( n ( RT ) ) = pV - PowerPoint PPT Presentation

∆ ( ∆ ∆ ∆ ∆ ∆ ∆ ∆ n ( RT ) ∆ ∆ ∆ ∆ ) = ∆ ∆ ∆ ∆ pV = = nRT nRT ( pV pV ) = n ( RT ) pV → → 2NO → → ∆ ∆ ∆ n = 0 ∆ ∆ ( ∆ ∆ ∆ 2 → → → → ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ N 2 + O 2 2NO n = 0 ( pV pV ) = 0 ) = 0 N 2 + O ∆ ∆ H= ∆ E ∆ T, ∆ ∆ ∆ ∆ H= ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ at const const T, E at → 2 NH → → → ∆ ∆ ∆ ∆ n = ∆ ∆ ∆ ∆ ( 2 → → → → ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ N 2 + 3 H 2 2 NH 3 n = - -2 2 (pV pV) = ) = - -2RT 2RT N 2 + 3 H 3 ∆ ∆ nRT ∆ ∆ ∆ ∆ ∆ ∆ nRT = = - -2 ( 8.314 ) ( 298 ) = 2 ( 8.314 ) ( 298 ) = - -4955.14 joules. 4955.14 joules. ∆ ∆ H = ∆ ∆ ∆ ∆ ∆ E + ∆ ∆ ∆ nRT ∆ ∆ ∆ ∆ ∆ ∆ H = ∆ ∆ ∆ ∆ E + ∆ ∆ ∆ ∆ nRT for reactions of ideal gases for reactions of ideal gases o ( NH ∆ ∆ ∆ ∆ H ∆ ∆ H ∆ ∆ ∆ ∆ ∆ ∆ = 2 ∆ ∆ ∆ ∆ H = 2 H f ( NH 3 ) = 2 ( - -46,110 ) = 46,110 ) = - -92,220 joules 92,220 joules fo 3 ) = 2 ( ∆ ∆ ∆ ∆ H = ∆ ∆ ∆ ∆ H = - -92,220 joules done at 92,220 joules done at const const T=298 K T=298 K ( i.e. initial and final state T’ ’s are same ) s are same ) ( i.e. initial and final state T ∆ ∆ E ∆ ∆ ∆ ∆ H ∆ ∆ - ∆ ∆ ∆ ∆ ∆ nRT ∆ ∆ ∆ ∆ = ∆ ∆ ∆ ∆ ∆ ∆ ∆ E = H - nRT = = - -92,220 92,220 - -( (- -4955) = 4955) = - -87,265 joules 87,265 joules

Example : Convert 1 mole of liquid H 2 O at 100 o o C into C into Example : Convert 1 mole of liquid H 2 O at 100 1 mole of H 2 O vapor at 100 o o C and p = 1 C and p = 1 atm atm. . 1 mole of H 2 O vapor at 100 → → → → H O ( l ) → → → → H 2 H 2 O ( vap vap ) ) H 2 O ( l ) 2 O ( Find q q p = +44,013.6 joules/mole ( heat absorbed ) Find p = +44,013.6 joules/mole ( heat absorbed ) ∆ ∆ H = ∆ ∆ ∆ ∆ ∆ ∆ H = q q p for process done at const const p p p for process done at ∴ ∴ ∴ ∴ ∆ ∆ ∆ H = ∆ ∆ ∴ ∴ ∴ ∴ ∆ ∆ ∆ H = 44,013.6 joules/mole 44,013.6 joules/mole ∆ ∆ E? What is ∆ ∆ ∆ ∆ ∆ ∆ E? What is ∆ ∆ ∆ ∆ H = ∆ E + ∆ ∆ ∆ ∆ ∆ ∆ ∆ ( ∆ ∆ E + p ∆ ∆ ∆ ∆ V ( ∆ ∆ ∆ ∆ ∆ ∆ H = ∆ ∆ ∆ ∆ E + ∆ ∆ ∆ ∆ ) = ∆ ∆ ∆ ∆ E + p ∆ ∆ ∆ ∆ ( pV pV ) = V (const const p) p) ∆ E = ∆ ∆ ∆ ∆ ∆ ∆ ∆ H ∆ ∆ V ∆ ∆ ∆ ∆ ∆ ∆ E = ∆ ∆ ∆ ∆ p ∆ ∆ ∆ ∆ H - - p V 1 mole H 2 O = 18 gm; volume 1 gm of H 2 O at 100 o o C is C is 1 mole H 2 O = 18 gm; volume 1 gm of H 2 O at 100 1.04 mL 1.04 mL/gm /gm v i = 18 ( 1.04 ) = 18.72 mL mL v i = 18 ( 1.04 ) = 18.72 v f = nRT nRT / p assume ideal gas / p assume ideal gas v f =

v f = v f = (1 mole)(82 mL mL- -atm atm/mole /mole- -deg deg)(373 )(373 deg deg)/1 )/1 atm atm (1 mole)(82 v f = ( 82 )( 373 ) mL mL = 30,586 = 30,586 mL mL v f = ( 82 )( 373 ) ∆ ∆ ∆ v = 30,586 ∆ ∆ ∆ ∆ ∆ v = 30,586 - - 19 = 30,567 19 = 30,567 mL mL = 30.57 liters = 30.57 liters → 101.4 joules/L → → → = 8.314 joules → → → → .082 L- -atm atm = 8.314 joules 101.4 joules/L- -atm atm .082 L p ∆ ∆ ∆ ∆ V = 30.57 L ∆ ∆ /mole = 30.57 × × 101.4 joules/ L × × × × ∆ ∆ × × V = 30.57 L- -atm atm/mole = 30.57 101.4 joules/ L- -atm atm = = p 3099.5 joules/mole 3099.5 joules/mole ∆ E = ∆ ∆ ∆ ∆ ∆ ∆ H ∆ ∆ ∆ ∆ ∆ V ∆ ∆ ∆ ∆ E = ∆ ∆ ∆ ∆ p ∆ ∆ ∆ ∆ H - - p V ∆ ∆ ∆ ∆ E = 44013.6 joules/mole ∆ ∆ ∆ ∆ E = 44013.6 joules/mole - - 3099.5 joules/mole 3099.5 joules/mole ∆ E = 40914.1 joules/mole ∆ ∆ ∆ ∆ ∆ ∆ H = 44013.6 joules/mole ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ E = 40914.1 joules/mole H = 44013.6 joules/mole

General Definitions of heat capacities (for 1 mole of gas): General Definitions of heat capacities (for 1 mole of gas): c p = dQ   (sub P means take derivative at constant P) (sub P means take derivative at constant P)   dT P c V = dQ   (sub V means take derivative at constant V) (sub V means take derivative at constant V)   V dT Substitute dQ dQ = = dE dE + + pdV pdV= = dE dE - - dw dw (1st Law) (1st Law) Substitute V = RT E = 3 2RT Remember (1 mole gas) Remember (1 mole gas) P d 3 d RT         2RT     c P = dE + P dV         P + P = =         dT dT dT dT P P     P P 2 R + PR 3 dT   = (3/2)R + R = (5/2)R = (5/2)R = (3/2)R + R   P dT P

Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

Thermochemi s t ry : Thermochemi s t ry : ∆ ∆ ∆ ∆ H directly by making a laboratory determination Can measure ∆ ∆ ∆ ∆ H directly by making a laboratory determination Can measure of q q p . Want to set up a table of enthalpy changes for chemical of p . Want to set up a table of enthalpy changes for chemical reactions reactions ° for a system ≡ ≡ ∆ ≡ ≡ ∆ ∆ ∆ ∆ H Standard Enthalpy Change ≡ ≡ ≡ ≡ ∆ ∆ ∆ ° ° ° H ° ° ° ° for a system Standard Enthalpy Change ° ≡ ∆ ∆ ∆ ∆ H ≡ enthalpy change in a chemical reaction for converting ≡ ≡ ≡ ∆ ∆ ∆ ∆ ° ≡ ≡ ≡ H ° ° ° ° ° ° enthalpy change in a chemical reaction for converting reactants in their standard state to products in their standard state. state. reactants in their standard state to products in their standard Standard State of a Substance ≡ ≡ ≡ That form of the substance which ≡ ≡ ≡ ≡ ≡ That form of the substance which Standard State of a Substance ° K and T = 298� ° ° ° ° ° ° ° is most stable at a pressure of 1 atm atm and T = 298� K is most stable at a pressure of 1

Exampl e s Exampl e s ° C, Stand. St. is graphite (not diamond or coal) ° ° ° , 25 ° ° ° ° Carbon : at 1 atm atm, 25 C, Stand. St. is graphite (not diamond or coal) Carbon : at 1 ° C ° ° ° , 25 ° ° ° ° Oxygen : 1 atm atm, 25 C Std. St. is O 2 gas Oxygen : 1 Std. St. is O 2 gas ° C ° ° ° , 25 ° ° ° ° Bromine : 1 atm atm, 25 C Std. St. is liquid Bromine Bromine : 1 Std. St. is liquid Bromine For a reaction: For a reaction: → → CO → → ( g ) → → → → C ( graphite ) + O 2 CO 2 ( g ) measure q q p in lab C ( graphite ) + O 2 ( g ) 2 ( g ) measure p in lab (Heat released when (Heat released when ° = ∆ ∆ ∆ ∆ H ∆ ∆ ∆ ∆ ° ° ° 298 ° ° ° ° H 298 = - - 393.52 393.52 kjoules kjoules/mole = /mole = q q p p burn graphite at const const p) p) burn graphite at This means evolve 393.52 393.52 kjoules kjoules of heat in converting 1 mole of heat in converting 1 mole This means evolve ° K. ° and 298 ° ° ° ° ° ° C graphite, 1 mole O 2 gas, into 1 mole CO 2 gas at 1 atm atm and 298 K. C graphite, 1 mole O 2 gas, into 1 mole CO 2 gas at 1 Can also burn CO in lab to produce CO 2 : Can also burn CO in lab to produce CO 2 : ( g ) → → → → → CO → → → CO ( g ) + 1/2 O 2 CO 2 ( g ) measure q q p in lab. CO ( g ) + 1/2 O 2 ( g ) 2 ( g ) measure p in lab. (Heat released when burn CO at (Heat released when burn CO at ° = ∆ H ∆ ∆ ∆ ∆ ∆ ∆ ∆ 298 ° ° ° ° ° ° ° H 298 = - - 282.98 282.98 kjoules kjoules/mole /mole const p) p) const

→ CO → ( g ) → → → → → → CO ( g ) + 1/2 O 2 CO 2 ( g ) Can measure Can measure q q p in lab. CO ( g ) + 1/2 O 2 ( g ) 2 ( g ) p in lab. Just burn CO in oxygen. Just burn CO in oxygen. → CO → ( g ) → → → → → → C ( graphite ) + O 2 CO 2 ( g ) C ( graphite ) + O 2 ( g ) 2 ( g ) Can measure q q p in lab. Just burn C in oxygen. Can measure p in lab. Just burn C in oxygen. C (graphite) + O 2 (g) C (graphite) + O 2 (g) → CO (g) + (1/2) O → → → → → → → CO (g) + (1/2) O 2 (g) 2 (g) Can’t measure in the lab. Can’t measure in the lab. QuickTime™ and a Video decompressor (Always get a little CO 2 ) (Always get a little CO 2 ) are needed to see this picture.