The reduction of aluminum oxide by hydrogen gas: Al2O3(s) + 3H2(g) ------- 2Al(s) + 3H2O(g) 2

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 kJ/mol

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

Step 4: Now, we add the two reactions together, cross-canceling chemicals where we can. In addition, we add the two dH of formations together:

Al2O3(s) + 3H2(g) ------- 2Al(s) + 3H2O(g) dHf = 950.6 kJ/mol

_______________________________________________________________

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) + 3H2O(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)

Solution:

We use the same procedure as we did for question (1) above.
Write the equations down. (Notice that the first equation was written down wrong. There are 3 moles of H2O, not 1.)

C2H5OH(l) + 3O2(g) -------> 2CO2(g) + 3H2O(l) delta H = -227 kcal
HC2H3O2(l) + 2O2(g) ------> 2CO2(g) + 2H2O(l) delta H = -209 kcal

Next, reverse the second equation and reverse the sign.

C2H5OH(l) + 3O2(g) -------> 2CO2(g) + 3H2O(l) delta H = -227 kcal
2CO2(g) + 2H2O(l) ------> HC2H3O2(l) + 2O2(g) delta H = +209 kcal

Finally, add the two equations together and do the math.

C2H5OH(l) + O2(g) -------> HC2H3O2(l) + H2O(l) delta H = -18 kcal

_______________________________________________________________

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

Solution:

This equation represents the energy released when 1 mole of H2(g) reacts with enough O2 to make liquid water. The question is asking for 1 kilomole. That's 1000 moles. Therefore, we just multiply the dH factor by 1000. The answer is -68320. Simple.

_______________________________________________________________

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

Response: We're not able to guide you through this problem because you have left off the partial reactions that we are to use for Hess' law. I don't think you should have too much difficulty with this question anyway. Just use the same procedure I've outlined in the first two examples above.