H sum h1 h 2 h n ch 4 g2o 2 g cog2h 2 og

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Unformatted text preview: and 1 atm: ∆H °f of CO2 = - 393.5 kJ / mol p.1112 • Standard enthalpy of formation of the most stable form of an element is zero. Enthalpies Reaction Enthalpies of Reaction ∆Horxn The standard enthalpy of reaction (∆Horxn ): the enthalpy of a reaction reaction carried out at 1 atm. atm aA + bB cC + dD → ∆H o = [c ⋅ ∆H o (C ) + d ⋅ ∆H o ( D )] − [a ⋅ ∆H o ( A) + b ⋅ ∆H o ( B)] rxn f f f f ∆H o = ∑ n∆H o (products) − ∑ m∆H o (reactants) rxn f f Hess’s Hess’s law: if a reaction is carried out in in a number of steps, ∆H for the overall overall reaction is the sum of ∆H for each sum individual step. ∆H sum = ∆H1 + ∆H 2 +...+∆H n CH 4 (g)+2O 2 (g) → CO(g)+2H 2 O(g) +1 2O 2 (g) ∆H = -607 kJ CO(g)+2H 2 O(g) +1 2O 2 (g) → CO 2 (g)+2H 2 O(l) ∆H = -283 kJ ______________________________________________ CH 4 (g)+2O 2 (g) → CO 2 (g)+2H 2 O(l) ∆H = -890 kJ Question: liquid Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of liquid benzene combusted? The standard enthalpy of formation of liquid benzene is 49.04 kJ/mol. C6 H 6 (l ) + O2 ( g ) → CO2 ( g ) + H 2O(l ) C6 H 6 (l ) + 15 O2 ( g ) 6CO2 ( g ) + 3H 2O(l ) 2 ∆H o = ∑ n∆H o (products) − ∑ m∆H o (reactants) rxn f f ° ∆H rxn = [6∆H °f (CO2 ) + 3∆H °f ( H 2O)] − [∆H °f (C6 H 6 )] ° ∆H rxn = [6 × (−393.5) + 3 × (−285.8)] − [(49.04)] = −3267.8 KJ C6 H 6 (l ) + 15 / 2O2 ( g ) 6CO2 ( g ) +...
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