Chemical Thermodynamics and Thermochemistry PDF

# Chemical Thermodynamics and Thermochemistry ## Intensive and Extensive Properties | Property | Intensive | Extensive | | ------------- | -------- | --------- | | Mass | E | E | | Volume | E | E | | Refractive index | I | I | | Densit...

# Chemical Thermodynamics and Thermochemistry ## Intensive and Extensive Properties | Property | Intensive | Extensive | | ------------- | -------- | --------- | | Mass | E | E | | Volume | E | E | | Refractive index | I | I | | Density of liquid or solid | I | I | | Pressure of gas |I | I | | Resistance | E | E | | Specific Resistance | I | I | | Spring Force | E | E | | Melting Point | E | E | | Boiling Point | E | E | | Temperature | I | I | ## Thermodynamic Functions - There are two types of thermodynamic functions. 1. **State Function:** For a certain path, this function will have the same value. For example: Internal energy 2. **Path Function:** For a certain path, this function will have a different value. For example: Heat and work ## Thermodynamic Processes ### Isochoric Process - In this process, the volume is constant, meaning the volume change is zero. - $$ΔV = 0$$ - $$V = Constant$$ - $$dV = 0$$ - $$P∝T (Gay-Lussac's Law)$$ ### Isobaric Process - In this process, the pressure is constant, meaning the pressure change is zero. - $$ΔP = 0$$ - $$ΔP= 0$$ - $$P = Constant$$ - $$V∝T (Charles's Law)$$ ### Isothermal Process - In this process, the temperature is constant, meaning the temperature change is zero. - $$ΔT= 0$$ - $$dT = 0$$ - $$T = Constant$$ - $$PV = Constant$$ - $$P∝ \frac{1}{V} (Boyle's Law)$$ ### Adiabatic Process - In this process, the heat is constant, meaning the heat change is zero. - $$V= Constant$$ - $$dq= 0$$ - $$Δq= 0$$ ### Cyclic Process - In this process, the initial and final states of the system are the same. - This process is represented by a closed curve. - $$ΔU= 0$$ ### Reversible Process - In this process, the system remains in a state of thermal, mechanical, and chemical equilibrium with its surroundings. - Graphically, this process has continuous curves. ### Irreversible Process - Only the initial and final states are in equilibrium. - In this process, the system only experiences external or constant pressure changes. - Graphically, it doesn't have continuous curves. ## First Law of Thermodynamics - $$dQ = dU + dW$$ - It is based on energy conservation. - $$dQ= dU + dW$$ - Where - $$dQ= Heat$$ - $$dU= Internal Energy$$ - $$dW= Work$$ ## Principle of Calorimetry - Heat Given = Heat Taken ## Expressing Heat In Terms of... - $$Q= m * s * ΔT$$ - Where: - $$m= mass of substance$$ - $$s= specific heat of substance$$ - $$ΔT= change in temperature$$ ## Specific heat of water - 1 Cal/gm°C ## Amount of heat for phase transfer - $$Φ=mL$$ - $$Φ= 92= mL (when solid -> liquid)$$ - $$Φ= 94= mL (when liquid -> Vapor) $$ - Where - $$L1= latent heat of fusion$$ - $$L4= latent heat of vaporization$$ ## Gases - $$dq= nCprocess dT$$ - Where - $$n= no. of moles of gas$$ - $$C= molar heat capacity of gas$$ - $$dT= change in temperature$$ ## Work $$dw = F.dx$$ $$dw= P_{gas} Adx$$ $$dw= P_{gas}dV$$ - Where: - $$P_{gas} A = Area under the curve of "P" vs "V" graph$$ - $$F = P$$ - This can be considered as the "Work by gas" and is a positive value. - This is the expansion of the gas, which is also a positive value. ## Internal Energy - Represented by “dU or dE” - It is the sum of all types of energy, including: - Kinetic Energy - Potential Energy - Rotational Energy - Vibrational Energy - In a cyclic process, work is the area of the closed loop. - In ideal gases, the internal energy is influenced by the temperature only. ## Thermodynamics ### Isochoric Process - The volume is constant, meaning the volume change is zero. - $$dV= 0$$ - The work here is considered the work done by the system, which is 0. - $$dW = Pdv= 0$$ - $$dQ= nCvdT$$ ### Isobaric Process - The pressure is constant, meaning the pressure change is zero. - $$dQ= nCpdt$$ - $$du = nCydT$$ - $$dW= Pdv$$ - $$PV=nRT$$ - $$Pdv+ vdp= nRdT$$ - $$Pdv = nRdT$$ - $$dQ = dV + dW$$ - $$nCpdt =nCydt + nRdT$$ - $$Cp=Cy+R (mayer's Relation)$$ - $$Cp>Cy$$ ### Adiabatic Process - The heat is constant, meaning the heat change is zero. - $$V = \frac{Cp}{Cv}$$. - $$Cp=Cy$$ - $$Cp = Cy+R$$ - $$YCv = Cv+R$$ - $$Cv(γ-1) = R$$ - $$Cv = \frac{R}{γ-1}$$ - $$Cp = \frac{γR}{γ-1}$$ ### Isothermal Process - The temperature is constant meaning the temperature change is zero. - $$dQ=nCprocess dT$$ - $$du = 0 (dT=0)$$ - $$dQ = dU + dW$$ - $$dQ = dW$$ - $$Cprocess = \frac{dQ}{ndT}$$ - $$Cprocess = 0$$ - $$dW= Pdv$$ - $$dW= nRT \frac{dV}{V}$$ - $$∫dW= nRT (∫\frac{dV}{V})$$ - $$W= nRT (lnV_2 - lnV_1)$$ - $$W= nRT ln(\frac{V_2}{V_1})$$ ### Adiabatic Process - In this process, the heat is constant, meaning the heat change is zero. - $$dQ = 0$$ - $$dQ = dW + dU$$ - $$0 = dW + dU$$ - $$dW = -dU$$ - $$dW = - nCvdT$$ - $$dW = -n(γ-1) (T_2 - T_1)$$ - $$dW = nRT_1 - nRT_2$$ - $$dW = P_1V_1 - P_2V_2$$ - For adiabatic processes, the following is also true: - $$dQ = nCprocess dT$$ - $$Cprocess = \frac{dQ}{ndT}=0$$ - $$Cprocess = 0$$ - $$dQ = dU+dW$$ - $$dU = -dW$$ - $$Pdv = - nCvdT$$ - $$nRT\frac{dV}{V}= - nCv dT$$ - $$\frac{dV}{V}= \frac{-Cv }{R} dT$$ - $$\frac{dV}{V}= \frac{-1}{γ-1}dT$$ - $$lnV = \frac{-1}{γ-1} lnT + lnc$$ - $$lnV + lnV^{-1} = lnT + lnc$$ - $$ln(V^{γ-1}T)= lnc$$ - $$V^{γ-1}T = constant$$ - $$PV = nRT$$ - $$PV^{γ}= constant$$ - $$nR^{γ}V^{γ-1}= constant$$ - $$pV^{γ}= constant$$ ## Thermochemistry - We calculate heat in terms of enthalpy. - **Enthalpy:** The heat at a constant pressure - Represented by: - ΔH or qp - **Note:** We can calculate the enthalpy of reaction, but not the enthalpy of any reactant or product individually. - For a reaction, enthalpy is an indication of its endo- or exothermic nature: - ΔH of reaction > 0 indicates an **endothermic** reaction. - ΔH of reaction < 0 indicates an **exothermic** reaction. **Relationship between enthalpy and internal energy based on the following reaction:** $$aA(s) + bB(g) + cC(g) → dD(g) + eE(g)$$ $$dw = P(V_2 - V_1)$$ $$dw = P(V_A+V_B-V_A-V_B-V_C)$$ $$dw = P(V_A-V_C)$$ - The volume of solids and liquids is negligible compared to gaseous materials in this case. - $$dw = P(V_B - P_VC)$$ $$dw = ΔRT - CRT$$ $$dw = (d-c)RT$$ $$dw = ΔηRT$$ - **Change in Enthalpy:** - "ΔΗ = ΔΕ + ΔηRT” - Δη = no. of moles of gaseous products - no. of moles of gaseous reactants - **Heat at constant volume:** - "Internal Energy" (nCvdT) - **Heat at constant pressure:** - "enthalpy" - **Example:** - **N2(g) + 3H2(g) → 2NH3(g)** - ΔH = ΔΕ + (2 - 4)RT - ΔH = ΔΕ - 2RT - **Another Example:** - **CaCO3(s) → CaO(s) + CO2(g)** - $$ΔH=ΔΕ + RT$$ ## Heat of Combustion - When one mol of substance is completely burnt in the presence of oxygen, the heat *evolved* is known as the heat of combustion. - Combustion generally refers to an **exothermic** reaction. - **Examples:** - $$CH4 + 2O2 -> CO2 + 2H2O$$ - $$C(graphite) + O2(g) -> CO2(g)$$ - $$H2(g) + 1/2O2(g) → H2O(l)$$ - **Note:** Heat of combustion should always be for one mole. ## Enthalpy of Neutralization - When one gram equivalent of an acid is completely neutralized with one gram equivalent of a base, the heat evolved is known as “heat of neutralization”. - **Example:** - Strong Acid + Strong Base → Salt + water - $$ΔH = (-13.7) K cal$$ ## Heat of formation - The heat of formation is calculated based on the number of moles neutralized. **Example:** - 1 mol HCL + 0.5 mol NaOH = 0.5 mol neutralized - $$ΔH = -13.7/2 = -6.85 kcal$$ **Factors impacting neutralization:** - If either the acid or the base is weak, the heat evolved from neutralization is less than 13.7 kcal. - This is because some heat is lost due to the ionization of the acid. ## The Heat of Combustion of Ethylene - The heat of combustion of ethylene, at 18°C and constant volume, is -335.8 Kcal when water is obtained in liquid form. - Calculate the heat of combustion at constant pressure. - $$C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)$$ - $$ΔE = -335.8 Kcal$$ - $$ΔE= (- 335.8-2 \times 2 \times 291) Kcal$$ - $$ΔE= (- 335.8 - 1164) Kcal$$ - $$ΔE = -335.8 -1164$$ - $$ΔE = -336.964 Kcal$$ ## Standard heat of formation - The standard heat of formation of methane gas at 300 K is -75.83 KJ. - Calculate the heat of formation at constant volume. - $$C(graphite) + 2 H2(g) → CH4(g)$$ - $$ΔH = ΔΕ + ΔηRT$$ - $$-75.83 = ΔΕ + (1-2)RT$$ - $$-75.83 = ΔΕ - RT$$ - $$ΔΕ = RT - 75.83$$ - $$ΔΕ = (8.31) × 300 - 75.83) KJ$$ - $$ΔΕ = (2493 - 75.83) KJ$$ - $$ΔΕ = -73.337 KJ$$ ## Thermodynamic Calculations - Standard heat of formation at 298k for the following substances: - $$CCl4(g) (-25.5 Kcal/mol)$$ - $$H2O(g) (-57.8 Kcal/mol)$$ - $$CO2(g) (-99.1 Kcal/mol)$$ - $$HCL(g) (-22.1 Kcal/mol)$$ - Calculate the heat of reaction for: - $$CCl4(g) + 2 H2O(g) → CO2(g) + 4HCL(g)$$ - $$ΔH°= [(-94.1) - 4 (-22.1)] - [2(-57.8) - 25.5]$$ - $$ΔH°= (-94.1 - 88.4 ) - ( -115.6 - 25.5)$$ - $$ΔH°=-182.5 + 141.1$$ - $$ΔH°= -41.4 Kcal$$ - **Note:** The molal heat of formation of $NH4NO3(s)$ is -367.5 KJ and that of $N2O(g)$ and $H2O(l)$ are 81.46 KJ and -285.78 KJ, respectively at 25°C and 1 atm. pressure. - Calculate $ΔH$ and $ΔΕ$ for the reaction: - $$NH4NO3(s) → N2O(g) + 2H2O(l)$$ - $$ΔH = (-367.5) - (-285.78 + 81.46)$$ - $$ΔH=(-367.5)-(-285.78+162.92)$$ - $$ΔH= -367.5 + 122.86$$ - $$ΔH= 144.64 KJ$$ - The heat of formation of the above reaction is 144.64 KJ: - $$ΔH = ΔΕ + ΔηRT$$ - $$144.64 = ΔΕ + 8.31 × 298$$ - $$ΔΕ = - 125 KJ$$ ## Bomb Calorimetry - Bomb calorimetry is used to burn a fuel. - If heat is added or evolved from a closed container, this heat is considered as $ΔΕ$. - $$m_1s_1t_1 + m_2s_2t_2 + m_3s_3t_3 ... m_ns_nt_n = (m_1+m_2+...) Sol J$$ - $$s= specific heat of water = 1g/ml$$ ## Calorific Value - This refers to heat of combustion per gram of fuel. ## Efficiency of Fuel - The heat of combustion for $ CH_4 $ and $C_4H_10$ are -890.3 KJ/mol and -2878KJ/mol, respectively. - Which one has greater efficiency as fuel per given? - Calorific fuel ∝ Efficiency of fuel - For $CH_4$: - 16 gm, ΔH=-89.3 KJ - 1 gm, ΔH = -890/16 = -55.56 KJ/gm - For $C_4H_10$: - 58 gm, ΔH= -2878.7 KJ - 1 gm, ΔH= -2878.7/58 = -99.6 KJ/gm - We compare the magnitudes of the results. The larger magnitude represents the larger heat evolved, leading to a larger calorific value. Therefore, $CH_4$ is a more efficient fuel source. ## Calculating the heat of formation of the substance - When 2 moles of $C_2H_6$ completely burn, 3129 KJ of heat is released. - The standard heat of formation of $CO_2$ and $H_2O$ is 395 kJ and 286 kJ, respectively. - Calculate the heat of formation of $C_2H_6$ - $$C_2H_6(g) + 7/2O_2(g) → 2CO_2(g) + 3H_2O$$ - $$ΔH = ΔH_{product} - ΔH_{reactant}$$ - $$ - 3129 = (2 \times -395 - 3 \times -286) - (HC_2H_6)$$ - $$ - 3129 = (-790 - 858) - H_{C_2H_6}$$ - $$H_{C_2H_6} = +1564.5 - 1648$$ - $$H_{C_2H_6} = -83.5 KJ$$ ## Calculating the amount of heat evolved for the burning of a gas - The standard heats of formation of $CH_4$, $CO_2$, and $H_2O$ are -76.2, -398.8, and -241.6 KJ/mole, respectively. - Calculate the amount of heat evolved by burning 1 cubic meter of $CH_4$ gas. - $$CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(g)$$ - $$ΔH= [2( - 241.6) - 398.8] - [-76.2]$$ - $$ΔH = - 483.2 -398.8 + 76.2 $$ - $$ΔH = 882.0 + 76.2$$ - $$ΔH = -805.8 KJ$$ - $$1 m^3 = 1000 L$$ - $$V = 1000L$$ - $$n = \frac{1000}{22.4} = 44.1$$ - $$ΔH = (-805.8 × 44.1) KJ$$ ## Heat Evolved From a Gas Mixture - A mixture of 3.67 L of ethylene ($C_2H_4$) and methane ($CH_4$) underwent combustion at 25°C, resulting in 6.11 L of $CO_2$. - Calculate the heat evolved from burning 1 L of the gas mixture. - The heat of combustion of ethylene and methane are -1423 and -891 KJ/mole, respectively, at 25°C - $$C_4H_4 + 2O_2 -> CO_2 + 2H_2O$$ - $$C_2H_4 + 3O_2 -> 2CO_2 + 2H_2O$$ - $$2V + 3.67 - V = 6.11$$ - $$V = 6.11 - 3.67 = 2.44 L$$ - $$Volume of C_2H_4 = 2.44 L$$ - $$Volume of CH_4 = 3.67 - 2.44 = 1.23 L$$ - Therefore, the volume of ethylene is 2/3 of the mixture, while the volume of methane is 1/3. - For 1 L of the mixture: - $C_2H_4 = 2/3 × 1 = 0.66 L$ - $CH_4 = 1/3 × 1 = 0.33 L$ - $$PV=nRT$$ - $$1 \times \frac{2}{3} = n \times \frac{1}{2} \times 298$$ - $$n =\frac{8}{298} = \frac{4}{149}$$ - $$PV=nRT$$ - $$1 \times \frac{1}{3} = n \times \frac{1}{2} \times 298$$ - $$n = \frac{2}{149}$$ - $$ΔH_{C_2H_4} = (\frac{4}{149} \times -1423) $$ - $$ΔH_{CH_4} = (\frac{2}{149} \times -891)$$ - **Total Heat Evolved:** - $$( \frac{4}{149} \times -1423) + ( \frac{2}{149} \times -891) KJ$$ - $$ ( \frac{8}{149} \times -1423 -891) KJ$$ - $$\frac{8}{149} \times -3737 = -50.81 KJ $$ ## Heat of Neutralization - Calculate the heat evolved for each reaction: 1. 0.5 mole of HCl is neutralized by 0.5 mol NaOH (0.5 mol neutralized) - Heat evolved = 13.7/2 Kcal 2. 100 ml of 0.2 M HCl is neutralized by 30 ml of 0.3 M NaOH (0.003 moles neutralized) - Heat evolved = 3 × 13.7 / 10 = 4.11 Kcal 3. 100 ml of 0.2 M HCl is neutralized by 0.3 M NaOH (0.02 moles neutralized) - Heat evolved = 13.7/50 Kcal 4. 400 ml of 0.2 M H2SO4 is neutralized by 600 ml of 0.7 M KOH (0.06 moles neutralized) - Heat evolved = (- 60/1000 × 13.7) Kcal ## Hydration - It refers to the amount of heat evolved when one mole of anhydrous salt is converted into its corresponding hydrated salt. - **Examples:** - $$CuSO4 + 5H2O → CuSO4.5H2O$$ - $$ΔH = -18.89 Kcal$$ - $$CaCl2 + 6H2O → CaCl2.6H2O$$ - $$ΔH = -18.8 Kcal$$ ## Enthalpy of Solution - When one mole of anhydrous salt, or a partially hydrated salt, is added to an excess amount of water, the heat evolved is referred to as the "heat of solution." - **Examples:** - $$CuSo4(s) + aq → CuSO4(aq)$$ - $$CaCl2 + aq → CaCl2(aq)$$ ## Change in enthalpy _1 Mol of CuSO4 is added to 5 moles of H2O, and the heat evolved is 18.69 Kcal. If 1 mol of CuSO4 is added to 1000 mol of H2O, 19.8 Kcal of heat is evolved. What is the change in enthalpy if a solution containing 5 moles of H2O and 1 mol of CuSO4 is added to 995 moles of water?_ - $$CuSO4.5H2O + 995H2O → CuSO4.1000H2O$$ - $$Change in enthalpy = -19.8 +18.69$$ - $$Change in enthalpy = -1.11 Kcal$$ ## Enthalpy of Ionization - The amount of heat absorbed when one mole of electrolyte dissociates into ions. - Example: $$AB -> A+ + B-$$ - ΔH is greater than zero. ## Enthalpy for Phase Change - **Enthalpy or Heat of Fusion:** - This happens when a mole of solid is converted to a liquid, and it is an endothermic reaction because heat is *absorbed* from the surroundings. - $$A_s(solid) →A_l(liquid)$$ - $$ΔH > 0$$ - The heat absorbed in this reaction is known as the "Heat of fusion." - **Enthalpy or Heat of Vaporization:** - This happens when a mole of liquid is converted to a gas, and it is also an endothermic reaction because heat is *absorbed* from the surroundings. - $$A_l(liquid) →A_g(gaseous)$$ - $$ΔH > 0$$ - The heat absorbed in this reaction is known as the "Heat of vaporization." - **Enthalpy or Heat of Sublimation:** - This happens when a mole of solid is converted to a gas. - This is also an endothermic reaction. - $$A_s(solid) →A_g(gaseous)$$ - $$ΔH > 0$$ - The heat absorbed in this reaction is known as the "Heat of sublimation." - $$ΔH_sub= ΔH_{fusion} + ΔH_{vap}$$ ## Bond Energy - The average amount of energy required to break all the bonds in a mole of a molecule. - Represented by "E". ## Resonance Energy - **Resonance energy** refers to the difference between the observed heat of formation and the actual heat of formation. ## Limitations of the First Law of Thermodynamics - It doesn't provide information on the direction in which a change will progress. - It doesn't indicate the extent to which a change takes place. **Examples:** 1. When a bullet strikes a black surface, the kinetic energy converts into heat. However, it doesn't explain why the heat cannot be converted back into kinetic energy to propel the bullet. 2. A vessel of water placed over fire receives heat. Why doesn't the heat flow back into the fire, leading to water cooling down and freezing? 3. It's not feasible to completely convert heat into work; some energy is always lost during the process. ## Second Law of Thermodynamics - All spontaneous processes are thermodynamically irreversible. - Without the help of an external agency, a spontaneous process cannot be reversed. - Heat cannot flow spontaneously from a colder body to a hotter body. - Converting heat entirely into work is impossible without some effect elsewhere. - Spontaneous processes are accompanied by a net increase in entropy of the universe. - Entropy acts as the arrow of time. - **Example:** - This law explains why a hot object doesn't spontaneously increase its temperature when it comes into contact with a cold object; it explains why heat always flows from hot to cold. - **Entropy:** The degree of randomness or disorder within a system. - It's represented by "S". - **Higher the arrangement of particles, the lower the randomness, hence the lower the entropy.** - It is a **state function**, meaning the change in its value only depends on the initial and final states, not the path taken. - It is an **extensive property**, meaning it depends on the amount of matter in the system. - The absolute entropy can be calculated for any substance: - $$ΔS= S_p - S_r$$ - **Entropy Changes based on the following reaction:** - $$aA + bB → cC + dD$$ - $$ΔS= (cS_c + dS_d) - (aS_a+ bS_b)$$ ## Entropy Calculations - **Entropy Calculations for Thermodynamic Processes:** - $$ΔS_{universe}=ΔS_{system } ΔS_{surroundings}$$ - $$ΔS = (\frac{dQ}{T})_{rev}$$ - $$dQ= dU + dW$$ - $$dQ = dU + dW$$ - $$dQ = \frac{nCvdT}{T} + \frac{PdV}{T}$$ - $$∫^{T2}_{T1} nCvdT → nCv ln(\frac{T2}{T1})$$ - $$PV = nRT$$ - $$\frac{PdV}{T} = \frac{nRdV}{V}$$ - $$∫^{V2}_{V1} \frac{nRdV}{V} → nR ln(\frac{V_2}{V_1})$$ - $$ΔS = nCvln(\frac{T_2}{T_1}) + nR ln(\frac{V_2}{V_1}) $$ - We know that: - $$Cv = Cp-R$$ - $$ΔS = (Cp-R)nln(\frac{T_2}{T_1})+nR ln (\frac{V_2}{V1})$$ - $$ΔS = nCp ln(\frac{T_2}{T_1}) - nR ln(\frac{T_2}{T_1})+nR ln (\frac{V_2}{V1})$$ - $$ΔS = nCp ln(\frac{T_2}{T_1})- nR[ln(\frac{T_2}{T1}) + ln (\frac{V2}{V1})]$$ - $$ΔS = nCp ln(\frac{T_2}{T_1}) + nR[ -ln(\frac{T_2}{T1}) + ln (\frac{V2}{V1})]$$ - $$P_1V_1 = P_2V_2$$ - $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} → P_2 = \frac{T_2}{T_1} P_1$$ - $$ΔS = nCp ln(\frac{T_2}{T_1}) + nR ln(\frac{P_2}{P_1}) $$ ## Entropy Calculations for Specific Processes ### Isochoric Process - $$ΔS = nCvln(\frac{T_2}{T_1}) + nR ln(1)$$ - $$ΔS= nCvln(\frac{T_2}{T_1})$$ ### Isobaric Process - $$P_1 = P_2$$ - $$ΔS = nCp ln (\frac{T_2}{T_1}) + nR ln(1)$$ - $$ΔS = nCp ln (\frac{T_2}{T_1})$$ ### Isothermal Process - $$dQ = dV + dW$$ - $$dQ = dW$$ - $$dQ = Pdv = nR dV$$ - $$\frac{dQ}{T} = nR \frac{dV}{V}$$ - $$∫\frac{dQ}{T} = nR ln(\frac{V_2}{V_1})$$ ### Adiabatic Process - **Note:** - $$dQ = 0$$ - The entropy is zero because the heat change is zero. - $$ΔS= 0$$ - $$ΔS = \frac{dQ}{T} = 0$$ ## Entropy Change During Phase Change ### Entropy of Fusion - This represents the change in entropy when one mol of a solid substance transforms into its liquid state at the melting point. - $$ΔS = \frac{ΔH_{fusion}}{T}$$ - $$T = Melting Point$$ ### Entropy of Vaporization - This represents the change in entropy when one mol of a liquid transforms into its gaseous state at the boiling point. - $$ΔS = \frac{ΔH_{vap}}{T}$$ ### Entropy of Sublimation - This represents the change in entropy when one mol of a solid substance transforms into its gaseous state. - $$ΔS = \frac{ΔH_{sub}}{T}$$ ### Entropy of Physical Transition - A representative example, transformation of graphite (Cgraphite) into diamond (Cdiamond): - $$ΔS = \frac{ΔH{transition}}{T{transition}}$$ ## Entropy Change During Chemical Reactions - **Example:** - $$CaCO3(s) -> CaO(s) + CO2(g)$$ - Δη> 0, indicating an increase in entropy. - Entropy increases because the randomness is higher in the products compared to reactants. - $$Δn < 0$$, indicating a decrease in entropy. - Entropy decreases because the reactants have higher randomness than the products. - $$Δn = 0$$, predicting a change in entropy is not possible. ## Gibbs Free Energy - $$ΔG = ΔH - TΔS$$ - $$ΔG = - TΔS_{universe} = ΔG$$ - Where: - $$ΔG = Gibbs's Free Energy / Maximum Useful Work$$ - $$ΔS = Entropy$$ - $$T = Temperature$$ - Based on the change in Gibbs free energy, we can figure out the spontaneity of a reaction/process: - $$ΔG < 0$$ = **Spontaneous** - $$ΔG> 0$$ = **Non-spontaneous** - $$ΔG = 0$$ = **At Equilibrium** - For a reaction to be considered spontaneous, ΔG must be negative. - $$ΔG = ΔH - TΔS$$ - **The reaction becomes spontaneous when:** - $$ΔH < 0$$ and $$ΔS > 0$$ - **When both ΔH and ΔS are negative, the reaction can be considered nonspontaneous (at lower temperatures) and spontaneous (at higher temperatures):** - $$ΔG = ΔH-TΔS$$ **For a reaction to be spontaneous at all temperatures:** - **ΔH < 0** and **ΔS > 0** **When both ΔH