{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

201-chapter_6_examples

201-chapter_6_examples - 2 g CS 2 l 3O 2 g ∆ H = 1072...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Ex: How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 ( s ) + 5O 2 ( g ) → P 4 O 10 ( s ) H = -3013 kJ/mol 266 g P 4 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ Ex: 2N 2 H 4 +N 2 O 4 → 3N 2 + 4H 2 O ∆H = -1049 KJ when two moles of N 2 H 4 and one mole of N 2 O 4 react. How much heat is evolved when 1.00 kg of N 2 O 4 is used? Moles of N 2 O 4 = 1000 g/(92.02 g/mol) = 10.9 moles q = 10.9 moles N 2 O 4 × (-1049 kJ/mol N 2 O 4 ) = -1.14 × 10 4 kJ 1.14 E4 kJ of heat was evolved (given off)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Ex: Calculate the standard enthalpy of formation of CS 2 ( l ) given that: C (graphite) + O 2 ( g ) CO 2 ( g ) H 0 = -393.5 kJ/mol rxn S (rhombic) + O 2 ( g ) SO 2 ( g ) H 0 = -296.1 kJ/mol rxn CS 2 ( l ) + 3O 2 ( g ) CO 2 ( g ) + 2SO 2 ( g ) H 0 = -1072 kJ/mol rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 ( l ) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 ( g ) CO 2 ( g ) H 0 = -393.5 kJ/mol 2S (rhombic) + 2O 2 ( g ) 2SO 2 ( g ) H 0 = -296.1 kJ/mol x 2 rxn CO 2 ( g ) + 2SO
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 ( g ) CS 2 ( l ) + 3O 2 ( g ) ∆ H = +1072 kJ/mol rxn + C (graphite) + 2S (rhombic) CS 2 ( l ) ∆ H = -393.5 + (2x-296.1) + 1072 = 86.3 kJ/mol rxn Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 ( l ) + 15O 2 ( g ) 12CO 2 ( g ) + 6H 2 O ( l ) ∆ H rxn n ∆ H (products) f = Σ m ∆ H (reactants) f Σ-∆ H 6 ∆ H (H O 12 ∆ H (CO + 2 ∆ H (C H rxn (H 2 O) f 2 ) f = [ ] -(C 6 H 6 ) f [ ] ∆ H rxn = [12x(–393.5)+ 6x(–187.6)] – [ 2x49.04 ] = -5946 kJ-5946 kJ 2 mol = - 2973 kJ/mol C 6 H 6...
View Full Document

{[ snackBarMessage ]}

Page1 / 3

201-chapter_6_examples - 2 g CS 2 l 3O 2 g ∆ H = 1072...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online