Chapter 4: Chemical Thermodynamics
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: For an isothermal process, the temperature is constant, so the change in internal energy ($\Delta U$) is zero ($\Delta U = 0$).
According to the first law, $\Delta U = q + W$. Since $\Delta U = 0$, the expression becomes:
$$0 = q + W \implies q = -W$$
Answer: Hess's law states that the overall enthalpy change for a chemical reaction is constant, regardless of whether the reaction takes place in one single step or in a series of multiple steps. (It is an application of the first law of thermodynamics).
Answer: A thermodynamic property whose value is independent of the quantity or size of matter present in the system is called an intensive property.
Example: Temperature ($T$), Pressure ($P$), or Density.
Answer: The relationship is given by the equation:
$$\Delta G^\circ = -2.303 RT \log_{10} K$$
2 Mark Questions (Short Answer-I)
Answer:
| Property | Isothermal Process | Adiabatic Process |
|---|---|---|
| 1. Temperature | Temperature of the system remains constant throughout the process ($\Delta T = 0$). | Temperature of the system changes (it may increase or decrease). |
| 2. Heat Exchange | Heat is exchanged between the system and surroundings ($q \neq 0$). | No heat is exchanged between the system and surroundings ($q = 0$). |
| 3. Internal Energy | Change in internal energy is zero ($\Delta U = 0$). | Change in internal energy is not zero ($\Delta U \neq 0$, $\Delta U = W$). |
| 4. Vessel Type | Carried out in a vessel with conducting walls (e.g., copper). | Carried out in a perfectly insulated vessel (e.g., thermos flask). |
Answer:
- State Function: A thermodynamic property whose value depends only on the current state (initial and final state) of the system and is entirely independent of the path followed to reach that state.
Examples: Internal Energy ($U$), Enthalpy ($H$), Entropy ($S$), Gibbs Free Energy ($G$), Pressure ($P$), Volume ($V$). - Path Function: A thermodynamic property whose value depends strictly on the path or mechanism followed during a process to go from the initial to the final state.
Examples: Work ($W$) and Heat ($q$).
Answer:
Gibbs free energy change ($\Delta G$) combines both enthalpy ($\Delta H$) and entropy ($\Delta S$) changes ($\Delta G = \Delta H - T\Delta S$) to predict spontaneity at constant temperature and pressure.
- If $\Delta G < 0$ (Negative): The reaction is spontaneous in the forward direction.
- If $\Delta G > 0$ (Positive): The reaction is non-spontaneous in the forward direction (but spontaneous in the reverse direction).
- If $\Delta G = 0$: The reaction is at equilibrium.
3 Mark Questions (Short Answer-II / Derivations)
Answer:
1. Consider $n$ moles of an ideal gas enclosed in a cylinder fitted with a frictionless, weightless, and movable piston. Let the gas expand isothermally and reversibly from an initial volume $V_1$ to a final volume $V_2$ at constant temperature $T$.
2. In a reversible process, the opposing external pressure ($P_{ex}$) is always infinitesimally smaller than the driving pressure of the gas ($P$).
Thus, $P_{ex} = P - dP$
3. The infinitesimal work done ($dW$) for a small volume change ($dV$) is:
$$dW = -P_{ex} \cdot dV$$
Substituting $P_{ex}$:
$$dW = -(P - dP) \cdot dV$$
$$dW = -P \cdot dV + dP \cdot dV$$
Since $dP$ and $dV$ are infinitesimally small quantities, their product $dP \cdot dV$ is negligible. Therefore:
$$dW = -P \cdot dV$$
4. The total maximum work ($W_{max}$) is obtained by integrating $dW$ from $V_1$ to $V_2$:
$$W_{max} = \int_{V_1}^{V_2} dW = \int_{V_1}^{V_2} -P \, dV$$
5. From the ideal gas equation, $P = \frac{nRT}{V}$. Substitute this in the integral:
$$W_{max} = -\int_{V_1}^{V_2} \frac{nRT}{V} dV$$
Since $n, R, T$ are constant for an isothermal process:
$$W_{max} = -nRT \int_{V_1}^{V_2} \frac{dV}{V}$$
$$W_{max} = -nRT [\ln V]_{V_1}^{V_2}$$
$$W_{max} = -nRT (\ln V_2 - \ln V_1) = -nRT \ln \left(\frac{V_2}{V_1}\right)$$
6. Converting natural logarithm to base-10 logarithm ($\ln x = 2.303 \log_{10} x$):
$$W_{max} = -2.303 nRT \log_{10} \left(\frac{V_2}{V_1}\right)$$
Answer:
Given Data:
Number of moles ($n$) $= 2 \text{ mol}$
Temperature ($T$) $= 300 \text{ K}$
Initial volume ($V_1$) $= 10 \text{ L}$
Final volume ($V_2$) $= 20 \text{ L}$
Universal gas constant ($R$) $= 8.314 \text{ J K}^{-1}\text{mol}^{-1}$
Formula:
$$W_{max} = -2.303 nRT \log_{10} \left(\frac{V_2}{V_1}\right)$$
Calculation:
$$W_{max} = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10} \left(\frac{20}{10}\right)$$
$$W_{max} = -2.303 \times 4988.4 \times \log_{10} (2)$$
We know $\log_{10} 2 = 0.3010$.
$$W_{max} = -11488.28 \times 0.3010$$
$$W_{max} = -3457.97 \text{ Joules}$$
Converting to kiloJoules (divide by 1000):
$$W_{max} = -3.458 \text{ kJ}$$
The negative sign indicates that the work is done BY the system (expansion work).
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 3-mark derivation/numerical combined with a 1-mark theory question).
(b) Define: Standard Enthalpy of Formation ($\Delta_f H^\circ$). [1 Mark] March 2019, March 2021
Answer (a): Derivation of $\Delta H = \Delta U + \Delta n_g RT$
1. Enthalpy ($H$) of a system is defined as the sum of internal energy ($U$) and pressure-volume work energy ($PV$).
$$H = U + PV$$
2. For a change of state at constant pressure ($P$), the change in enthalpy is:
$$\Delta H = \Delta U + P\Delta V$$
$$\Delta H = \Delta U + P(V_2 - V_1)$$
$$\Delta H = \Delta U + PV_2 - PV_1 \quad \text{--- (Equation 1)}$$
3. Consider a gaseous reaction taking place at constant temperature ($T$) and constant pressure ($P$).
Let $n_1$ be the number of moles of gaseous reactants occupying volume $V_1$.
Let $n_2$ be the number of moles of gaseous products occupying volume $V_2$.
4. Applying the ideal gas equation ($PV = nRT$) for reactants and products:
$PV_1 = n_1RT$
$PV_2 = n_2RT$
5. Substituting these values into Equation 1:
$$\Delta H = \Delta U + n_2RT - n_1RT$$
$$\Delta H = \Delta U + (n_2 - n_1)RT$$
6. Let $\Delta n_g = n_2 - n_1$ (which is the difference between the number of moles of gaseous products and gaseous reactants).
$$\Delta H = \Delta U + \Delta n_g RT$$
This is the required relationship.
Answer (b): Theory
Standard Enthalpy of Formation ($\Delta_f H^\circ$): It is defined as the enthalpy change accompanying the formation of exactly one mole of a pure compound from its constituent elements, with all substances being in their standard states (1 bar pressure and usually 298 K temperature).
(b) What is an isolated system? [1 Mark] Oct 2013, March 2023
Answer (a): Numerical on Spontaneity ($\Delta G$)
Given Data:
Enthalpy change ($\Delta H$) $= -224 \text{ kJ} = -224,000 \text{ Joules}$
Entropy change ($\Delta S$) $= -153 \text{ J K}^{-1}$
Temperature ($T$) $= 298 \text{ K}$
Formula:
Gibbs-Helmholtz equation: $$\Delta G = \Delta H - T\Delta S$$
Calculation:
$$\Delta G = -224,000 - (298 \times -153)$$
$$\Delta G = -224,000 - (-45,594)$$
$$\Delta G = -224,000 + 45,594$$
$$\Delta G = -178,406 \text{ Joules}$$
Converting back to kiloJoules:
$$\Delta G = -178.406 \text{ kJ}$$
Prediction of Spontaneity:
Since the value of Gibbs free energy change ($\Delta G$) is negative, the reaction is spontaneous at 298 K.
Answer (b): Theory
Isolated System: A system which can exchange neither mass (matter) nor energy (heat or work) with its surroundings is called an isolated system. (e.g., hot tea kept in a perfectly insulated thermos flask).
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