Physical chemistr Chapter 5 Chapter 5 Standard Thermodynamic Functions of reaction
Chapter 5 Standard Thermodynamic Functions of Reaction Physical Chemistry Chapter 5
1 hysical Chemistry Chapter 5 aStandard Thermodynamic Functions of Reaction Standard states of pure substances Standard state p=l bar Solid or liquid: P=l bar, T Gas: P=l bar. T' gas ideal The standard-state pressure is denoted by P≡1bar.T The standard state of a substance at a specified temperature is its pure form at l bar
Standard States of Pure Substances Standard state P ＝ 1 bar Gas: P = 1 bar, T, gas ideal The standard-state pressure is denoted by P o : P o 1 bar, T Solid or liquid：P ＝ 1 bar, T The standard state of a substance at a specified temperature is its pure form at 1 bar. Standard Thermodynamic Functions of Reaction Physical Chemistry Chapter 5
Physical Chemistry Chapter 5 Standard Enthalpy of reaction For a chemical reaction aA+bB→cC+dD Standard enthalpy(change)of reaction AHT ( Sometime,△, ht is used for△H) AHT=CHm(C)+dHm(D)-aHm(A)-bHm7(B) Molar enthalpy of substance(C)in standard state at T Hm(C T For the general reaction(Eq (4.94) 0→>v;A en AHP=∑H (5.3)
Standard Enthalpy of Reaction For a chemical reaction Then → i 0 i Ai aA + bB → cC + dD Molar enthalpy of substance (C) in standard state at T ( ) ( ) ( ) ( ) H cH , C dH , D aH , A bH , B o m T o m T o m T o m T o T + − − Standard enthalpy (change) of reaction o HT (Sometime, is used for ) o HT o r HT For the general reaction (Eq. (4.94)) (5.3)* i i Hm,T,i o o HT ( ) H , C o m T Physical Chemistry Chapter 5
Physical Chemistry Chapter 5 Standard Enthalpy of reaction 一→ For example 2CH6)+1502⑧)→12C02(g)+6HO() AHT =12Hm7(C02,8)+HMT(H2O, 7)-2HMT(CH6, 1)-15Hm7(O2, g) AHo depends on how the reaction is written C6H6()+7.502(g)→6C02(g)+3H2O( △H=6Hmn7(CO2,g)+3Hm(H2O,1)-Hm7(C6H6,1)-7.5Hm(O2,g) Then △H0=2△H
Standard Enthalpy of Reaction For example Then Hodepends on how the reaction is written 12 ( , ) 6 ( , ) 2 ( , ) 15 ( , ) H H , CO2 g H , H2 O l H , C6 H6 l H , O2 g o m T o m T o m T o m T o T = + − − 2 ' o T o HT = H 2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l) C6H6 (l) + 7.5 O2 (g) → 6 CO2 (g) + 3 H2O (l) ' 6 ( , ) 3 ( , ) ( , ) 7.5 ( , ) H H , CO2 g H , H2 O l H , C6 H6 l H , O2 g o m T o m T o m T o m T o T = + − − Physical Chemistry Chapter 5
Physical Chemistry Chapter 5 Standard Enthalpy of formation The reference form(reference phase) of an element at T' is usually taken as the form of the element that is most stable at T and l-bar pressure For example, A H307. H, co(&) for H2 Co(g) at 307K C(graphite, 307 K, po )+H2(ideal gas, 307 K, Po)+1202 (ideal gas, 307K, Po)>H,CO(ideal gas, 307K, Po) For an element in its reference form,A ho is zero For example, A H307graphite =0 C(graphite, 307K, Po)>c(graphite, 307K, Po) Nothing happens in this process however, boz. diamond to(a h3ozdiamond=1.9k/mol) C( graphite, 307K, Po )>C(diamond, 307 K, Po
Standard Enthalpy of Formation The reference form (reference phase) of an element at T is usually taken as the form of the element that is most stable at T and 1-bar pressure. For example, for H2CO(g) at 307 K C (graphite, 307 K, Po )＋H2 (ideal gas, 307 K, Po )＋½O2 (ideal gas, 307 K, Po ) → H2CO (ideal gas, 307 K, Po ) o f H307,H CO(g) 2 For an element in its reference form, is zero. For example, ＝0。 o f HT o f H307,graphite C (graphite, 307 K, P o ) → C (diamond, 307 K, P o ) C (graphite, 307 K, P o ) → C (graphite, 307 K, P o ) Nothing happens in this “process” however, 0 ( ) 307, o f H diamond H kJ mol o f diamond 1.9 / 307, = Physical Chemistry Chapter 5