Lect. 4 chemical potential of an ideal gas vant hoff reaction isotherm,vant hoff equation
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This presentation includes variation of chemical potential of an ideal gas. Vant Hoff reaction isotherm and Vant Hoff equation
![CHEMICAL POTENTIAL OF AN IDEAL GAS IN A GASEOUS MIXTURE
For an ideal gas, we know that
PV = nRT ….(i)
As the system consist of 𝑛1, 𝑛2, 𝑛3, etc. moles of the different constituents,
𝑛 = 𝑛1 + 𝑛2 + 𝑛3+ ……..+ 𝑛 𝑖 + ⋯ … . (ii)
Substituting this value in equation (i), we get,
PV = (𝑛1 + 𝑛2 + 𝑛3+ ……..+𝑛 𝑖 + ⋯ ) RT …(iii)
𝑉 = (𝑛1 + 𝑛2 + 𝑛3 + ⋯ + 𝑛 𝑖 + ⋯ )
RT
P
… (iv)
Differentiating this equation with respect to ni keeping all other n’s constant as well as
temperature and pressure constant, we get,
𝜕𝑉
𝜕𝑛 𝑖 𝑇,𝑃,𝑛1,𝑛2,𝑛3 ,
=
𝑅𝑇
𝑃
… (v)
Vi =
RT
P
Since
𝜕𝑉
𝜕𝑛 𝑖
= Vi] … . . (vi)
But we know that,
𝜕𝐺
𝜕𝑃 𝑇,𝑁
= 𝑉
Differentiating both side by
𝜕
𝜕𝑛 𝑖
we get ,
𝜕
𝜕𝑛 𝑖
𝜕𝐺
𝜕𝑃 𝑇,𝑁
=
𝜕𝑉
𝜕𝑛 𝑖
𝜕
𝜕𝑃
𝜕𝐺
𝜕𝑛 𝑖 𝑇,𝑁
= Vi Since
𝜕𝑉
𝜕𝑛 𝑖
= Vi ]
𝜕µi
𝜕𝑃
= Vi Since
𝜕𝐺
𝜕𝑛 𝑖
= µi
] … vii
1-September -20 Dr. Yogita Sahebrao Thakare](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Similaire à Lect. 4 chemical potential of an ideal gas vant hoff reaction isotherm,vant hoff equation
Similaire à Lect. 4 chemical potential of an ideal gas vant hoff reaction isotherm,vant hoff equation (16)
Plus de Shri Shivaji Science College Amravati
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Lect. 4 chemical potential of an ideal gas vant hoff reaction isotherm,vant hoff equation
- 1. Dr. Y. S. THAKARE M.Sc. (CHE) Ph D, NET, SET Assistant Professor in Chemistry, Shri Shivaji Science College, Amravati Email: yogitathakare_2007@rediffmail.com B Sc- II Year SEM-III PAPER-III PHYSICAL CHEMISTRY UNIT- V -THERMODYNAMICS AND EQUILIBRIUM Topic- Chemical potential of an ideal gas in a gaseous mixture and Vant Hoff Reaction Isotherm 1-September -20 1
- 2. CHEMICAL POTENTIAL OF AN IDEAL GAS IN A GASEOUS MIXTURE For an ideal gas, we know that PV = nRT ….(i) As the system consist of 𝑛1, 𝑛2, 𝑛3, etc. moles of the different constituents, 𝑛 = 𝑛1 + 𝑛2 + 𝑛3+ ……..+ 𝑛 𝑖 + ⋯ … . (ii) Substituting this value in equation (i), we get, PV = (𝑛1 + 𝑛2 + 𝑛3+ ……..+𝑛 𝑖 + ⋯ ) RT …(iii) 𝑉 = (𝑛1 + 𝑛2 + 𝑛3 + ⋯ + 𝑛 𝑖 + ⋯ ) RT P … (iv) Differentiating this equation with respect to ni keeping all other n’s constant as well as temperature and pressure constant, we get, 𝜕𝑉 𝜕𝑛 𝑖 𝑇,𝑃,𝑛1,𝑛2,𝑛3 , = 𝑅𝑇 𝑃 … (v) Vi = RT P Since 𝜕𝑉 𝜕𝑛 𝑖 = Vi] … . . (vi) But we know that, 𝜕𝐺 𝜕𝑃 𝑇,𝑁 = 𝑉 Differentiating both side by 𝜕 𝜕𝑛 𝑖 we get , 𝜕 𝜕𝑛 𝑖 𝜕𝐺 𝜕𝑃 𝑇,𝑁 = 𝜕𝑉 𝜕𝑛 𝑖 𝜕 𝜕𝑃 𝜕𝐺 𝜕𝑛 𝑖 𝑇,𝑁 = Vi Since 𝜕𝑉 𝜕𝑛 𝑖 = Vi ] 𝜕µi 𝜕𝑃 = Vi Since 𝜕𝐺 𝜕𝑛 𝑖 = µi ] … vii 1-September -20 Dr. Yogita Sahebrao Thakare
- 3. From equation (vi) and (vii) 𝝏µ 𝐢 𝝏𝑷 = 𝐑𝐓 𝐏 … … . (𝐯𝐢𝐢𝐢) This equation can be written as 𝒅µ 𝐢 = 𝐑𝐓 𝐏 𝐝𝐏 = 𝐑𝐓 𝐝 𝐥𝐧𝐏 … 𝐢𝐱 In a mixture of gaseous, if Pi is the partial pressure of the constituent, we have 𝐏𝐢 𝐕 = 𝐧 𝐢 𝐑𝐓 … . . (𝐱) PV=nRT …(xi) From(x) and (xi), we have 𝐏𝐢 𝐏 = 𝐧 𝐢 𝐧 OR 𝐏𝐢 = 𝐧 𝐢 𝐧 𝐏 = 𝐊. 𝐏. Where, 𝑲 = 𝐧 𝐢 𝐧 is a constant. Taking logarithm on both sides 𝐥𝐧𝐏𝐢 = 𝒍𝒏 𝐊 + 𝐥𝐧𝐏 Hence on differentiation 𝐝𝐥𝐧𝐏𝐢 = 𝒅𝐥𝐧𝐏 𝐒𝐢𝐧𝐜𝐞 𝐝𝒍𝒏 𝐊=0 Substituting this value in equation (ix), we get, 𝒅µ 𝐢 = 𝐑𝐓 𝐝 𝐥𝐧𝐏𝐢 Integrating this equation, we get µ 𝐢 = µ 𝒊 𝟎 + 𝐑𝐓 𝐝 𝐥𝐧𝐏𝐢 … (𝐱𝐢𝐢) Where, µ 𝒊 𝟎 is the integration constant, the value of which depends upon the nature of the gas and also on the temperature. From equation (xii), it is obvious that when Pi =1 atm, µ 𝐢 = µ 𝒊 𝟎 Thus, µ 𝒊 𝟎 is the chemical potential of the constituent when its partial pressure is unity in the mixture of ideal gases. 1-September -20 Dr. Yogita Sahebrao Thakare
- 4. VAN’T HOFF REACTION ISOTHERM Let us consider following gaseous reaction at temperature T K. 𝐚𝐀 𝐠 + 𝐛𝐁(𝐠) 𝐓𝐊 𝐥𝐋(𝐠) + 𝐦𝐌(𝐠) Free energy changes (∆G) for this reaction is given by ∆𝐆 = 𝐆 𝐏𝐫𝐨𝐝𝐮𝐭 - 𝐆 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭 ∆𝐆 = 𝐥µ 𝐋 + 𝐦µ 𝐌 − (𝐚µ 𝐀 + 𝐛µ 𝐁) Where, µ 𝐋, µ 𝐌, µ 𝐀, µ 𝐁 are the chemical potential of various species in the reaction mixture at temperature T and are given by, µ 𝐋 = µ 𝐋 𝟎 + 𝐑 𝐓 𝐥𝐧𝐏 𝐋 µ 𝐌 = µ 𝐌 𝟎 + 𝐑 𝐓 𝐥𝐧𝐏 𝐌 µ 𝐀 = µ 𝐀 𝟎 + 𝐑 𝐓 𝐥𝐧𝐏 𝐀 µ 𝐁 = µ 𝐁 𝟎 + 𝐑 𝐓 𝐥𝐧𝐏 𝐁 In the above equations , µ 𝐋 𝟎 , µ 𝐌 𝟎 , µ 𝐀 𝟎 , µ 𝐁 𝟎 are chemical potential in standered state (partial pressure=1) and 𝐏 𝐋, 𝐏 𝐌, 𝐏 𝐀, 𝐏 𝐁 are the partial pressure at temperature T. Substituting the values of chemical potentialsfrom equation (ii) in equation (i), we get, ∆𝐆 = 𝐥(µ 𝐋 𝟎 + 𝐑 𝐓 𝐥𝐧𝐏 𝐋) + 𝐦(µ 𝐌 𝟎 + 𝐑 𝐓 𝐥𝐧𝐏 𝐌) − 𝐚(µ 𝐀 𝟎 + 𝐑 𝐓 𝐥𝐧𝐏 𝐀) + 𝐛(µ 𝐁 𝟎 + 𝐑 𝐓 𝐥𝐧𝐏 𝐁) On rearranging, we get ∆𝐆 = 𝐥µ 𝐋 𝟎 + 𝐦µ 𝐌 𝟎 − (𝐚µ 𝐀 𝟎 + 𝐛µ 𝐁 𝟎 ) + 𝐑𝐓 𝐥 𝐥𝐧𝐏 𝐋 + 𝐦 𝐥𝐧𝐏 𝐌 – ( 𝐚 𝐥𝐧𝐏 𝐀 + 𝐛 𝐥𝐧𝐏 𝐁) ∆𝐆 = ∆𝐆 𝟎 + 𝐑𝐓 𝐥 𝐥𝐧𝐏 𝐋 + 𝐦 𝐥𝐧𝐏 𝐌 – ( 𝐚 𝐥𝐧𝐏 𝐀 + 𝐛 𝐥𝐧𝐏 𝐁) ∆𝐆 = ∆𝐆 𝟎 + 𝐑𝐓 𝐥𝐧 (𝐏 𝐋) 𝐥 + 𝐥𝐧 (𝐏 𝐌) 𝐦 − 𝐥𝐧( 𝐏 𝐀) 𝐚 + 𝐥𝐧 (𝐏 𝐁) 𝐛 ∆𝐆 = ∆𝐆 𝟎 + 𝐑𝐓 { 𝐥𝐧 (𝐏 𝐋) 𝐥 × 𝐏 𝐌 𝐦 − 𝐥𝐧 𝐏 𝐀 𝐚 × 𝐏 𝐁 𝐛 } ∆𝐆 = ∆𝐆 𝟎 + 𝐑𝐓 𝐥𝐧 (𝐏 𝐋) 𝐥 × (𝐏 𝐌) 𝐦 (𝐏 𝐀) 𝐚 × (𝐏 𝐁) 𝐛 = 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧 𝐪𝐮𝐨𝐭𝐢𝐞𝐧𝐭 … … (𝐢𝐢𝐢) 1-September -20 Dr. Yogita Sahebrao Thakare
- 5. Thus Van’t Hoff reaction isotherm gives free energy changes (∆𝐆) for a chemical reaction in terms of standered free energy changes (∆𝐆 𝟎 ) and the partial pressure of the reactants and products at a given temperature (T). At equilibrium, ∆𝐆 = 𝟎. Hence, the reaction isotherm in the equilibrium state of the chemical reaction will be. 𝟎 = ∆𝐆 𝟎 + 𝐑𝐓 𝐥𝐧 (𝐏𝐋)𝐥 × (𝐏 𝐌) 𝐦 (𝐏 𝐀) 𝐚 × (𝐏 𝐁) 𝐛 𝐞𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 OR −∆𝐆 𝟎 = 𝐑𝐓 𝐥𝐧 (𝐏𝐋)𝐥 × (𝐏 𝐌) 𝐦 (𝐏 𝐀) 𝐚 × (𝐏 𝐁) 𝐛 𝐞𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 … . (𝐢𝐯) OR −∆𝐆 𝟎 𝑹𝑻 = 𝐥𝐧 (𝐏𝐋)𝐥 × (𝐏 𝐌) 𝐦 (𝐏 𝐀) 𝐚 × (𝐏 𝐁) 𝐛 𝐞𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 OR 𝒆 −∆𝐆 𝟎 𝑹𝑻 = (𝐏𝐋)𝐥 × (𝐏 𝐌) 𝐦 (𝐏 𝐀) 𝐚 × (𝐏 𝐁) 𝐛 𝐞𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 Since, ∆𝐆 𝟎 represents standered free energy change of the reaction, it must be constant at a given temperature. Also, R is gas constant. i.e. −∆𝐆 𝟎 𝑹𝑻 = 𝒄𝒐𝒏𝒔𝒕𝒏𝒂𝒕1-September -20 Dr. Yogita Sahebrao Thakare
- 6. Hence, 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 = (𝐏𝐋) 𝐥 × (𝐏 𝐌) 𝐦 (𝐏 𝐀) 𝐚 × (𝐏 𝐁) 𝐛 𝐞𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 𝐊 𝐩 = (𝐏𝐋)𝐥 × (𝐏 𝐌) 𝐦 (𝐏 𝐀) 𝐚 × (𝐏 𝐁) 𝐛 𝐞𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 . . . . 𝐯 Where, 𝐊 𝐩 is the equilibrium constant of the reaction at a given temperature. Therefore equation (iv) may also be written as ∆𝐆 𝟎 = −𝐑𝐓 𝐥𝐧𝐊 𝐩 ∆𝐆 𝟎 = −𝟐. 𝟑𝟎𝟑𝐑𝐓 𝐥𝐧𝐊 𝐩 … . (𝐢𝐯) Derivation of equation (v) and (vi) is thus a thermodynamic derivation of the law of chemical equilibrium. 1-September -20 Dr. Yogita Sahebrao Thakare
- 7. VAN’T HOFF EQUATION (Temperature Dependence of Equilibrium Constant) Let us consider following reaction at equilibrium 𝐚𝐀 𝐠 + 𝐛𝐁(𝐠) 𝐓𝐊 𝐥𝐋(𝐠) + 𝐦𝐌(𝐠) ……(i) Applying Van’t Hoff reaction at equilibrium, we get ∆𝐆 𝟎 = −𝐑𝐓𝐥𝐧𝐊 𝐩 On differentiating above equation with respect to temperature, we have, 𝒅(∆𝑮 𝟎 ) 𝒅𝑻 = −𝑹 𝑻 𝒅(𝒍𝒏𝑲 𝒑) 𝒅𝑻 + 𝒍𝒏𝑲 𝒑 𝐝(∆𝐆 𝟎 ) 𝐝𝐓 = −𝐑𝐓 𝐝(𝐥𝐧𝐊 𝐩) 𝐝𝐓 − 𝐑𝐥𝐧𝐊 𝐩 Multiplying by T on both sides 𝐓 𝐝(∆𝐆 𝟎 ) 𝐝𝐓 = −𝐑𝐓 𝟐 𝐝(𝐥𝐧𝐊 𝐩) 𝐝𝐓 − 𝐑 𝐓 𝐥𝐧𝐊 𝐩 But, ∆𝐆 𝟎 = −𝐑𝐓𝐥𝐧𝐊 𝐩 Hence, 𝐓 𝐝(∆𝐆 𝟎 ) 𝐝𝐓 = −𝐑𝐓 𝟐 𝐝(𝐥𝐧𝐊 𝐩) 𝐝𝐓 + ∆𝐆 𝟎 OR ∆𝐆 𝟎 = 𝐑𝐓 𝟐 𝐝(𝐥𝐧𝐊 𝐩) 𝐝𝐓 + 𝐓 𝐝(∆𝐆 𝟎 ) 𝐝𝐓 … . . (𝐢𝐢𝐢) According to Gibb’s Helmholtz equation, ∆𝐆 𝟎 = ∆𝐇 𝟎 + 𝐓 𝐝(∆𝐆 𝟎 ) 𝐝𝐓 … … (𝐢𝐯) From (iii) and (iv), we get 𝐑𝐓 𝟐 𝐝 𝐥𝐧𝐊 𝐩 𝐝𝐓 = ∆𝐇 𝟎1-September -20 Dr. Yogita Sahebrao Thakare
- 8. OR 𝐝 𝐥𝐧𝐊 𝐩 𝐝𝐓 = ∆𝐇 𝟎 𝐑𝐓 𝟐 . . . (𝐯) Equation (v) is known as Van’t Hoff equation. It has been experimentally observed that the enthalpy change ∆𝐇 𝟎 accompanying a chemical reaction does not change appreciably with the change in partial pressure of reactant or products. Thus we may replaced ∆𝐇 𝟎 𝐛𝐲 ∆𝐇 OR Van’t Hoff equation can be written as 𝐝 𝐥𝐧𝐊 𝐩 𝐝𝐓 = ∆𝐇 𝐑𝐓 𝟐 … . . (𝐯𝐢) 1-September -20 Dr. Yogita Sahebrao Thakare
- 9. INTEGRATED FORM OF VAN’T HOFF EQUATION Equation (vi) may be written as 𝐝 𝐥𝐧𝐊 𝐩 = ∆𝐇 𝐑𝐓 𝟐 𝐝𝐓 … (𝐯𝐢𝐢) Let 𝐊 𝐩 𝟏 is the equilibrium constant at T1K and 𝐊 𝐩 𝟐 is the equilibrium constant at T2K. Further, let us assume that ∆𝐇 remains constant between the temperature T1 and T2 K. On integrating equation (vii), we get 𝐝 𝐥𝐧𝐊 𝐩 𝐊 𝐩 𝟐 𝐊 𝐩 𝟏 = ∆𝐇 𝐑 𝟏 𝐓 𝟐 𝐝𝐓 𝐓 𝟐 𝐓 𝟏 𝐥𝐧𝐊 𝐩 𝟏 𝐊 𝐩 𝟏 𝐊 𝐩 𝟐 = ∆𝐇 𝐑 − 𝟏 𝐓 𝐓 𝟏 𝐓 𝟐 𝐥𝐧 𝐊 𝐩 𝟐 − 𝐥𝐧𝐊 𝐩 𝟏 = ∆𝐇 𝐑 − 𝟏 𝐓 𝟐 − (− 𝟏 𝐓 𝟏 ) 𝐥𝐧 𝐊 𝐩 𝟐 𝐊 𝐩 𝟏 = ∆𝐇 𝐑 𝟏 𝐓 𝟏 − 𝟏 𝐓 𝟐 𝐥𝐧 𝐊 𝐩 𝟐 𝐊 𝐩 𝟏 = ∆𝐇 𝐑 𝐓 𝟐 − 𝐓 𝟏 𝐓 𝟏 𝐓 𝟐 OR 𝐥𝐨𝐠 𝐊 𝐩 𝟐 𝐊 𝐩 𝟏 = ∆𝐇 𝟐. 𝟑𝟎𝟑𝐑 𝐓 𝟐 − 𝐓 𝟏 𝐓 𝟏 𝐓 𝟐 … (𝐯𝐢𝐢𝐢) Above equation (𝐯𝐢𝐢𝐢) is the integrated form of Van’t Hoff equation. 1-September -20 Dr. Yogita Sahebrao Thakare
- 10. Applications of Van’t Hoff equation I) Calculation of equilibrium constant (𝐊 𝐩) at any desired temperature from the knowledge of equilibrium constant at a known temperature and heat of reaction (∆𝐇). II) Calculations of enthalpy change (∆𝐇) accompanying the chemical reaction from the knowledge of equilibrium constant at two different temperature. 1-September -20 Dr. Yogita Sahebrao Thakare
- 11. Important Formulae 1. ∆𝐺 = ∆𝐻 + 𝑇 𝑑 ∆𝐺 𝑑𝑇 𝑃 2. ∆𝐴 = ∆𝐸 + 𝑇 𝑑 ∆𝐴 𝑑𝑇 𝑉 3. G0 = - RT In KP OR G = -2.303 RT log KP 4. log Kp2 Kp1 = ∆H 2.303R T2−T1 T1T2 5. R = Gas constant= 8.314 J K-1 mol-1 6. T = Absolute temperature (00C =273K) 1-September -20 Dr. Yogita Sahebrao Thakare