2. Hess’s Law &thermochemical equations can be combined algebraically to determine enthalpy changes for a different chemical reaction &Example 1 &C(s) + O 2 (g) ==> CO 2 (g) + 393 kJ &CO 2 (g) + 283 kJ ==> 1/2 O 2 (g) + CO(g) &-1/2 O 2 and CO 2 (g) produces this &C(s) + 1/2 O 2 (g) ==> CO(g) + 110kJ

3. Example 2 #1 2H 2 O(g) + 483.2 kJ ==> 2H 2 (g) + O 2 (g) #2 H 2 (g) + 1/2O 2 (g) ==> H 2 O(l) +285.5 kJ Given the above 2 equations complete the thermo- chemical equation for: #3 H 2 O(l) ==> H 2 O(g) Step 1 - recognize H 2 O(l) is a reactant (on the left) in equation 3. Notice it is found in equation 2 as a product so flip equation 2. H 2 O(l) +285.5 kJ ==> H 2 (g) + 1/2O 2 (g)flipped 2 Step 2 -recognize H 2 O(g) is a product in equation 3 and in equation 1 it is a reactant and there are 2H 2 O(g) so flip equation 1 and divide it by 2. H 2 (g) + 1/2 O 2 (g) ==> H 2 O(g) + 241.6kJ flipped 1, /2 H 2 O(l) + 43.9 kJ ==> H 2 O(g) final answer

4. Example 3 #1 S(s) + O 2 (g) ==> SO 2 (g)  H o  = - 295.8 kJ #2 S(s) + 3/2 O 2 (g) ==> SO 3 (g)  H o  = - 394.8 kJ Given the above 2 equations complete the thermo chemical equation for: #3SO 2 (g) +1/2 O 2 (g) =>SO 3 (g) Step 1 -recognize that SO 2 (g) is a reactant in equation 3 but a product in equation 1. Flip equation 1 and write the quantity of heat inside the equation. SO 2 (g) + 295.8 kJ ==> S(s) + O 2 (g) flipped 1 Step 2 -recognize that SO 3 (g) is a product in equation 3 so leave #2 alone but place the quantity of heat inside it. S(s) + 3/2 O 2 (g) ==> SO 3 (g) + 394.8 kJ unchanged Step 3 -subtract 295.8 kJ from each side. SO 2 (g) + 1/2 O 2 (g) ==>SO 3 (g) + 99 kJ final answer