2. Standard Heats of Reaction The value of  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)

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

4. 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  o f. For example the equation for the formation of 1.0 mole of sulfuric acid is:

5. H 2 SO 4 H2H2 +1/8S 8 +2O 2 H 2 CO 3 H2H2 +C+3/2O 2 Write a heat of formation equation for perchloric acid. 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.

6. HClO 4 1/2H 2 + 1/2Cl 2 +2O 2 Now using the heat of formation equations for CO 2 (g), H 2 O (l) and H 2 CO 3 (aq) find the heat of reaction for : H 2 O (l) + CO 2 (g) => H 2 CO 3 (aq) Use the heat of formation tables in your textbook.

7. 1.C(s) + O 2 (g)  CO 2 (g) + 394 kJ 2.H 2 (g) + ½ O 2  H 2 O(l) + 286 kJ 3.H 2 + C + 3/2 O 2  H 2 CO 3 + 699.65 kJ H 2 O (l) + CO 2 (g) => H 2 CO 3 (aq) Since H 2 O (l) is on the left flip equation 2. Since CO 2 (g) is on the left flip equation 1. and since H 2 CO 3 (aq) is on the right leave equation 3 alone.

8. 1. CO 2 (g) + 394 kJ  C(s) + O 2 (g) 2. H 2 O(l) + 286 kJ  H 2 (g) + ½ O 2 3.H 2 + C + 3/2 O 2  H 2 CO 3 + 699.65 kJ H 2 O (l) + CO 2 (g) => H 2 CO 3 (aq)+ 20 kJ Is there a way to use the  H f o values to arrive at the  H r o for the reaction above.

9.  r o =  f o products -  f o reactants Use this equation to find the  r o for the combustion of methane (CH 4 ) and complete the thermochemical equation. First write the balanced chemical equation: Assume products are CO 2 (g) and H 2 O(l)

10. CH 4 (g) + O 2 (g) --> CO 2 (g) + H 2 O (l) 2 Now look up all the of  f o and apply the formula  r o =  f o products -  f o reactants  r o = [( -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 2

11. CH 4 (g) +O 2 (g) -->CO 2 (g) +2H 2 O (l) +891 kJ Now apply the same technique, using the tables to complete the thermochemical equations for the complete combustions of: Ethene - C 2 H 4 (g) Ethane -C 2 H 6 (g) Butane - C 4 H 10 (g) Assume all reactions produce CO 2 (g) and H 2 O (l) as products

12. C 2 H 4 (g) +3O 2 (g) -->2CO 2 (g) +2H 2 O (l) Now look up all the of  f o and apply the formula  r o =  f o products -  f o reactants  r o = [( -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 + 1411.1 kJ

13. C 2 H 6 (g) +7O 2 (g) -->2CO 2 (g) +3H 2 O (l) Now look up all the of  f o and apply the formula  r o =  f o products -  f o reactants  r o = [( -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 + 2 1560.6 kJ

14. C 4 H 10 (g) +13O 2 (g) ->4CO 2 (g) +5H 2 O (l) Now look up all the of  f o and apply the formula  r o =  f o products -  f o reactants  r o = [( -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 + 2877.4 kJ 2