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Enathalpy of the reaction

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1.) Calculate delta H for the following reaction:
The reduction of aluminum oxide by hydrogen gas:
Al2O3(s) + 3H2(g) ------- 2Al(s) + 3H2O(g)
2.) Solve for entalapy using Hess's Law.
The standard heat of combustion of liquid ethyl alcohol is -227 kcal/mol and that of acetic acid is -209 kcal/mol. The equations are:
C2H5OH(l) + 3O2(g) ------- 2CO2(g) +H2O(l) delta H = -227 kcal
HC2H3O2(l) + 2O2(g) ------2CO2(g) + 2H2O(l) delta H = -209 kcal
What is the heat of reaction for the oxidation of ethyl alcohol to acetic acid?
C2H5OH(l) + O2(g) -------- HC2H3O2(l) + H2O(l)
3.) How much heat energy is given off when one kilomole of hydrogen gas (H2) at 25 degrees Celsius and 1 atm is combined with enough oxygen (O2) to make liquid water at 25 degrees Celsius?
H2(g) + 1/2 O2(g) ------ H2O(l) delta H = -68.320
4.) Calculate the heat of combustion for the reaction in which ethane combines with oxygen to give carbon dioxide and water vapor. C2H6 + 7/2 O2 ----- 2CO2 +3H2O
a.) 2 C(s) + 3H2(g) ---- C2H6(g) delta H = -20.2 Kcal.

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The solution provides several examples of calculating the enathalpy of the reactions.

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1. Calculate delta H for the following reaction:

The reduction of aluminum oxide by hydrogen gas:

Al2O3(s) + 3H2(g) ------- 2Al(s) + 3H2O(g)

Solution:

Step 1: First, we write out the reactions for the heat of formation for Al2O3 and H2O:

2Al(s) + 3/2 O2(g) ---> Al2O3(s) dHf = -1676 kJ/mol

H2(g) + 1/2 O2(g) ----> H2O(g) dHf = -241.8 kJ/mol

Step 2: Next, we need to multiply the second reaction by 3 so that it fits the overall pattern in the initial reduction of aluminum oxide. In so doing, we must multiply the overall delta H of formation by 3:

2Al(s) + 3/2 O2(g) ---> Al2O3(s) dHf = -1676 kJ/mol

3 H2(g) + 3/2 O2(g) ----> 3 H2O(g) dHf = -725.4 kJ/mol

Step 3: We now reverse the direction of the first reaction, reversing the sign of the dHf in the process:

Al2O3(s) ---> 2Al(s) + 3/2 O2(g) dHf = 1676 ...

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