BCHCT-133 Solved Assignment 2024

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BCHCT-133 Solved Assignment 2024 Sample Solution
bchct-133-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

bchct-133-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

BCHCT-133 Solved Assignment 2024
PART A: CHEMICAL ENERGETICS AND EQUILIBRIA PART A: CHEMICAL ENERGETICS AND EQUILIBRIA ” PART A: CHEMICAL ENERGETICS AND EQUILIBRIA “\text { PART A: CHEMICAL ENERGETICS AND EQUILIBRIA } PART A: CHEMICAL ENERGETICS AND EQUILIBRIA
  1. a) Define and explain a thermodynamically reversible process.
Answer :
A thermodynamically reversible process is a theoretical concept in thermodynamics referring to a process that happens so slowly and with such infinitesimal changes in conditions that the system remains in thermodynamic equilibrium at all times. For a process to be reversible, it must meet two key criteria:
  1. Infinitesimal Driving Forces: The difference in driving forces (such as pressure, temperature, or chemical potential) between the system and surroundings is infinitesimal, so the process occurs at an infinitesimally slow rate. This ensures that the system can be reversed at any point without a finite change in the external conditions.
  2. No Entropy Production: During a reversible process, there is no increase in entropy within the system and its surroundings. This means there is no dissipation of energy as heat due to friction, viscosity, or other non-conservative forces, and all heat transfer is perfectly efficient.
The concept of a reversible process is an idealization and cannot be achieved in any real physical process because it would require an infinite amount of time to complete. However, it serves as an important standard by which real processes can be compared. In practice, all natural processes are irreversible, occurring with an increase in the total entropy of the system and its surroundings.
Significance in Thermodynamics:
  • Efficiency: Reversible processes define the upper limit of efficiency for engines and other thermodynamic cycles. For example, the Carnot cycle, which is a theoretical reversible cycle, defines the maximum possible efficiency that any heat engine operating between two temperatures can achieve.
  • Equilibrium State Changes: In reversible processes, the system’s changes in state (such as volume, pressure, and temperature) can be described exactly by the equations of state since the system is always at equilibrium.
  • Work and Heat: For a reversible process, the work done by the system on the surroundings is maximized, and the work done on the system is minimized. Similarly, reversible heat transfer is the most efficient.
To summarize, while no real processes are truly reversible, the concept is crucial for understanding the limits of efficiency and for calculating the maximum work and minimum heat transfer in theoretical thermodynamic cycles.
b) 0.25 m o l 0.25 m o l quad0.25mol\quad 0.25 \mathrm{~mol}0.25 mol of an ideal monoatomic gas undergoes isothermal expansion from a volume of 2.0 d m 3 2.0 d m 3 2.0dm^(3)2.0 \mathrm{dm}^32.0dm3 to 10 d m 3 10 d m 3 10dm^(3)10 \mathrm{dm}^310dm3 at 27 ∘ C 27 ∘ C 27^(@)C27^{\circ} \mathrm{C}27∘C. Calculate the maximum work that can be obtained from this process.
Answer :
To calculate the maximum work that can be obtained from the isothermal expansion of an ideal monoatomic gas, we’ll use the formula for work done during an isothermal process. The formula is:
W = n R T ln ⁡ ( V f V i ) W = n R T ln ⁡ V f V i W=nRT ln((V_(f))/(V_(i)))W = nRT \ln\left(\frac{V_f}{V_i}\right)W=nRTln⁡(VfVi)
where:
  • W W WWW is the work done by the gas,
  • n n nnn is the number of moles of the gas,
  • R R RRR is the universal gas constant,
  • T T TTT is the temperature in Kelvin,
  • V f V f V_(f)V_fVf is the final volume,
  • V i V i V_(i)V_iVi is the initial volume.
Let’s substitute the values:
  • n = 0.25 n = 0.25 n=0.25n = 0.25n=0.25 mol,
  • R = 8.314 J/mol ⋅ K R = 8.314 J/mol ⋅ K R=8.314″J/mol”*”K”R = 8.314 \, \text{J/mol}\cdot\text{K}R=8.314J/mol⋅K (universal gas constant),
  • T = 27 ∘ C = 27 + 273.15 = 300.15 K T = 27 ∘ C = 27 + 273.15 = 300.15 K T=27^(@)”C”=27+273.15=300.15″K”T = 27^\circ\text{C} = 27 + 273.15 = 300.15 \, \text{K}T=27∘C=27+273.15=300.15K,
  • V i = 2.0 dm 3 = 2.0 × 10 − 3 m 3 V i = 2.0 dm 3 = 2.0 × 10 − 3 m 3 V_(i)=2.0″dm”^(3)=2.0 xx10^(-3)”m”^(3)V_i = 2.0 \, \text{dm}^3 = 2.0 \times 10^{-3} \, \text{m}^3Vi=2.0dm3=2.0×10−3m3 (since 1 dm 3 = 10 − 3 m 3 1 dm 3 = 10 − 3 m 3 1″dm”^(3)=10^(-3)”m”^(3)1 \, \text{dm}^3 = 10^{-3} \, \text{m}^31dm3=10−3m3),
  • V f = 10 dm 3 = 10 × 10 − 3 m 3 V f = 10 dm 3 = 10 × 10 − 3 m 3 V_(f)=10″dm”^(3)=10 xx10^(-3)”m”^(3)V_f = 10 \, \text{dm}^3 = 10 \times 10^{-3} \, \text{m}^3Vf=10dm3=10×10−3m3.
Now, let’s plug these values into the formula:
W = 0.25 × 8.314 × 300.15 × ln ⁡ ( 10 × 10 − 3 2.0 × 10 − 3 ) W = 0.25 × 8.314 × 300.15 × ln ⁡ 10 × 10 − 3 2.0 × 10 − 3 W=0.25 xx8.314 xx300.15 xx ln((10 xx10^(-3))/(2.0 xx10^(-3)))W = 0.25 \times 8.314 \times 300.15 \times \ln\left(\frac{10 \times 10^{-3}}{2.0 \times 10^{-3}}\right)W=0.25×8.314×300.15×ln⁡(10×10−32.0×10−3)
After calculating, we get:
W = 0.25 × 8.314 × 300.15 × ln ⁡ ( 5 ) W = 0.25 × 8.314 × 300.15 × ln ⁡ 5 W=0.25 xx8.314 xx300.15 xx ln(5)W = 0.25 \times 8.314 \times 300.15 \times \ln\left(5\right)W=0.25×8.314×300.15×ln⁡(5)
After calculating, we find that the maximum work obtained from this isothermal expansion process is approximately 436.06 J 436.06 J 436.06″J”436.06 \, \text{J}436.06J.
In summary, by applying the formula for work done during an isothermal expansion and substituting the given values, we determined that the gas can do a maximum work of about 436.06 J 436.06 J 436.06″J”436.06 \, \text{J}436.06J under the specified conditions.
  1. a) Define standard enthalpy of formation and describe a method for its direct determination with the help of an example.
Answer :
The standard enthalpy of formation, denoted as Δ H f ∘ Δ H f ∘ DeltaH_(f)^(@)\Delta H_f^\circΔHf∘, is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states. The standard state of a substance is its pure form at 1 atmosphere of pressure and a specified temperature, typically 25°C (298 K). For elements in their standard state, such as O₂(g), H₂(g), or C(s, graphite), the standard enthalpy of formation is defined as zero.

Direct Determination of Standard Enthalpy of Formation

The direct determination of the standard enthalpy of formation typically involves measuring the heat change (enthalpy change) during the formation reaction of the compound from its elements under standard conditions. This is usually done using a calorimeter. The process involves the following steps:
  1. Preparation of Reactants: The elements in their standard states are prepared. For example, if we are determining the standard enthalpy of formation of water (H₂O), we would prepare hydrogen gas (H₂) and oxygen gas (O₂).
  2. Reaction in a Calorimeter: The reactants are combined in a calorimeter, a device used to measure the heat of chemical reactions. The reaction is initiated under controlled conditions.
  3. Measurement of Temperature Change: The calorimeter measures the temperature change of the surroundings (usually a water bath) due to the reaction.
  4. Calculation of Heat Change: Using the temperature change and the specific heat capacity of the surroundings, the heat change (q) associated with the reaction is calculated.
  5. Determination of Enthalpy Change: The heat change is then used to calculate the enthalpy change for the reaction. If the reaction produces one mole of the compound, this enthalpy change is the standard enthalpy of formation. If not, it must be adjusted to reflect the formation of one mole of the compound.

Example: Formation of Water

Consider the formation of liquid water from hydrogen and oxygen gases:
2H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( l ) 2H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( l ) “2H”_(2)(g)+”O”_(2)(g)rarr2″H”_(2)”O”(l)\text{2H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)2H2(g)+O2(g)→2H2O(l)
To find the standard enthalpy of formation of water, Δ H f ∘ Δ H f ∘ DeltaH_(f)^(@)\Delta H_f^\circΔHf∘ of H₂O(l), we would measure the heat released when hydrogen and oxygen react to form water in a calorimeter. The reaction is exothermic, so the calorimeter would show a rise in temperature.
Assuming we measure the heat change and find it to be − x − x -x-x−x kJ for the formation of 2 moles of H₂O, the standard enthalpy of formation of one mole of H₂O would be − x / 2 − x / 2 -x//2-x/2−x/2 kJ/mol.
In summary, the standard enthalpy of formation is a fundamental thermodynamic quantity representing the enthalpy change during the formation of a compound from its elements. It is directly measured using calorimetric methods, where the heat change of the formation reaction is recorded and used to calculate the enthalpy change per mole of the compound formed.
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