Thermodynamics for JEE Adv.

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### Introduction to Thermodynamics Thermodynamics is the study of energy transformations between heat and other forms, particularly work. It helps predict the feasibility and direction of chemical and physical processes. Key concepts include: - **System:** The part of the universe under observation (e.g., a chemical reaction in a beaker). - **Surroundings:** Everything outside the system. - **Boundary:** Separates the system from its surroundings. - **State Functions:** Properties that depend only on the current state of the system, not on how that state was reached (e.g., temperature, pressure, volume, internal energy, enthalpy, entropy, Gibbs free energy). Represented by uppercase letters (e.g., U, H, S, G). - **Path Functions:** Properties that depend on the path taken between two states (e.g., heat (Q), work (W)). Represented by lowercase letters (e.g., q, w). ### Zeroth Law of Thermodynamics If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law establishes the concept of temperature. - **Concept:** Allows for the definition and measurement of temperature. If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A and C are also in thermal equilibrium, meaning they have the same temperature. ### First Law of Thermodynamics: Statements The First Law is the principle of conservation of energy applied to thermodynamic systems. #### 1. Energy Conservation Energy can neither be created nor destroyed; it can only be converted from one form to another. #### 2. Internal Energy Change The change in internal energy ($\Delta U$) of a system is equal to the heat supplied to the system ($Q$) plus the work done *on* the system ($W$). - **Formula:** $\Delta U = Q + W$ - $\Delta U$: Change in **Internal Energy** of the system (a state function). It represents the total energy possessed by the system (kinetic and potential energies of molecules). Its value depends only on initial and final states. - $Q$: **Heat** absorbed *by* the system (a path function). - $Q > 0$: Heat flows into the system (endothermic process). - $Q 0$: Work is done *on* the system by the surroundings (e.g., compression). - $W ### Work (W) Formulae Work ($W$) is a path function, meaning its value depends on the specific way a process is carried out. In chemistry, we often consider "pressure-volume" work. #### 1. Work done in Reversible Isothermal Expansion/Compression A process is **reversible** if it can be reversed without leaving any net change in the surroundings. **Isothermal** means constant temperature. - **Formula (Work done *on* the system):** $W = -nRT \ln\left(\frac{V_2}{V_1}\right) = -nRT \ln\left(\frac{P_1}{P_2}\right)$ - $n$: Number of moles of the ideal gas. - $R$: Ideal gas constant ($8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ or $0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}$). - $T$: Absolute temperature in Kelvin (constant for isothermal process). - $V_1$: Initial volume. - $V_2$: Final volume. - $P_1$: Initial pressure. - $P_2$: Final pressure. - **Sign Convention:** A negative value for $W$ means work is done *by* the system (expansion). A positive value means work is done *on* the system (compression). #### Example 1 (Good): 2 moles of an ideal gas undergo reversible isothermal expansion from 10 L to 20 L at 300 K. Calculate the work done *by* the gas. ($R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$) * **Given:** * $n = 2 \text{ mol}$ * $V_1 = 10 \text{ L}$ * $V_2 = 20 \text{ L}$ * $T = 300 \text{ K}$ * $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ * **Calculate work done *on* the system ($W$):** * $W = -nRT \ln\left(\frac{V_2}{V_1}\right)$ * $W = -2 \text{ mol} \times 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K} \times \ln\left(\frac{20 \text{ L}}{10 \text{ L}}\right)$ * $W = -2 \times 8.314 \times 300 \times \ln(2) \text{ J}$ * $W \approx -4988.4 \times 0.693 \text{ J} \approx -3456 \text{ J}$ * **Work done *by* the gas:** Since $W$ (work on system) is $-3456 \text{ J}$, the work done *by* the gas is $+3456 \text{ J}$. #### 2. Work done in Irreversible Expansion/Compression (against constant external pressure) An **irreversible** process is a real-world process that is not reversible, often characterized by a sudden change or expansion/compression against a constant external pressure. - **Formula (Work done *on* the system):** $W = -P_{ext}\Delta V = -P_{ext}(V_2 - V_1)$ - $P_{ext}$: Constant external pressure against which the system is expanding or compressing. - $\Delta V$: Change in volume ($V_{final} - V_{initial}$). - $V_1$: Initial volume. - $V_2$: Final volume. - **Sign Convention:** Same as above. #### Example (Good): A gas expands from 5 L to 15 L against a constant external pressure of 1 atm. Calculate the work done *by* the gas in Joules. (1 L atm = 101.3 J) * **Given:** * $V_1 = 5 \text{ L}$ * $V_2 = 15 \text{ L}$ * $P_{ext} = 1 \text{ atm}$ * **Calculate work done *on* the system ($W$):** * $W = -P_{ext}(V_2 - V_1)$ * $W = -1 \text{ atm} \times (15 \text{ L} - 5 \text{ L}) = -1 \times 10 \text{ L atm} = -10 \text{ L atm}$ * Convert to Joules: $W = -10 \text{ L atm} \times 101.3 \text{ J/L atm} = -1013 \text{ J}$ * **Work done *by* the gas:** The work done *by* the gas is $+1013 \text{ J}$. #### 3. Work done in Adiabatic Reversible Expansion/Compression An **adiabatic** process is one where no heat is exchanged with the surroundings ($Q=0$). - **Formula (Work done *on* the system):** $W = \Delta U = nC_V\Delta T = nC_V(T_2 - T_1)$ - $n$: Number of moles of the ideal gas. - $C_V$: Molar heat capacity at constant volume of the gas. - $\Delta T$: Change in temperature ($T_{final} - T_{initial}$). - $T_1$: Initial absolute temperature. - $T_2$: Final absolute temperature. - **Adiabatic Relations (for ideal gases):** These relate pressure, volume, and temperature changes in an adiabatic process. - $PV^\gamma = \text{constant}$ - $TV^{\gamma-1} = \text{constant}$ - $P^{1-\gamma}T^\gamma = \text{constant}$ - $\gamma$: The adiabatic index, defined as the ratio of molar heat capacities, $\gamma = C_p/C_V$. - Monatomic ideal gas (e.g., He, Ne): $\gamma = 5/3$. - Diatomic ideal gas (e.g., $N_2, O_2$): $\gamma = 7/5$. #### Example (Typical JEE): 1 mole of an ideal monatomic gas ($\gamma = 5/3$) at 300 K and 1 atm pressure is compressed reversibly and adiabatically to a final pressure of 4 atm. Calculate the final temperature and the work done on the gas. ($R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$, $C_V$ for monatomic gas $= 3R/2$) * **Given:** * $n = 1 \text{ mol}$ * $T_1 = 300 \text{ K}$ * $P_1 = 1 \text{ atm}$ * $P_2 = 4 \text{ atm}$ * $\gamma = 5/3$ * $C_V = 3R/2$ * **Calculate final temperature ($T_2$):** Use the adiabatic relation $P^{1-\gamma}T^\gamma = \text{constant}$, which can be written as $\left(\frac{T_2}{T_1}\right) = \left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}$. * First, calculate the exponent: $\frac{\gamma-1}{\gamma} = \frac{5/3 - 1}{5/3} = \frac{2/3}{5/3} = 2/5 = 0.4$. * $T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{0.4} = 300 \text{ K} \times \left(\frac{4 \text{ atm}}{1 \text{ atm}}\right)^{0.4}$ * $T_2 = 300 \times 4^{0.4} = 300 \times (2^2)^{0.4} = 300 \times 2^{0.8}$ * $T_2 \approx 300 \times 1.741 \approx 522.3 \text{ K}$ * **Calculate work done *on* the gas:** $W = nC_V(T_2 - T_1)$. * $W = 1 \text{ mol} \times \frac{3}{2} \times 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times (522.3 \text{ K} - 300 \text{ K})$ * $W = 1.5 \times 8.314 \times 222.3 \text{ J}$ * $W \approx 2772 \text{ J}$ #### 4. Work in Other Simple Processes - **Isochoric (constant volume):** Work done ($W$) = 0 (since $\Delta V = 0$, no expansion or compression). - **Isobaric (constant pressure, irreversible):** $W = -P\Delta V$ (this is the same as work against constant external pressure, where $P_{ext}$ is simply the constant system pressure $P$). ### Enthalpy (H) Enthalpy ($H$) is a state function that is particularly useful for processes occurring at constant pressure, which is common in chemical reactions. #### 1. Definition of Enthalpy - **Formula:** $H = U + PV$ - $H$: **Enthalpy** of the system (a state function). It represents the total heat content of a system at constant pressure. - $U$: Internal Energy of the system. - $P$: Pressure of the system. - $V$: Volume of the system. #### 2. Change in Enthalpy ($\Delta H$) - **Formula for general processes:** $\Delta H = \Delta U + \Delta(PV)$ - $\Delta H$: Change in enthalpy. At constant pressure, $\Delta H = Q_p$ (heat exchanged at constant pressure). - $\Delta(PV)$: Change in the product of pressure and volume. - **Formula for chemical reactions involving gases (at constant temperature):** $\Delta H = \Delta U + \Delta n_g RT$ - $\Delta n_g$: Change in the number of moles of gaseous products minus the number of moles of gaseous reactants. (i.e., $\Delta n_g = \sum n_{g, products} - \sum n_{g, reactants}$). - $R$: Ideal gas constant. - $T$: Absolute temperature in Kelvin. #### Example 1 (Good): For the reaction $2H_2(g) + O_2(g) \to 2H_2O(l)$ at 298 K, if the change in internal energy ($\Delta U$) is -564 kJ, calculate the change in enthalpy ($\Delta H$). ($R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$) * **Given:** * $\Delta U = -564 \text{ kJ}$ * $T = 298 \text{ K}$ * $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ * **Determine $\Delta n_g$ from the balanced equation:** * Moles of gaseous products = 0 (since $H_2O$ is liquid). * Moles of gaseous reactants = 2 (from $H_2$) + 1 (from $O_2$) = 3 mol. * $\Delta n_g = n_{g, products} - n_{g, reactants} = 0 - 3 = -3 \text{ mol}$. * **Calculate $\Delta n_g RT$:** * $\Delta n_g RT = (-3 \text{ mol}) \times (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \times (298 \text{ K})$ * $\Delta n_g RT = -7436.4 \text{ J} = -7.436 \text{ kJ}$ * **Calculate $\Delta H$:** * $\Delta H = \Delta U + \Delta n_g RT$ * $\Delta H = -564 \text{ kJ} + (-7.436 \text{ kJ})$ * $\Delta H = -571.436 \text{ kJ}$ #### Example 2 (Typical JEE): What is the relationship between $\Delta H$ and $\Delta U$ for the combustion of benzene $C_6H_6(l)$ at 300 K? $2C_6H_6(l) + 15O_2(g) \to 12CO_2(g) + 6H_2O(l)$ * **Given:** Reaction and temperature $T = 300 \text{ K}$. * **Determine $\Delta n_g$ from the balanced equation:** * Moles of gaseous products = 12 (from $CO_2$). * Moles of gaseous reactants = 15 (from $O_2$). (Note: $C_6H_6$ and $H_2O$ are liquids, so they aren't counted for gaseous moles). * $\Delta n_g = n_{g, products} - n_{g, reactants} = 12 - 15 = -3 \text{ mol}$. * **Relationship:** * $\Delta H = \Delta U + \Delta n_g RT$ * Substitute $\Delta n_g = -3$: * $\Delta H = \Delta U + (-3)RT$ * $\Delta H = \Delta U - 3RT$ ### Heat Capacities Heat capacities quantify how much heat is required to change the temperature of a substance. #### 1. Molar Heat Capacity at Constant Volume ($C_V$) - **Definition:** The amount of heat required to raise the temperature of 1 mole of a substance by 1 Kelvin at constant volume. - **Formula (for any substance):** $C_V = \left(\frac{\partial U}{\partial T}\right)_V$ - $\partial U / \partial T$: Partial derivative of internal energy with respect to temperature. - Subscript $V$: Indicates that volume is held constant. - **Relationship with $\Delta U$ (for ideal gas):** $\Delta U = nC_V\Delta T$ - $n$: Number of moles. - $C_V$: Molar heat capacity at constant volume. - $\Delta T$: Change in temperature ($T_{final} - T_{initial}$). #### 2. Molar Heat Capacity at Constant Pressure ($C_p$) - **Definition:** The amount of heat required to raise the temperature of 1 mole of a substance by 1 Kelvin at constant pressure. - **Formula (for any substance):** $C_p = \left(\frac{\partial H}{\partial T}\right)_P$ - $\partial H / \partial T$: Partial derivative of enthalpy with respect to temperature. - Subscript $P$: Indicates that pressure is held constant. - **Relationship with $\Delta H$ (for ideal gas):** $\Delta H = nC_p\Delta T$ - $n$: Number of moles. - $C_p$: Molar heat capacity at constant pressure. - $\Delta T$: Change in temperature. #### 3. Relation between $C_p$ and $C_V$ (Mayer's Formula) For an ideal gas, there's a simple relationship between $C_p$ and $C_V$. - **Formula:** $C_p - C_V = R$ - $R$: Ideal gas constant. #### 4. Ratio of specific heats ($\gamma$) - **Formula:** $\gamma = C_p/C_V$ - This ratio is important in adiabatic processes. - **Typical values for ideal gases:** - Monatomic ideal gas (e.g., He, Ar): $C_V = 3R/2$, $C_p = 5R/2 \implies \gamma = 5/3 \approx 1.67$. - Diatomic ideal gas (e.g., $N_2, O_2$): $C_V = 5R/2$, $C_p = 7R/2 \implies \gamma = 7/5 = 1.4$. - Polyatomic linear ideal gas (e.g., $CO_2$): $C_V = 5R/2$, $C_p = 7R/2 \implies \gamma = 7/5 = 1.4$. - Polyatomic non-linear ideal gas (e.g., $CH_4, H_2O$): $C_V = 3R$, $C_p = 4R \implies \gamma = 4/3 \approx 1.33$. #### Example 1 (Good): Calculate the heat required to raise the temperature of 2 moles of a monatomic ideal gas from 27°C to 127°C at constant volume. ($R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$) * **Given:** * $n = 2 \text{ mol}$ * $T_{initial} = 27°C = 300 \text{ K}$ * $T_{final} = 127°C = 400 \text{ K}$ * For a monatomic ideal gas, $C_V = \frac{3}{2}R$. * **Calculate $\Delta T$:** $\Delta T = 400 \text{ K} - 300 \text{ K} = 100 \text{ K}$. * **Heat at constant volume ($Q_V = \Delta U$):** * $Q_V = nC_V\Delta T = n \times \left(\frac{3}{2}R\right) \times \Delta T$ * $Q_V = 2 \text{ mol} \times \left(\frac{3}{2} \times 8.314 \text{ J mol}^{-1} \text{ K}^{-1}\right) \times 100 \text{ K}$ * $Q_V = 3 \times 8.314 \times 100 \text{ J} = 2494.2 \text{ J}$ #### Example 2 (Typical JEE): Two moles of an ideal diatomic gas are heated from 300 K to 500 K under constant pressure. Calculate the change in internal energy ($\Delta U$) and change in enthalpy ($\Delta H$). ($R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$) * **Given:** * $n = 2 \text{ mol}$ * $T_{initial} = 300 \text{ K}$ * $T_{final} = 500 \text{ K}$ * For a diatomic ideal gas, $C_V = \frac{5}{2}R$ and $C_p = \frac{7}{2}R$. * **Calculate $\Delta T$:** $\Delta T = 500 \text{ K} - 300 \text{ K} = 200 \text{ K}$. * **Change in Internal Energy ($\Delta U$):** * $\Delta U = nC_V\Delta T = 2 \text{ mol} \times \left(\frac{5}{2}R\right) \times \Delta T$ * $\Delta U = 2 \times \frac{5}{2} \times 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 200 \text{ K}$ * $\Delta U = 5 \times 8.314 \times 200 \text{ J} = 8314 \text{ J}$ * **Change in Enthalpy ($\Delta H$):** * $\Delta H = nC_p\Delta T = 2 \text{ mol} \times \left(\frac{7}{2}R\right) \times \Delta T$ * $\Delta H = 2 \times \frac{7}{2} \times 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 200 \text{ K}$ * $\Delta H = 7 \times 8.314 \times 200 \text{ J} = 11639.6 \text{ J}$ * **Check with Mayer's formula:** $\Delta H - \Delta U = 11639.6 - 8314 = 3325.6 \text{ J}$. Also, $nR\Delta T = 2 \times 8.314 \times 200 = 3325.6 \text{ J}$. The values are consistent. ### Enthalpy of Reaction ($\Delta H_{rxn}$) The enthalpy change ($\Delta H_{rxn}$) refers to the heat absorbed or released during a chemical reaction at constant pressure. #### 1. Standard Enthalpy of Formation ($\Delta H_f^\circ$) - **Definition:** The enthalpy change when 1 mole of a compound is formed from its elements in their most stable states (standard conditions: 298 K, 1 atm pressure for gases, 1 M concentration for solutions). - **Formula:** $\Delta H_{rxn}^\circ = \sum (\text{coeff}) \Delta H_f^\circ (\text{products}) - \sum (\text{coeff}) \Delta H_f^\circ (\text{reactants})$ - $\Delta H_{rxn}^\circ$: Standard enthalpy change of the reaction. - $\Delta H_f^\circ (\text{products})$: Standard enthalpy of formation for each product. - $\Delta H_f^\circ (\text{reactants})$: Standard enthalpy of formation for each reactant. - (coeff): Stoichiometric coefficient of each product/reactant in the balanced chemical equation. - **Important:** The standard enthalpy of formation for an element in its most stable form (e.g., $O_2(g)$, $C(graphite)$, $H_2(g)$) is defined as zero. #### Example 1 (Good): Calculate the standard enthalpy of combustion of methane ($CH_4(g)$) given the following standard enthalpies of formation ($\Delta H_f^\circ$): $\Delta H_f^\circ (CH_4(g)) = -74.8 \text{ kJ/mol}$ $\Delta H_f^\circ (CO_2(g)) = -393.5 \text{ kJ/mol}$ $\Delta H_f^\circ (H_2O(l)) = -285.8 \text{ kJ/mol}$ * **Balanced Combustion Reaction:** $CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$ * **Apply the formula:** * $\Delta H_{comb}^\circ = [1 \times \Delta H_f^\circ (CO_2(g)) + 2 \times \Delta H_f^\circ (H_2O(l))] - [1 \times \Delta H_f^\circ (CH_4(g)) + 2 \times \Delta H_f^\circ (O_2(g))]$ * Recall that $\Delta H_f^\circ (O_2(g)) = 0$ (element in standard state). * $\Delta H_{comb}^\circ = [1 \times (-393.5 \text{ kJ/mol}) + 2 \times (-285.8 \text{ kJ/mol})] - [1 \times (-74.8 \text{ kJ/mol}) + 2 \times (0 \text{ kJ/mol})]$ * $\Delta H_{comb}^\circ = [-393.5 - 571.6] - [-74.8]$ * $\Delta H_{comb}^\circ = -965.1 - (-74.8) = -965.1 + 74.8 = -890.3 \text{ kJ/mol}$ #### 2. Enthalpy of Reaction from Bond Energies ($\Delta H_{bond}$) This method provides an estimation of reaction enthalpy based on the energy required to break bonds in reactants and the energy released when new bonds are formed in products. - **Formula:** $\Delta H_{rxn} = \sum (\text{Bond energies of reactants}) - \sum (\text{Bond energies of products})$ - Alternatively: $\Delta H_{rxn} = (\text{Energy required to break bonds}) - (\text{Energy released when bonds form})$ - $\sum (\text{Bond energies of reactants})$: Sum of the average bond enthalpies of all bonds broken in the reactant molecules (an endothermic process, positive energy). - $\sum (\text{Bond energies of products})$: Sum of the average bond enthalpies of all bonds formed in the product molecules (an exothermic process, negative energy contribution, hence subtracted). - **Important:** Bond energies are typically average values and thus provide approximate $\Delta H_{rxn}$ values. #### Example (Typical JEE): Using bond enthalpies, calculate the enthalpy of hydrogenation of ethene: $C_2H_4(g) + H_2(g) \to C_2H_6(g)$ Given average bond enthalpies (kJ/mol): $C=C = 614$, $C-C = 348$, $C-H = 413$, $H-H = 436$. * **Reactant bonds to be broken:** * In $C_2H_4$: 1 $C=C$ bond, 4 $C-H$ bonds. * In $H_2$: 1 $H-H$ bond. * Total energy to break bonds = $(1 \times 614) + (4 \times 413) + (1 \times 436)$ * $= 614 + 1652 + 436 = 2702 \text{ kJ}$ * **Product bonds to be formed:** * In $C_2H_6$: 1 $C-C$ bond, 6 $C-H$ bonds. * Total energy to form bonds = $(1 \times 348) + (6 \times 413)$ * $= 348 + 2478 = 2826 \text{ kJ}$ * **Enthalpy of reaction:** * $\Delta H_{rxn} = (\text{Energy to break bonds}) - (\text{Energy to form bonds})$ * $\Delta H_{rxn} = 2702 \text{ kJ} - 2826 \text{ kJ} = -124 \text{ kJ}$ ### Second Law of Thermodynamics: Statements The Second Law describes the direction of spontaneous processes and introduces the concept of entropy. #### 1. Entropy Increase (Clausius Statement) The entropy of an isolated system always increases in a spontaneous process and remains constant in a reversible process. - **Formula:** $\Delta S_{universe} \ge 0$ - $\Delta S_{universe}$: Change in entropy of the universe (system + surroundings). - `>` sign for spontaneous (irreversible) processes. - `=` sign for reversible (equilibrium) processes. - **Concept:** This implies that the universe is constantly moving towards a state of greater disorder or randomness. An isolated system will spontaneously proceed towards states of higher entropy. #### 2. Kelvin-Planck Statement (Heat Engines) It is impossible to construct a device which operates in a cycle and produces no effect other than the extraction of heat from a single thermal reservoir and the production of an equivalent amount of work. - **Concept:** This means no heat engine can convert 100% of the heat it absorbs into work. Some heat must always be expelled to a colder reservoir. #### 3. Efficiency of Heat Engines The maximum possible efficiency of a heat engine is given by the Carnot efficiency. - **Formula:** $\text{Efficiency} (\eta) = 1 - \frac{T_C}{T_H}$ - $\eta$: Thermodynamic efficiency of the heat engine. Always less than 1 (or 100%). - $T_C$: Absolute temperature of the cold reservoir (sink) in Kelvin. - $T_H$: Absolute temperature of the hot reservoir (source) in Kelvin. - **Concept:** This formula sets the theoretical upper limit for the efficiency of any heat engine operating between two given temperatures. Reversible (Carnot) engines achieve this efficiency; real engines are always less efficient. ### Entropy (S) Entropy ($S$) is a state function that measures the degree of randomness or disorder in a system. The more ways a system can arrange its particles or energy, the higher its entropy. #### 1. Change in Entropy ($\Delta S$) - **Formula (General, for reversible process):** $\Delta S = \int \frac{dQ_{rev}}{T}$ - $dQ_{rev}$: Infinitesimal amount of heat exchanged reversibly. - $T$: Absolute temperature in Kelvin. - **Formula (For reversible isothermal process):** $\Delta S = \frac{Q_{rev}}{T}$ - $Q_{rev}$: Heat exchanged reversibly at constant temperature. - **Formula (For phase transitions, e.g., melting, boiling):** $\Delta S_{transition} = \frac{\Delta H_{transition}}{T_{transition}}$ - $\Delta H_{transition}$: Enthalpy change for the phase transition (e.g., $\Delta H_{fusion}$, $\Delta H_{vaporization}$). - $T_{transition}$: Absolute temperature at which the phase transition occurs (melting point, boiling point). - **Units:** Joules per Kelvin ($J/K$) or kilojoules per Kelvin ($kJ/K$). #### 2. Entropy Change for an Ideal Gas - **Formula (General for reversible process):** * $\Delta S = nC_V \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right)$ * $\Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right)$ - $n$: Number of moles of ideal gas. - $C_V$: Molar heat capacity at constant volume. - $C_p$: Molar heat capacity at constant pressure. - $R$: Ideal gas constant. - $T_1, T_2$: Initial and final absolute temperatures. - $V_1, V_2$: Initial and final volumes. - $P_1, P_2$: Initial and final pressures. - **Note:** These formulas are for processes of an ideal gas and can be simplified for specific conditions (e.g., isothermal, isobaric, isochoric). #### Example 1 (Good): Calculate the entropy change when 1 mole of ice at 0°C melts to form water at 0°C. Given: Standard enthalpy of fusion ($\Delta H_{fusion}$) for ice = $6.01 \text{ kJ/mol}$. * **Given:** * $n = 1 \text{ mol}$ * $\Delta H_{fusion} = 6.01 \text{ kJ/mol} = 6010 \text{ J/mol}$ * $T_{transition} = 0°C = 273 \text{ K}$ (Melting occurs isothermally at the melting point). * **Apply the formula for phase transitions:** * $\Delta S = \frac{\Delta H_{fusion}}{T_{transition}}$ * $\Delta S = \frac{6010 \text{ J/mol}}{273 \text{ K}} \approx 22.01 \text{ J K}^{-1} \text{ mol}^{-1}$ #### Example 2 (Typical JEE): 1 mole of an ideal gas expands isothermally and reversibly from an initial volume of 10 L to 100 L at 300 K. Calculate the entropy change of the gas. ($R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$) * **Given:** * $n = 1 \text{ mol}$ * $V_1 = 10 \text{ L}$ * $V_2 = 100 \text{ L}$ * $T = 300 \text{ K}$ (isothermal, so $T_1 = T_2$). * **Apply the ideal gas entropy change formula:** * $\Delta S = nC_V \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right)$ * Since the process is isothermal, $T_1 = T_2$, so $\ln(T_2/T_1) = \ln(1) = 0$. The first term becomes zero. * Therefore, the simplified formula for isothermal expansion/compression is $\Delta S = nR \ln\left(\frac{V_2}{V_1}\right)$. * $\Delta S = 1 \text{ mol} \times 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times \ln\left(\frac{100 \text{ L}}{10 \text{ L}}\right)$ * $\Delta S = 8.314 \times \ln(10) \text{ J K}^{-1}$ * Using $\ln(10) \approx 2.303$: * $\Delta S \approx 8.314 \times 2.303 \text{ J K}^{-1} \approx 19.14 \text{ J K}^{-1}$ ### Third Law of Thermodynamics The Third Law defines the absolute zero of entropy. - **Statement:** The entropy of a perfect crystalline substance is zero at absolute zero temperature (0 K). - **Concept:** This law provides a reference point for determining absolute entropy values ($S^\circ_T$) for substances. A perfect crystal at 0 K has maximum order and minimal thermal motion, hence zero entropy. For non-perfect crystals or mixtures, entropy is not zero at 0 K. ### Gibbs Free Energy (G) and Helmholtz Free Energy (A) These are thermodynamic potentials that combine enthalpy/internal energy, temperature, and entropy to predict the spontaneity and maximum useful work obtainable from a process. #### 1. Gibbs Free Energy (G) Gibbs Free Energy is most commonly used in chemistry, as many reactions occur at constant temperature and pressure. - **Definition:** $G = H - TS$ - $G$: **Gibbs Free Energy** (a state function). Represents the maximum amount of non-PV work that can be extracted from a thermodynamically closed system at constant temperature and pressure. - $H$: Enthalpy. - $T$: Absolute temperature in Kelvin. - $S$: Entropy. - **Change in Gibbs Free Energy ($\Delta G$):** * **Formula (for isothermal process):** $\Delta G = \Delta H - T\Delta S$ - $\Delta G$: Change in Gibbs Free Energy. - $\Delta H$: Change in Enthalpy. - $T$: Absolute temperature (constant). - $\Delta S$: Change in Entropy of the system. * **Criterion for Spontaneity (at constant T, P):** - $\Delta G 0$: Process is non-spontaneous in the forward direction (spontaneous in the reverse direction). * **Relationship to non-PV work:** $\Delta G = -W_{non-PV, max}$ - $W_{non-PV, max}$: Maximum useful (non-pressure-volume) work obtainable from the system (e.g., electrical work in a battery) at constant T and P. - **Units:** Joules (J) or kilojoules (kJ). #### 2. Relationship between $\Delta G^\circ$ and Equilibrium Constant ($K_{eq}$) The standard Gibbs Free Energy change ($\Delta G^\circ$) of a reaction is linked to its equilibrium constant. - **Formula:** $\Delta G^\circ = -RT \ln K_{eq}$ - $\Delta G^\circ$: **Standard Gibbs Free Energy change** (at standard conditions, typically 298 K and 1 atm/1 M). - $R$: Ideal gas constant ($8.314 \text{ J mol}^{-1} \text{ K}^{-1}$). - $T$: Absolute temperature in Kelvin (must be the same as the temperature at which $K_{eq}$ is defined). - $K_{eq}$: Equilibrium constant for the reaction. - **Meaning:** This formula allows calculation of $K_{eq}$ from thermodynamic data or vice versa. A large negative $\Delta G^\circ$ corresponds to a very large $K_{eq}$ (products favored at equilibrium). A large positive $\Delta G^\circ$ corresponds to a very small $K_{eq}$ (reactants favored at equilibrium). #### 3. Helmholtz Free Energy (A) Helmholtz Free Energy is used for processes occurring at constant temperature and volume. - **Definition:** $A = U - TS$ - $A$: **Helmholtz Free Energy** (a state function). Represents the maximum total work (PV + non-PV) that can be extracted from a system at constant temperature and volume. - $U$: Internal Energy. - $T$: Absolute temperature in Kelvin. - $S$: Entropy. - **Change in Helmholtz Free Energy ($\Delta A$):** * **Formula (for isothermal process):** $\Delta A = \Delta U - T\Delta S$ * **Criterion for Spontaneity (at constant T, V):** - $\Delta A 0$: Process is non-spontaneous. * **Relationship to total work:** $\Delta A = -W_{total, max}$ - $W_{total, max}$: Maximum total work obtainable from the system at constant T and V. #### Example 1 (Good): For a reaction, $\Delta H = 200 \text{ kJ/mol}$ and $\Delta S = 400 \text{ J K}^{-1} \text{ mol}^{-1}$ at 298 K. Determine if the reaction is spontaneous at 298 K under standard conditions. * **Given:** * $\Delta H = 200 \text{ kJ/mol} = 200000 \text{ J/mol}$ * $\Delta S = 400 \text{ J K}^{-1} \text{ mol}^{-1}$ * $T = 298 \text{ K}$ * **Calculate $\Delta G$:** * $\Delta G = \Delta H - T\Delta S$ * $\Delta G = 200000 \text{ J/mol} - (298 \text{ K} \times 400 \text{ J K}^{-1} \text{ mol}^{-1})$ * $\Delta G = 200000 \text{ J/mol} - 119200 \text{ J/mol}$ * $\Delta G = 80800 \text{ J/mol} = 80.8 \text{ kJ/mol}$ * **Conclusion:** Since $\Delta G$ is positive ($\Delta G > 0$), the reaction is **non-spontaneous** in the forward direction at 298 K. #### Example 2 (Typical JEE): The standard Gibbs free energy change ($\Delta G^\circ$) for a reaction is $-115 \text{ kJ/mol}$ at 298 K. Calculate the equilibrium constant ($K_{eq}$) for the reaction. ($R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$) * **Given:** * $\Delta G^\circ = -115 \text{ kJ/mol} = -115000 \text{ J/mol}$ * $T = 298 \text{ K}$ * $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ * **Apply the formula:** $\Delta G^\circ = -RT \ln K_{eq}$ * $-115000 \text{ J/mol} = -(8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \times (298 \text{ K}) \times \ln K_{eq}$ * $\ln K_{eq} = \frac{-115000}{-8.314 \times 298} = \frac{115000}{2477.572} \approx 46.416$ * To find $K_{eq}$, take the exponential of both sides: $K_{eq} = e^{46.416}$ * $K_{eq} \approx 1.2 \times 10^{20}$ * **Conclusion:** The equilibrium constant is very large, indicating that the reaction proceeds almost entirely to completion, forming products at equilibrium. ### Criterion for Spontaneity Spontaneity refers to the tendency of a process to occur without continuous external intervention. #### 1. In terms of Entropy of the Universe ($\Delta S_{universe}$) The most fundamental criterion for spontaneity, applicable to any process. - **Formula:** $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$ - $\Delta S_{system}$: Change in entropy of the reacting system. - $\Delta S_{surroundings}$: Change in entropy of the surroundings. For a process occurring at constant temperature, $\Delta S_{surroundings} = -\frac{Q_{system}}{T} = -\frac{\Delta H_{system}}{T}$ (if constant P). - **Criterion:** * Spontaneous (Irreversible) Process: $\Delta S_{universe} > 0$ * Equilibrium (Reversible) Process: $\Delta S_{universe} = 0$ * Non-spontaneous Process: $\Delta S_{universe} 0$ #### 3. In terms of Helmholtz Free Energy Change ($\Delta A$) Criterion for processes at constant temperature and volume. - **Criterion (at constant T, V):** * Spontaneous: $\Delta A 0$ #### Temperature Dependence of Spontaneity The term $-T\Delta S$ in the Gibbs Free Energy equation ($\Delta G = \Delta H - T\Delta S$) means that temperature can significantly influence the spontaneity of a reaction. - **Case 1: $\Delta H 0$ (Entropy Increases)** * $\Delta G$ will always be negative ($\Delta H$ is negative, $-T\Delta S$ is negative). * **Always Spontaneous** at all temperatures. - **Case 2: $\Delta H > 0$ (Endothermic) and $\Delta S 0$ (Endothermic) and $\Delta S > 0$ (Entropy Increases)** * $\Delta G = (\text{positive value}) - T(\text{positive value})$. * For spontaneity ($\Delta G |\Delta H|$). #### Example (Good): A reaction has $\Delta H = -100 \text{ kJ/mol}$ and $\Delta S = -200 \text{ J K}^{-1} \text{ mol}^{-1}$. At what temperature range will the reaction be spontaneous? * **Given:** * $\Delta H = -100 \text{ kJ/mol} = -100000 \text{ J/mol}$ * $\Delta S = -200 \text{ J K}^{-1} \text{ mol}^{-1}$ * **Condition for spontaneity:** $\Delta G