Question

Calculate AH° for each of the following reactions, which occur in the atmosphere. (a) C2H4(g) + O3(g) → CH3CHO(g) + O2(g) СkJ
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Answer #1

Following is the - complete Answer -&- Explanation: for the given: Question: in...typed format...

\RightarrowAnswer:

  1. part - (a): \Delta Horxn. = - 366.0 kJ / mol
  2. part - (b):    \Delta Horxn. =  - 200.09 kJ / mol
  3. part - (c):    \Delta Horxn. =  - 132.5 kJ / mol
  4. part - (d): \Delta Horxn. =  - 114.18 kJ / mol

\RightarrowExplanation:

Following is the complete : Explanation: for the above: Answer...

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\RightarrowPart - (a):

  • Given:
  1. balanced chemical equation: C2H4 (g) + O3 (g)  \rightleftharpoons CH3CHO (g) + O2 (g) -------------Equation - 1
  • ​​​​​​​Step - 1:

​​​​​​​Now, we know the following standard enthalpy of formation: of the given: reactants and products:

  1. \DeltaHfo [ C2H4 (g) ] = + 52.3 kJ / mol
  2. \DeltaHfo [ O3 (g) ] = + 143.0 kJ / mol
  3. \DeltaHfo [ CH3CHO(g) ] = -170.7  kJ / mol
  4. \DeltaHfo [O2(g) ] = 0.0 kJ / mol
  • ​​​​​​​Step - 2:

​​​​​​​Therefore, according to Equation - 1: the value of the standard change of enthalpy: of Equation - 1, will be the following;

\Rightarrow\DeltaHorxn. = [ ( 1.0 mol ) x ( \Delta Hfo [ CH3CHO(g)) ] ) + (1.0 mol ) x (\DeltaHfo [O2(g) ] ) ] - [ ( \Delta Hfo[ C2H4 (g) ] + \Delta Hfo[ O3 (g) ) ]

\Rightarrow \Delta Horxn. = [ ( 1.0 mol ) x (-170.7  kJ / mol ) ] ) + (1.0 mol ) x (0.0 kJ/mol ) ] - [ ( 1.0 mol) x (52.3 kJ / mol ] + (1.0 mol ) x ( 143.0 kJ / mol ) ] = - 366.0 kJ / mol

\Rightarrow \Delta Horxn. = - 366.0 kJ / mol

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\RightarrowPart - (b):

  • Given:
  1. ​​​​​​​balanced chemical equation: O3 (g) + NO (g)  \rightleftharpoons NO2 (g) + O2 (g) -------------Equation - 2
  • ​​​​​​​Step - 1:

​​​​​​​Now, we know the following standard enthalpy of formation: of the given: reactants and products:

  1. \DeltaHfo [ NO2 (g) ] = + 33.2 kJ / mol
  2. \DeltaHfo [ O3 (g) ] = + 143.0 kJ / mol
  3. \DeltaHfo [NO(g) ] = +90.29 kJ / mol
  4. \DeltaHfo [O2(g) ] = 0.0 kJ / mol
  • ​​​​​​​Step - 2:

​​​​​​​Therefore, according to Equation - 1: the value of the standard change of enthalpy: of Equation - 1, will be the following;

\Rightarrow\DeltaHorxn. = [ ( 1.0 mol ) x ( \Delta Hfo [ NO2(g)) ] ) + (1.0 mol ) x (\DeltaHfo [O2(g) ] ) ] - [ ( \Delta Hfo[ O3 (g) ] + \Delta Hfo[ NO(g) ) ]

\Rightarrow \Delta Horxn. = [ ( 1.0 mol ) x (+ 33.2 kJ / mol ) ] ) + (1.0 mol ) x (0.0 kJ/mol ) ] - [ ( 1.0 mol) x (+ 143.0 kJ / mol   ] + (1.0 mol ) x ( +90.29 kJ / mol ) ] = - 200.09 kJ / mol

\Rightarrow \Delta Horxn. = - 200.09 kJ / mol

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\RightarrowPart - (c):

  • Given:
  1. ​​​​​​​balanced chemical equation: SO3 (g) + H2O (l)  \rightleftharpoons H2SO4 (aq) -------------Equation - 3
  • ​​​​​​​Step - 1:

​​​​​​​Now, we know the following standard enthalpy of formation: of the given: reactants and products:

  1. \DeltaHfo [H2SO4 (aq) ] = -814.0 kJ / mol
  2. \DeltaHfo [ SO3 (g) ] = −395.7 kJ / mol
  3. \DeltaHfo [H2O (l) ] = −285.8 kJ / mol
  • ​​​​​​​Step - 2:

​​​​​​​Therefore, according to Equation - 1: the value of the standard change of enthalpy: of Equation - 1, will be the following;

\Rightarrow\DeltaHorxn. = [ ( 1.0 mol ) x ( \Delta Hfo [H2SO4 (aq) ] ) ] - [ ( \Delta Hfo[ SO3 (g) ] + \Delta Hfo[ H2O (l) ) ]

\Rightarrow \Delta Horxn. = [ ( 1.0 mol ) x (-814.0 kJ / mol ) ] ) ] - [ ( 1.0 mol) x (−395.7 kJ / mol ) + (1.0 mol ) x (−285.8 kJ / mol ) ]

   = - 132.5 kJ / mol

\Rightarrow \Delta Horxn. = - 132.5 kJ / mol

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\RightarrowPart - (d):

  • Given:
  1. ​​​​​​​balanced chemical equation: 2 NO (g) + O2 (g)  \rightleftharpoons 2 NO2 (g) -------------Equation - 3
  • ​​​​​​​Step - 1:

​​​​​​​Now, we know the following standard enthalpy of formation: of the given: reactants and products:

  1. \DeltaHfo [ NO2 (g) ] = + 33.2 kJ / mol
  2. \DeltaHfo [NO(g) ] = +90.29 kJ / mol
  3. \DeltaHfo [O2(g) ] = 0.0 kJ / mol
  • ​​​​​​​Step - 2:

​​​​​​​Therefore, according to Equation - 1: the value of the standard change of enthalpy: of Equation - 1, will be the following;

\Rightarrow\DeltaHorxn. = [ ( 2.0 mol ) x ( \Delta Hfo [NO2 (g) ] ) ] - [ ( 2.0 mol ) x ( \Delta Hfo[ NO(g) ] + ( 1.0 mol ) x (\DeltaHfo[ O2 (g)) ]

\Rightarrow \Delta Horxn. = [ ( 2.0 mol ) x (+ 33.2 kJ / mol ) ] ) ] - [ (2.0 mol) x (+90.29 kJ / mol ) + (1.0 mol ) x (0.0 kJ/mol) ]

   = - 114.18 kJ / mol

\Rightarrow \Delta Horxn. = - 114.18 kJ / mol

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