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H-H H + H 430 kJ + Breaking bond always requires E Endothermic: - PDF document

Jan 22, 2023 •216 likes •260 views

H-H H + H 430 kJ + Breaking bond always requires E Endothermic: heat = R Topic 9.2 Heat of reaction, heat of formation, Table I E absorbed H + H H- H + 430 kJ Forming bonds always releases E Exothermic: heat = P E

  1. H-H � H + H 430 kJ + Breaking bond always requires E • Endothermic: heat = R Topic 9.2 Heat of reaction, heat of formation, Table I • E absorbed H + H � H- H + 430 kJ Forming bonds always releases E • Exothermic: heat = P • E released Types of Heats of Reaction ( Δ H) Δ H for combustion rxn Heat of formation = Δ H for 1 mole of comp’d formed from elements in standard conditions (25°C and 1 atm) H 2 (g) + H 2 (g) 1 2 O 2 (g) � H 2 O (g) H 2 O (g) Heats of formation Heats of solution = Δ H for the dissolution of a soluble comp’d NaCl ( s ) � N a NaCl ( s ) � N a NaCl ( s ) � (aq) (aq) + Cl Heats of solution for + + � (aq) ionic compounds Δ H for neutralization rxn Formation of liquid water 1) What is the heat of formation for 0.5 mole H 2 O (l) ? 2H 2 ( g ) + O 2( g ) � 2H 2 O ( l ) + 571.6 kJ From table: Δ H = -571.6 kJ 2 H 2 ( g ) + O 2( g ) � 2 H 2 O ( l ) l ) � H = � 571 . 6 kJ Divide Δ H by 4 Decomposition of liquid water 2H 2 O ( l ) � 2H 2 ( g ) + O 2( 571.6 kJ + 1 2H 2 ( g ) + 1 4 O 2( g ) � 1 g ) 2 H 2 O ( l ) l ) � H = � 142 . 9 kJ Δ H = +571.6 kJ What ever you do to coefficients do to Δ H If rxn reversed, change sign for Δ H 1

  2. 3) What is the heat of reaction for the combustion of 2) What is the heat of formation for 0.2 mole H 2 O (g) ? 8 g mole C 6 H 12 O 6 ? From table: From table: � H = � 2804 kJ C 6 H 12 O 6 + 6O 2( g ) � 6 CO 2( g ) + 6H 2 O ( l ) 2H 2 ( g ) + O 2(g) � 2H 2 O ( l ) � H = � 483.6 kJ g ) � 8 g 1 mole C 6 H � 12 O 6 � � � = 0.044 mole C 6 H 12 O 6 Divide Δ H 8 g � = � 180 g by 10 � � � l ) � H = � 48.3 kJ � H = (0.044)( )( � 2804) ) = � 123 kJ 5) In which reactions do the products have a lower 4) One mole of which of the following compounds energy content than the reactants? would react with oxygen to release the most energy? C 8 H 18( l ) C 6 H 12 O 6( s ) Exothermic C ( s ) a) Combustion of sugar 2C 6 H 12 O 6 + 6O 2 � 6CO 2 + 6H 2 O � H = - 2804 kJ H 2( g ) b) Formation of NO 2 2 C 8 H 18 + 25 O 2 � 16 CO 2 + 18 H 2 O O � H = -10,943 kJ N 2 + 2O 2 � 2NO 2 � H = + 66.4 kJ 2 C 6 H 12 O 6 + 6 O 2 � 6 CO 2 + 6 H 2 O O � H = - 2804 kJ c) Decomposition of liquid water 2H 2 O ( l ) � 2H 2 ( g ) + O 2( g ) � H = + 571.6 kJ C + O 2 � CO 2 � H = - 393 kJ 2 2 d) Dissolution of NaOH 2 H 2 + O 2 � 2 H 2 O O � H = - 483 kJ H 2 O NaOH (s)2 � Na + (aq) + OH � (aq) � H = -44.5 kJ � � � Using Δ H formation to calculate Δ H reaction: 7) Calculate the Δ H reaction for the combustion of ethyne 6) Calculate the Δ H reaction for the reaction of nitrogen monoxide with oxygen to yield nitrogen dioxide. balanced equation: balanced equation: H 2 O ( l ) C 2 H 2( g ) + 2.5 O 2( g ) � CO 2( g ) + 2 2 NO ( g ) + 2 NO + O 2( g ) � � 2 2 NO 2( g ) Find Δ H f values Find Δ H f values from table I: +182.6/2 kJ +66.4/2 kJ 2( ) 1/2( ) -571.6 kJ 227.4 kJ 0 kJ -393.5 kJ for 1 mole from table I: The Δ H f for any element in standard state = 0 kJ � H rxn = � H f prod � � H f react � H rxn = � H f prod . prod. � � H f react. for 2 mol = + 66.4 � 182.6 = � 116.2 kJ � H rxn = 2( � 393.5) ) + 1/ 2( � 571.6) ) � 1(227.4) ) = � 1300.2 kJ NO 2 for 1 mol 6 = � 58.1 kJ NO 2 2

  3. Calculate the Δ H reaction for the 8) combustion of C 2 H 4(g) 9) vaporization of liquid water 3

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