a container at 298 K . Calculate the values of its heat capacities at constant pressure and at constant volume conditions.
   (ii) 1 mole of an ideal gas is allowed to undergo isothermal reversible expansion at ${25}^{\circ }\mathrm{C}$${25}^{\circ }\mathrm{C}$25^(@)C25^{\circ} \mathrm{C}${25}^{\circ }\mathrm{C}$ from a volume of $10{\mathrm{dm}}^3$$10{\mathrm{dm}}^3$10dm^(3)10 \mathrm{dm}^3$10{\mathrm{dm}}^3$ to $20{\mathrm{dm}}^3$$20{\mathrm{dm}}^3$20dm^(3)20 \mathrm{dm}^3$20{\mathrm{dm}}^3$. Calculate the maximum amount of work done by the gas on the surroundings.

3. (a) (i) State Hess’ law of constant heat summation and give its significance
   (ii) Using the bond enthalpy datafrom Table 3.2 of (Unit 3; p, 76), calculate the enthalpy of hydrogenation of 1-propene.

4. (a) (i) What is residual entropy? What kind of systems show residual entropy?
   (ii) Explain the difference between enthalpy driven and entropy driven reactions.

5. (a) (i) Define degree of hydrolysis and derive its relationship with hydrolysis constant
   (ii) Explain the effect of addition of inert gas to a gaseous equilibrium reaction under the conditions of i) constant volume and ii) constant pressure.
   (b) (i) Calculate the pH of 0.1 M aqueous solution of sodium formate at 298 K . [Given: $\mathrm{Ka}(\mathrm{HCOOH})=1.7\times {10}^{-4}$$\mathrm{Ka}(\mathrm{HCOOH})=1.7\times {10}^{-4}$Ka(HCOOH)=1.7 xx10^(-4)\mathrm{Ka}(\mathrm{HCOOH})=1.7 \times 10^{-4}$\mathrm{Ka}(\mathrm{HCOOH})=1.7\times {10}^{-4}$ at 298 K
   (ii) Define solubility product constant and derive an expression for the solubility product constant for a sparingly soluble salt of ${\mathrm{M}}_2{\mathrm{A}}_3$${\mathrm{M}}_2{\mathrm{A}}_3$M_(2)A_(3)\mathrm{M}_2 \mathrm{~A}_3${\mathrm{M}}_2{\mathrm{A}}_3$ type having a solubility of S $\mathrm{mol}\mathrm{dm}{}^{-3}$$\mathrm{mol}\mathrm{dm}{}^{-3}$moldm^(-3)\mathrm{mol} \mathrm{dm}{ }^{-3}$\mathrm{mol}\mathrm{dm}{}^{-3}$.

PART B: FUNCTIONAL GROUP ORGANIC CHEMISTRY-I

6. (a) Both 1,3,5-cycloheptatrienyl cation and 1,3,5-cycloheptatriene have $6\pi$$6\pi$6pi6 \pi$6\pi$ elections, comments on the aromaticity of these compounds.
   (b) Discuss the role of Lewis acids in halogenations and alkylation reactions of benzene. Write the limitations of Freidel Crafts alkylation reaction.
7. (a) Nitro group is meta directing, explain.
   (b) What would be the final products of the reaction of HI with diethyl ether and anisol? Write the mechanism of both the reactions.
8. (a) What is aldol condensation? Write its mechanism.
   (b) Arrange the following in order of their increasing reactivity towards ${\mathrm{S}}_{\mathrm{N}}2$${\mathrm{S}}_{\mathrm{N}}2$S_(N)2\mathrm{S}_{\mathrm{N}} 2${\mathrm{S}}_{\mathrm{N}}2$ substitution reactions: bromomethane, 1-bromo-propane, 2-bromopropane, 2 -bormo-2-methyl propane. Justify your answer.
9. (a) Write their representative reaction for the following reactions.
   (b) Sandmeyer reaction
   (c) Reimer-Tiemann reaction
   (d) Mannich reaction
   (e) Willgerodt reaction
   (f) Knoevenagel reaction

Answer:

1. (a) (i) Define chemical thermodynamics and outline its significance.

Answer:

1. Predicting Reaction Feasibility: It helps in determining whether a chemical reaction will occur spontaneously under specific conditions.
2. Understanding Energy Transfer: It provides insights into how energy is absorbed or released during chemical reactions, essential for designing efficient processes in industries.
3. Equilibrium Analysis: Chemical thermodynamics aids in understanding the conditions under which a chemical equilibrium is reached, guiding the optimization of industrial reactions.
4. Calculating Work and Heat: It enables the calculation of work done by or on a system and the heat exchanged, crucial for processes like engines, refrigeration, and chemical manufacturing.
5. Design of Chemical Processes: The principles of thermodynamics are key in designing processes like distillation, refrigeration, and power plants, ensuring they are energy-efficient and cost-effective.
6. Enzyme and Biological Reactions: It helps explain biochemical processes in living organisms, from metabolic pathways to the functioning of enzymes.

1. (a) (ii) Explain the concept of thermodynamic reversibility with the help of a suitable example.

Answer:

Example: Reversible Isothermal Expansion of an Ideal Gas

• Process description:
  • The gas is allowed to expand slowly by pushing the piston up.
  • At every point during the expansion, the gas is in thermal equilibrium with its surroundings (since heat is exchanged to maintain temperature).
  • The system’s pressure and volume change continuously and uniformly, ensuring no entropy is generated.
• Reversibility:
  • If you reverse the process, you can compress the gas back to its original volume and pressure by lowering the piston very slowly, again ensuring equilibrium at every stage.
  • No energy is lost in the process, and both the system and surroundings are returned to their original states with no net change in entropy.

1. (b) (i) 223 J of heat was supplied to a thermodynamic system and its internal energy was found to decrease by 92 J. Calculate the amount of associated work and state whether the work was done by the system or done on the system?

Answer:

• $\mathrm{\Delta }U$$\mathrm{\Delta }U$Delta U\Delta U$\mathrm{\Delta }U$ is the change in internal energy of the system.
• $Q$$Q$QQ$Q$ is the heat supplied to the system.
• $W$$W$WW$W$ is the work done by the system.

Given:

• Heat supplied, $Q=223\text{J}$$Q=223\text{J}$Q=223″J”Q = 223 \, \text{J}$Q=223\text{J}$
• Change in internal energy, $\mathrm{\Delta }U=-92\text{J}$$\mathrm{\Delta }U=-92\text{J}$Delta U=-92″J”\Delta U = -92 \, \text{J}$\mathrm{\Delta }U=-92\text{J}$ (The internal energy decreases, so it’s negative.)

Interpretation:

1. (b) (ii) Derive the relationship between the temperature and pressure for a reversible adiabatic process.

Answer:

Deriving the Relationship between Temperature and Pressure for a Reversible Adiabatic Process

Step 1: Start with the First Law of Thermodynamics

Step 2: Substitute Ideal Gas Equation

Step 3: Integrate to Find the Relationship

Step 4: Apply Logarithmic Properties

Step 5: Substitute the Specific Heat Ratio $\gamma$$\gamma$gamma\gamma$\gamma$

Step 6: Deriving the Temperature-Pressure Relationship

2. (a) (i) 0.5 mole of an ideal gas is taken in a container at 298 K. Calculate the values of its heat capacities at constant pressure and at constant volume conditions.

Answer:

Step 1: Understanding the given information

• Number of moles of the gas, $n=0.5\text{mol}$$n=0.5\text{mol}$n=0.5″mol”n = 0.5 \, \text{mol}$n=0.5\text{mol}$
• Temperature, $T=298\text{K}$$T=298\text{K}$T=298″K”T = 298 \, \text{K}$T=298\text{K}$

Step 2: Use of heat capacities for ideal gases

• $C_V=\frac{f}{2}R$$C_V=\frac{f}{2}R$C_(V)=(f)/(2)RC_V = \frac{f}{2} R$C_V=\frac{f}{2}R$, where $f$$f$ff$f$ is the number of degrees of freedom of the gas molecules.
• $C_P=C_V+R$$C_P=C_V+R$C_(P)=C_(V)+RC_P = C_V + R$C_P=C_V+R$, where $R$$R$RR$R$ is the universal gas constant ($R=8.314\text{J/mol·K}$$R=8.314\text{J/mol·K}$R=8.314″J/mol·K”R = 8.314 \, \text{J/mol·K}$R=8.314\text{J/mol·K}$).

Step 3: Assumption about the type of gas

• $C_V=\frac{5}{2}R$$C_V=\frac{5}{2}R$C_(V)=(5)/(2)RC_V = \frac{5}{2} R$C_V=\frac{5}{2}R$
• $C_P=C_V+R=\frac{5}{2}R+R=\frac{7}{2}R$$C_P=C_V+R=\frac{5}{2}R+R=\frac{7}{2}R$C_(P)=C_(V)+R=(5)/(2)R+R=(7)/(2)RC_P = C_V + R = \frac{5}{2} R + R = \frac{7}{2} R$C_P=C_V+R=\frac{5}{2}R+R=\frac{7}{2}R$

Step 4: Calculate the heat capacities

For $C_V$$C_V$C_(V)C_V$C_V$:

For $C_P$$C_P$C_(P)C_P$C_P$:

Step 5: Calculate for 0.5 mole of gas

For $C_V$$C_V$C_(V)C_V$C_V$ for 0.5 moles:

For $C_P$$C_P$C_(P)C_P$C_P$ for 0.5 moles:

Final Answer:

• The heat capacity at constant volume ($C_V$$C_V$C_(V)C_V$C_V$) is $\boxed{{\displaystyle 10.39\text{J/K}}}$$\boxed{{\displaystyle 10.39\text{J/K}}}$10.39″J/K”\boxed{10.39 \, \text{J/K}}$\boxed{{\displaystyle 10.39\text{J/K}}}$.
• The heat capacity at constant pressure ($C_P$$C_P$C_(P)C_P$C_P$) is $\boxed{{\displaystyle 14.55\text{J/K}}}$$\boxed{{\displaystyle 14.55\text{J/K}}}$14.55″J/K”\boxed{14.55 \, \text{J/K}}$\boxed{{\displaystyle 14.55\text{J/K}}}$.

2. (a) (ii) 1 mole of an ideal gas is allowed to undergo isothermal reversible expansion at ${25}^{\circ }\mathrm{C}$${25}^{\circ }\mathrm{C}$25^(@)C25^{\circ} \mathrm{C}${25}^{\circ }\mathrm{C}$ from a volume of $10{\mathrm{dm}}^3$$10{\mathrm{dm}}^3$10dm^(3)10 \mathrm{dm}^3$10{\mathrm{dm}}^3$ to $20{\mathrm{dm}}^3$$20{\mathrm{dm}}^3$20dm^(3)20 \mathrm{dm}^3$20{\mathrm{dm}}^3$. Calculate the maximum amount of work done by the gas on the surroundings.

Answer:

• $W$$W$WW$W$ is the work done by the gas.
• $n$$n$nn$n$ is the number of moles of the gas.
• $R$$R$RR$R$ is the universal gas constant ($R=8.314\text{J/mol·K}$$R=8.314\text{J/mol·K}$R=8.314″J/mol·K”R = 8.314 \, \text{J/mol·K}$R=8.314\text{J/mol·K}$).
• $T$$T$TT$T$ is the temperature in Kelvin.
• $V_1$$V_1$V_(1)V_1$V_1$ and $V_2$$V_2$V_(2)V_2$V_2$ are the initial and final volumes, respectively.

Step 1: Convert all given values to proper units

• Number of moles of gas, $n=1\text{mol}$$n=1\text{mol}$n=1″mol”n = 1 \, \text{mol}$n=1\text{mol}$
• Temperature, $T={25}^{\circ }\text{C}=25+273.15=298.15\text{K}$$T={25}^{\circ }\text{C}=25+273.15=298.15\text{K}$T=25^(@)”C”=25+273.15=298.15″K”T = 25^\circ \text{C} = 25 + 273.15 = 298.15 \, \text{K}$T={25}^{\circ }\text{C}=25+273.15=298.15\text{K}$
• Initial volume, $V_1=10{\text{dm}}^3=10\times {10}^{-3}{\text{m}}^3=0.01{\text{m}}^3$$V_1=10{\text{dm}}^3=10\times {10}^{-3}{\text{m}}^3=0.01{\text{m}}^3$V_(1)=10″dm”^(3)=10 xx10^(-3)”m”^(3)=0.01″m”^(3)V_1 = 10 \, \text{dm}^3 = 10 \times 10^{-3} \, \text{m}^3 = 0.01 \, \text{m}^3$V_1=10{\text{dm}}^3=10\times {10}^{-3}{\text{m}}^3=0.01{\text{m}}^3$
• Final volume, $V_2=20{\text{dm}}^3=20\times {10}^{-3}{\text{m}}^3=0.02{\text{m}}^3$$V_2=20{\text{dm}}^3=20\times {10}^{-3}{\text{m}}^3=0.02{\text{m}}^3$V_(2)=20″dm”^(3)=20 xx10^(-3)”m”^(3)=0.02″m”^(3)V_2 = 20 \, \text{dm}^3 = 20 \times 10^{-3} \, \text{m}^3 = 0.02 \, \text{m}^3$V_2=20{\text{dm}}^3=20\times {10}^{-3}{\text{m}}^3=0.02{\text{m}}^3$

Step 2: Apply the formula for work

Final Answer:

2. (b) (i) Derive the relationship between ${\mathrm{\Delta }}_{r}U$${\mathrm{\Delta }}_{r}U$Delta _(r)U\Delta_r U${\mathrm{\Delta }}_{r}U$ and ${\mathrm{\Delta }}_{s}H$${\mathrm{\Delta }}_{s}H$Delta _(s)H\Delta_s H${\mathrm{\Delta }}_{s}H$.

Answer:

Step 1: Recall the first law of thermodynamics

Step 2: Define Enthalpy

Step 3: Relating ${\mathrm{\Delta }}_{r}U$${\mathrm{\Delta }}_{r}U$Delta _(r)U\Delta_r U${\mathrm{\Delta }}_{r}U$ and ${\mathrm{\Delta }}_{r}H$${\mathrm{\Delta }}_{r}H$Delta _(r)H\Delta_r H${\mathrm{\Delta }}_{r}H$ for a Reversible Process

Step 4: Generalize for Reactions Involving Heat Transfer

Step 5: Special Case for Ideal Gases

Final Answer:

2. (b) (ii) Describe the method for experimental determination of energy changes accompanying chemical reactions under constant volume conditions.

Answer:

Method for Experimental Determination of Energy Changes

Step 1: Set Up a Bomb Calorimeter

• The bomb is designed to withstand high pressures, especially when the reaction involves combustion.
• The system is isolated to ensure no heat exchange with the external environment (adiabatic conditions).

Step 2: Place the Reactants in the Bomb

• The reactants (e.g., a known mass of a substance like a fuel) are placed inside the bomb, typically in the form of a solid or liquid.
• The bomb is filled with pure oxygen to ensure complete combustion or reaction in the case of a combustion reaction.

Step 3: Ignite the Reaction

• The reaction is initiated (e.g., by using an electrically heated wire) to ensure it proceeds completely under controlled conditions.
• Since the bomb is sealed and the volume is constant, no expansion of the gases is allowed, and the system is effectively at constant volume.

Step 4: Measure the Temperature Change

• After the reaction is initiated, the heat released (or absorbed) by the reaction is transferred to the surrounding water.
• The temperature of the water is carefully monitored using temperature sensors (often thermocouples or resistance thermometers).

Step 5: Calculate the Heat Released (or Absorbed)

• The heat change ($q_{\text{water}}$$q_{\text{water}}$q_(“water”)q_{\text{water}}$q_{\text{water}}$) in the surrounding water is calculated using the formula:

• $m_w$$m_w$m_(w)m_w$m_w$ = mass of the water surrounding the bomb.
• $C_w$$C_w$C_(w)C_w$C_w$ = specific heat capacity of water.
• $\mathrm{\Delta }T$$\mathrm{\Delta }T$Delta T\Delta T$\mathrm{\Delta }T$ = temperature change of the water.

Step 6: Determine the Energy Change ($\mathrm{\Delta }U$$\mathrm{\Delta }U$Delta U\Delta U$\mathrm{\Delta }U$)

• Under constant volume, the heat absorbed or released by the system is equal to the change in internal energy ($\mathrm{\Delta }U$$\mathrm{\Delta }U$Delta U\Delta U$\mathrm{\Delta }U$) of the system. This is because no work is done (since $\mathrm{\Delta }V=0$$\mathrm{\Delta }V=0$Delta V=0\Delta V = 0$\mathrm{\Delta }V=0$, and $W=P\mathrm{\Delta }V$$W=P\mathrm{\Delta }V$W=P Delta VW = P \Delta V$W=P\mathrm{\Delta }V$).
• The energy change ($\mathrm{\Delta }U$$\mathrm{\Delta }U$Delta U\Delta U$\mathrm{\Delta }U$) for the reaction at constant volume is thus:

Step 7: Account for Heat Capacity of the Calorimeter

Step 8: Analysis and Calculation

• The energy change $\mathrm{\Delta }U$$\mathrm{\Delta }U$Delta U\Delta U$\mathrm{\Delta }U$ is determined based on the temperature change measured in the water and the known calorimetric constants.
• If the reaction involves a change in temperature, volume, or involves gases, additional corrections may be required to account for the properties of the reactants and products.

Final Answer: