Hydrogen is a very appealing fuel, in part because burning it produces only non-polluting water. One of the challenges that researchers face in making hydrogen fuel a reality is how to produce hydrogen economically. Researchers are investigating methods of producing hydrogen indirectly. The following series of equations represent one such method.
3FeCl2(s) + 4H2O(g) → Fe3O4(s) + 6HCl(g) + H2(g) ΔH° = 318 kJ
Fe3O4(s) + 3/2Cl2(g) + 6HCl(g) → 3FeCl3(s) + 3H2O(g) + 1/2O2(g) ΔH° = −249 kJ
3FeCl3(s) → 3FeCl2(s) + 3/2Cl2(g) ΔH° = 173 kJ
(a) Show that the net result of the three reactions is the decomposition of water to produce hydrogen and oxygen.
(b) Use Hess’s law and the enthalpy changes for the reactions to determine the enthalpy change for the decomposition of one mole of water. Check your answer, using the enthalpy of formation of water.
how do i show the deomcposition of water and for b) how do i start?
Add the equations and cancel molecules that are the same on both sides.(It may b necssary to reverse some of the equations.) For example, you have 3 FeCl2 in equation 1 on the left side and 3 FeCl2 in equation 3 on the right side. They cancel. Continue and see what is left.I didn't go through all of it but the problem suggests that everything except H2O ==> H2 + O2 cancels. This is not balanced as is but the coefficients probably will balance in the end.
For the b part, add the equations (reverse some if necessary--change the sign of delta H if reversed), then add all of the delta H values together to arrive at a final delta H for the series.
Thank you, helped, the equations do balance out, and there are more than one ways to solve this either by multiplying the first equation by 2 and then manipulating further as well as just canceling out and adding all by hess's law.
(a) To show that the net result of the three reactions is the decomposition of water to produce hydrogen and oxygen, we can add the three equations together, taking into account the stoichiometric coefficients and canceling out any common species on both sides of the equation:
3FeCl2(s) + 4H2O(g) → Fe3O4(s) + 6HCl(g) + H2(g)
Fe3O4(s) + 3/2Cl2(g) + 6HCl(g) → 3FeCl3(s) + 3H2O(g) + 1/2O2(g)
3FeCl3(s) → 3FeCl2(s) + 3/2Cl2(g)
Adding these equations, we get:
3FeCl2(s) + 4H2O(g) + 3FeCl3(s) + 3/2Cl2(g) + 6HCl(g) → Fe3O4(s) + 6HCl(g) + 3FeCl2(s) + 3H2O(g) + 1/2O2(g)
Simplifying further, the common species on both sides cancel out:
4H2O(g) + 3/2Cl2(g) → Fe3O4(s) + 3H2O(g) + 1/2O2(g)
This can be further simplified as:
4H2O(g) + 3/2Cl2(g) → Fe3O4(s) + 3/2O2(g)
We can rewrite the equation to show the decomposition of water:
4H2O(g) → 3/2O2(g) + 4H2(g)
Thus, the net result of the three reactions is indeed the decomposition of water to produce hydrogen and oxygen.
(b) To determine the enthalpy change for the decomposition of one mole of water using Hess's law, we need to manipulate the given reactions to have the desired reaction of water decomposition. First, we'll manipulate the given equations as follows:
equation 1: 3FeCl2(s) + 4H2O(g) → Fe3O4(s) + 6HCl(g) + H2(g)
equation 2: Fe3O4(s) + 3/2Cl2(g) + 6HCl(g) → 3FeCl3(s) + 3H2O(g) + 1/2O2(g)
equation 3: 3FeCl3(s) → 3FeCl2(s) + 3/2Cl2(g)
Now, we can manipulate these equations to obtain the desired equation:
3FeCl2(s) + 4H2O(g) + 3FeCl3(s) → Fe3O4(s) + 6HCl(g) + 3/2Cl2(g) + 3H2O(g) + 1/2O2(g)
Notice that the water molecules on the left side of the equation cancel out with the water molecules on the right side of the equation:
3FeCl2(s) + 3FeCl3(s) → Fe3O4(s) + 6HCl(g) + 3/2Cl2(g) + 1/2O2(g)
Finally, we calculate the enthalpy change for the desired reaction:
ΔH° = (ΔH°1) + (ΔH°2) + (ΔH°3)
ΔH° = (318 kJ) + (-249 kJ) + (173 kJ)
ΔH° = 242 kJ
Therefore, the enthalpy change for the decomposition of one mole of water is 242 kJ. To check this answer, you can also use the enthalpy of formation of water, which is -286 kJ/mol.
To show that the net result of the three reactions is the decomposition of water to produce hydrogen and oxygen, we need to combine the equations in a way that cancels out the other species and leaves us with water, hydrogen, and oxygen.
Here's how we can do it:
1. Multiply the first equation by 2 and reverse the second equation:
2FeCl2(s) + 8H2O(g) → 2Fe3O4(s) + 12HCl(g) + 2H2(g) ΔH° = 636 kJ
-2Fe3O4(s) - 3Cl2(g) - 6HCl(g) → -6FeCl3(s) - 6H2O(g) - O2(g) ΔH° = 249 kJ
2. Multiply the third equation by 2 and reverse it:
6FeCl3(s) → 6FeCl2(s) + 3Cl2(g) ΔH° = -346 kJ
3. Add the three equations together:
2FeCl2(s) + 8H2O(g) - 6FeCl3(s) → 2Fe3O4(s) + 12HCl(g) + 2H2(g) -6FeCl2(s) - 3Cl2(g) - 6HCl(g) - O2(g)
Simplifying it further, we have:
8H2O(g) - 6FeCl3(s) → 2Fe3O4(s) + 12HCl(g) + 2H2(g) -6FeCl2(s) - 3Cl2(g) - 6HCl(g) - O2(g)
Now, we can see that water (H2O) is the reactant on the left side and hydrogen (H2) and oxygen (O2) are the products on the right side. Therefore, the net result is the decomposition of water to produce hydrogen and oxygen.
For part b) of the question, let's use Hess's Law to determine the enthalpy change for the decomposition of one mole of water.
Given the following reactions and their enthalpy changes:
1. 2H2(g) + O2(g) → 2H2O(g) (ΔH° = -572 kJ)
2. 2Fe(Cl2)3(s) → 2FeCl3(s) + 3Cl2(g) (ΔH° = -2 x (-173 kJ) = 346 kJ)
3. 3FeCl2(s) + 3/2O2(g) → 3Fe(Cl2)3(s) (ΔH° = 3 x (-249 kJ) = -747 kJ)
We can reverse equation 1 and multiply it by 2:
-2H2O(g) → -4H2(g) - 2O2(g) (ΔH° = 2 x 572 kJ = 1144 kJ)
Now, let's add equations 2 and 3 to get the overall reaction:
2Fe(Cl2)3(s) + 3FeCl2(s) + 3/2O2(g) → 2FeCl3(s) + 3Cl2(g) + 3Fe(Cl2)3(s) (ΔH° = 346 kJ -747 kJ = -401 kJ)
Finally, add the overall reaction to the reversed equation of water decomposition:
-2H2O(g) + 2Fe(Cl2)3(s) + 3FeCl2(s) + 3/2O2(g) → -4H2(g) - 2O2(g) + 2FeCl3(s) + 3Cl2(g) + 3Fe(Cl2)3(s) (ΔH° = 1144 kJ - 401 kJ = 743 kJ)
Therefore, the enthalpy change for the decomposition of one mole of water is 743 kJ.
To check our answer, we can calculate the enthalpy change using the enthalpy of formation of water:
ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants)
ΔH° = [(2 x 0 kJ/mol) - (2 x (-285.8) kJ/mol)] - [(2 x 0 kJ/mol) + (-241.8 kJ/mol)] = 571.6 kJ
Comparing this value with our previous result of 743 kJ, we can see that they are not the same. Therefore, there seems to be an error in the calculations and we should go back and double-check our steps.