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Standard Heats of Reaction

Standard Heats of Reaction. The value of D H for a reaction depends on the temperature and pressure so scientists have agreed upon a set of reference conditions for the comparison of enthalpy data. These conditions are 25 o C and 100. kPa. (SATP).

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Standard Heats of Reaction

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  1. Standard Heats of Reaction The value of DH for a reaction depends on the temperature and pressure so scientists have agreed upon a set of reference conditions for the comparison of enthalpy data. These conditions are 25oC and 100. kPa. (SATP)

  2. Whenever standard conditions are used the symbol for enthalpy change becomes DHo. The degree sign means “under standard conditions”. The units used are kJ/mol or kcal/mol where 4.18 J = 1 cal.

  3. Standard Heats of Formation When one mole of a compound is produced from its elements in their most stable state the quantity of energy involved is called the heat of formation and is given the symbol DHof. For example the equation for the formation of 1.0 mole of sulfuric acid is:

  4. H2 + 1/8S8 + 2O2 H2SO4 H2 + C + 3/2O2 H2CO3 Write a heat of formation equation for carbonic acid. Please note the standard heat of formation of all elements in their standard elemental states is zero. Write a heat of formation equation for perchloric acid.

  5. + + 2O2 1/2Cl2 1/2H2 HClO4 Now using the heat of formation equations for CO2(g),H2O(l) and H2CO3(aq) find the heat of reaction for : H2O(l) + CO2(g) => H2CO3(aq) Use the heat of formation tables in your textbook.

  6. H2O(l) + CO2(g) => H2CO3(aq) 1. C(s) + O2 (g)  CO2(g) + 394 kJ 2. H2(g) + ½ O2  H2O(l) + 286 kJ 3. H2 + C + 3/2 O2  H2CO3 + 699.65 kJ Since H2O(l)is on the left flip equation 2. Since CO2(g) is on the left flip equation 1. and since H2CO3(aq) is on the right leave equation 3 alone.

  7. 1. CO2(g) + 394 kJ C(s) + O2 (g) 2. H2O(l) + 286 kJ  H2(g) + ½ O2 3. H2 + C + 3/2 O2  H2CO3 + 699.65 kJ H2O(l) + CO2(g) => H2CO3(aq)+ 20 kJ Is there a way to use the DHfo values to arrive at the DHro for the reaction above.

  8. DHro = SDHfoproducts-SDHforeactants Use this equation to find the DHro for the combustion of methane (CH4) and complete the thermochemical equation. First write the balanced chemical equation: Assume products are CO2(g) and H2O(l)

  9. CH4(g) + O2(g) --> CO2 (g) + H2O (l) 2 2 Now look up all the of DHfo and apply the formula DHro = SDHfoproducts-SDHforeactants DHro = [(-394 kJ/mol x 1 mol) + (-286 kJ/mol x 2 mol)]- [(-74.9 kJ/mol x 1 mol) + ( 0 kJ/mol x 2 mol) ] = - 966 kJ - (-74.9 kJ) = -891 kJ

  10. CH4(g) +O2(g) -->CO2 (g) +2H2O(l)+891 kJ Now apply the same technique, using the tables to complete the thermochemical equations for the complete combustions of: Ethene - C2H4(g) Ethane -C2H6 (g) Butane - C4H10(g) Assume all reactions produce CO2(g) and H2O(l) as products

  11. C2H4(g)+3O2(g) -->2CO2(g) +2H2O(l) + 1411.1 kJ Now look up all the of DHfo and apply the formula DHro = SDHfoproducts-SDHforeactants DHro = [(-393.5 kJ/mol x 2mol) + (-285.8 kJ/mol x 2 mol)]- [(52.5 kJ/mol x 1 mol) + ( 0 kJ/mol x 3mol) ] = - 1358.6 kJ - (52.5 kJ) = - 1411.1 kJ

  12. C2H6(g)+7O2(g) -->2CO2(g) +3H2O(l) 1560.6 kJ + 2 Now look up all the of DHfo and apply the formula DHro = SDHfoproducts-SDHforeactants DHro = [(-394 kJ/mol x 2mol) + (-286kJ/mol x 3 mol)]- [(-83.8 kJ/mol x 1 mol) + ( 0 kJ/mol x 3.5mol) ] = - 1560.6 kJ

  13. C4H10(g)+13O2(g) ->4CO2(g) +5H2O(l) + 2877.4 kJ 2 Now look up all the of DHfo and apply the formula DHro = SDHfoproducts-SDHforeactants DHro = [(-393.5 kJ/mol x 4mol) + (-285.8kJ/mol x 5 mol)]- [(-125.6 kJ/mol x 1 mol) + ( 0 kJ/mol x 6.5mol) ] = - 2877.4 kJ

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