Thermochemical Equations 热化学方程式 1.Enthalpy:an extensive property: 广度性质 Given:H2O(s)→H20(0 △H=6.01kJ stoichiometric coefficients 计量系数 2H20(s)→2H20(0△H=？ △H=2×6.01 CH4(g)+2O2(g)=C02(g+2H2O(g)△H=-802k 2CH4(g)+402(g)=2C02(g)+4H2O(g)△H=-1604kJ
H2O (s) H2O (l) ∆H = 6.01 kJ Thermochemical Equations 2H2O (s) 2H2O (l) 1. Enthalpy： an extensive property： ∆H = ? Given: stoichiometric coefficients 热化学方程式 计量系数 广度性质 2H2O (s) 2H2O (l) ∆H = ? ∆H = 2 ×6.01 CH (g)+2O (g) = CO (g)+2H O(g) H = -802 kJ 4 2 2 2 ∆ CH (g)+ O (g) = CO (g) 4 2 2 2 2 4 2 + H O(g) H 4 16 ∆ = - J 04 k
Thermochemical Equations 热化学方程式 2.Change in enthalpy (AH)depends on physical state: State function H2OI)→HO(g) △H=44.0kJ HO(g)→H2O() △H=? △H=-44.0kJ 2H,O(g)→2H,O() △H=-2×44.0kJ Ifyou reverse a reaction,the sign of AH changes
2. Change in enthalpy ( 2. Change in enthalpy (∆H) depends on ) depends on physical state: Thermochemical Equations 热化学方程式 2 2 H O l H O g ( ) ( ) H=44.0 kJ → ∆ State function If you reverse a reaction, the sign of ∆H changes 2 2 H O g H O l ( ) ( ) H= → ∆ ? ∆H= 4- 4.0 kJ 2 2 2 2 - 2 H O g H O l ( ) ( ) H= → ∆ × 44.0 kJ
Exercise 2:Given the changes of enthalpy of reaction (1)&(2) Calculate how much heat is involyed in the 3rd reaction? (1)2H2(g)+O2(8)=2H0(g)△H=-483.9kJ/mo1 (2)2HO(g)→2H,O()△H=-88kJ1mol (3)2H2(g)+O2(g)→2HOI)△H=?kJ1mol (1) 2H2(g)+O2(g)=2HO(g) +(2) 2H,O(g)=2H,O() Hess's Law (3) 2H2(g)+O2(g)=2HO() △H3=△H1+△H2=-438.9kJ-88kJ=-572kJ AH of overall reaction is the sum of the AH for each individual step. △H=△H1+△H2+△H
2 2 (2) 2 ( ) 2 ( H O H O H kJ mo g l → ∆ = ) 88 / − l 2 2 2 (1) 2 ( ) ( ) = 2H O(g) H 483.9 / H O mol g + g ∆ = − kJ 2 2 2 (1) 2 ( ) ( ) = 2H O( ) H O g g + g = 2 2 2 (3) 2 ( ) ( ) 2 ( ) / H O H g + = g → O mo l ? ∆H kJ l Exercise 2: Given the changes of enthalpy of reaction (1) & (2). Calculate how much heat is involved in the 3rd reaction? 2 2 2 2 2 +(2) 2 ( ) 2 ( ) (3) 2 ( ) ( ) = 2 ( ) H O H O H O H O g l g g l = + 3 1 2 ∆H = ∆ + ∆ H H = − = − 438.9 88 572 kJ kJ − kJ ∆H of overall reaction is the sum of the ∆H for each individual step. ∆ ∆ ∆ ∆ H = H + H + H 1 2 3 Hess’s Law
Exercise 3:Given the changes of enthalpy of reaction (1)&(2). Calculate how much heat is involyed in the 3rd reaction? N2g+202g→2N02g H1=66.4kJ 2N0(g+02g)→2N028 H2=-114.2kJ N2(g+O2g→2N0(g H3=？ N2(g)+22g)→2e2(g △H=66.4kJ +2O,(g)→2NO(g)+Dg) △H,=+114.2kJ N2(g)+O2(g)→2NO(g) △H,=(66.4+114.2)kJ Hss'sLaw:△H=△H1+△H2+△H
2 2 2 N (g) + 2O (g) 2NO (g) → → N2(g) + 2O 2(g) → 2NO 2(g) ∆H1 = 66.4 kJ 2 NO (g) + O 2(g) → 2NO 2(g) ∆H2 = - 114.2 kJ N2(g) + O 2(g) → 2NO(g) ∆H3 = ? 1 H = 66.4 kJ ∆ ∆ Exercise 3: Given the changes of enthalpy of reaction (1) & (2). Calculate how much heat is involved in the 3rd reaction? 2 2 + 2NO (g) 2 NO 2 2 N (g) + O (g) 2 (g) + O (g) N ) O(g → → ' 2 3 +114.2 kJ (66.4+114.2)kJ H = H = ∆ ∆ 1 2 3 Hess s Law ' : H = H + H + H ∆ ∆ ∆ ∆
Exercise 4:Given the changes of enthalpy of reaction (1),(2)&(3) Calculate how much heat is involved in the 4th reaction? N0g+O3g→NO2g+02g 4H1=-198.9kJ 203(8→3028g H2=-284.6kJ 028g→208g) H3=495.0kJ NOg+O(g→N02,g H=? NO(g)+(g)NO,(g)+e.(g) △H=-198.9kJ ,(g)→g) △H2=-(-284.6)/2kJ 0g)→6，(g) AH,=-(495.0)/2kJ N0(g)+O(g)→NO2(g)△H4=-198.9kJ-(-284.6)/2kJ-(495.0)/2kJ =-304.1kJ Hess'sLaw:△H=△H1+△H2+AH3
NO(g) + O (g) NO (g) + O (g) H = -198.9 kJ → ∆ NO(g) + O 3(g) → NO 2(g) + O2(g) ∆H1 = -198.9 kJ 198.9 kJ 2 O3(g) → 3 O 2(g) ∆H2 = -284.6 kJ 284.6 kJ O 2(g) → 2O(g) ∆H3 = 495.0 kJ NO(g) + O(g) → NO 2(g) ∆H4= ? Exercise 4: Given the changes of enthalpy of reaction (1), (2) & (3) Calculate how much heat is involved in the 4th reaction? 3 ' 3 2 2 1 2 ' 2 3 2 NO(g) + O (g) NO (g) + O (g) H = -198.9 kJ O (g) O (g) = (-284.6) kJ O(g) O (g) = (495.0) kJ 3 H - 2 2 1 H - 2 2 → ∆ → → ∆ ∆ NO(g) + O(g) NO (g) H = -198.9 kJ (-284. 2 4 6) kJ (495.0) kJ = - 2 - 2 -304.1kJ → ∆ 1 2 3 Hess s Law ' : H = H + H + H ∆ ∆ ∆ ∆