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

1. a) Define and explain a thermodynamically reversible process.

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.

• 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.

• $W$$W$WW$W$ is the work done by the gas,
• $n$$n$nn$n$ is the number of moles of the gas,
• $R$$R$RR$R$ is the universal gas constant,
• $T$$T$TT$T$ is the temperature in Kelvin,
• $V_f$$V_f$V_(f)V_f$V_f$ is the final volume,
• $V_i$$V_i$V_(i)V_i$V_i$ is the initial volume.

• $n=0.25$$n=0.25$n=0.25n = 0.25$n=0.25$ mol,
• $R=8.314\text{J/mol}\cdot \text{K}$$R=8.314\text{J/mol}\cdot \text{K}$R=8.314″J/mol”*”K”R = 8.314 \, \text{J/mol}\cdot\text{K}$R=8.314\text{J/mol}\cdot \text{K}$ (universal gas constant),
• $T={27}^{\circ }\text{C}=27+273.15=300.15\text{K}$$T={27}^{\circ }\text{C}=27+273.15=300.15\text{K}$T=27^(@)”C”=27+273.15=300.15″K”T = 27^\circ\text{C} = 27 + 273.15 = 300.15 \, \text{K}$T={27}^{\circ }\text{C}=27+273.15=300.15\text{K}$,
• $V_i=2.0{\text{dm}}^3=2.0\times {10}^{-3}{\text{m}}^3$$V_i=2.0{\text{dm}}^3=2.0\times {10}^{-3}{\text{m}}^3$V_(i)=2.0″dm”^(3)=2.0 xx10^(-3)”m”^(3)V_i = 2.0 \, \text{dm}^3 = 2.0 \times 10^{-3} \, \text{m}^3$V_i=2.0{\text{dm}}^3=2.0\times {10}^{-3}{\text{m}}^3$ (since $1{\text{dm}}^3={10}^{-3}{\text{m}}^3$$1{\text{dm}}^3={10}^{-3}{\text{m}}^3$1″dm”^(3)=10^(-3)”m”^(3)1 \, \text{dm}^3 = 10^{-3} \, \text{m}^3$1{\text{dm}}^3={10}^{-3}{\text{m}}^3$),
• $V_f=10{\text{dm}}^3=10\times {10}^{-3}{\text{m}}^3$$V_f=10{\text{dm}}^3=10\times {10}^{-3}{\text{m}}^3$V_(f)=10″dm”^(3)=10 xx10^(-3)”m”^(3)V_f = 10 \, \text{dm}^3 = 10 \times 10^{-3} \, \text{m}^3$V_f=10{\text{dm}}^3=10\times {10}^{-3}{\text{m}}^3$.

2. a) Define standard enthalpy of formation and describe a method for its direct determination with the help of an example.

Direct Determination of Standard Enthalpy of Formation

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