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.

Question

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) + 12O2(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?

Answer 1

See your other post above.

Answer 2

(a) To show that the net result of the three reactions is the decomposition of water to produce hydrogen and oxygen, we need to cancel out the common compounds on both sides of the equations and add the equations together.

First, let's cancel out the FeCl2 and FeCl3 compounds in equations 1 and 3:

3FeCl2(s) + 4H2O(g) → Fe3O4(s) + 6HCl(g) + H2(g) ΔH° = 318 kJ
3FeCl3(s) → 3FeCl2(s) + 3/2Cl2(g) ΔH° = 173 kJ

By adding these two equations together, we cancel out the FeCl2 compound:

3FeCl2(s) + 4H2O(g) + 3FeCl3(s) → Fe3O4(s) + 3FeCl2(s) + 6HCl(g) + 3/2Cl2(g) + H2(g) ΔH° = 491 kJ

Next, let's cancel out the Cl2 compound in equation 2 and the 3FeCl2 and H2O compounds in equation 1:

Fe3O4(s) + 3/2Cl2(g) + 6HCl(g) + 3FeCl2(s) → 3FeCl3(s) + 3H2O(g) + 12O2(g) ΔH° = −249 kJ
3FeCl2(s) + 4H2O(g) → Fe3O4(s) + 6HCl(g) + H2(g) ΔH° = 318 kJ

By adding these two equations together, we cancel out the Cl2 and FeCl2 compounds:

Fe3O4(s) + 6HCl(g) + 3/2Cl2(g) + 3H2O(g) → 3FeCl3(s) + 3H2O(g) + 12O2(g) + H2(g) ΔH° = 69 kJ

Simplifying the equation by canceling out the water (H2O) and rearranging the terms, we get:

Fe3O4(s) + 6HCl(g) + 3/2Cl2(g) → 3FeCl3(s) + 12O2(g) + H2(g)

This equation shows that the net result of the three reactions is the decomposition of water to produce hydrogen (H2) and oxygen (O2).

(b) To determine the enthalpy change for the decomposition of one mole of water, we need to use Hess's Law and the enthalpy changes for the reactions.

The goal is to obtain an equation for the decomposition of one mole of water (H2O → H2 + 1/2O2), so we need to rearrange the equations in a way that cancels out the common compounds.

First, let's manipulate equation 1 so that we have H2O on the reactant side:

3FeCl2(s) + 4H2O(g) → Fe3O4(s) + 6HCl(g) + H2(g) ΔH° = 318 kJ

Next, let's manipulate equation 2 so that we have Fe3O4 and HCl on the reactant side:

Fe3O4(s) + 3/2Cl2(g) + 6HCl(g) + 3H2O(g) → 3FeCl3(s) + 12O2(g) + H2(g) ΔH° = -249 kJ

Lastly, we can manipulate equation 3 so that we have FeCl3 on the product side:

3FeCl3(s) → 3FeCl2(s) + 3/2Cl2(g) ΔH° = 173 kJ

Now, we can add these three manipulated equations together, cancel out the common compounds, and see if the net equation corresponds to the decomposition of water:

3FeCl2 + 4H2O + Fe3O4 + 3/2Cl2 + 6HCl + 3H2O + 3FeCl3 →
Fe3O4 + 6HCl + 3/2Cl2 + 3FeCl3 + 12O2 + H2

Cancelling out the common compounds gives:

3H2O → 6HCl + 12O2 + H2

Now it is clear that the net equation represents the decomposition of water (H2O) to produce hydrogen (H2) and oxygen (O2).

To calculate the enthalpy change for the decomposition of one mole of water, we just need to use the value of ΔH° for the net equation. In this case, ΔH° = 69 kJ.

To check our answer, we can compare the enthalpy change with the enthalpy of formation of water (ΔH° = -286 kJ/mol). The enthalpy of formation of water represents the enthalpy change when one mole of water is formed from its constituent elements (hydrogen and oxygen).

Since the enthalpy change for the decomposition of one mole of water (69 kJ) and the enthalpy of formation of water (-286 kJ/mol) have opposite signs, they should sum up to zero according to Hess's Law.

69 kJ + (-286 kJ) = -217 kJ

The result is not zero, which suggests that there might be an error in the calculations or the given enthalpy changes.

Answer 3

(a) To show that the net result of the three reactions is the decomposition of water to produce hydrogen and oxygen, we need to add the three reactions together and cancel out any common species on both sides of the equations.

First, let's write the balanced equations for the three reactions given:

1. 3FeCl2(s) + 4H2O(g) → Fe3O4(s) + 6HCl(g) + H2(g)
2. Fe3O4(s) + 3/2Cl2(g) + 6HCl(g) → 3FeCl3(s) + 3H2O(g) + 12O2(g)
3. 3FeCl3(s) → 3FeCl2(s) + 3/2Cl2(g)

Now, let's add these equations together:

3FeCl2(s) + 4H2O(g) + Fe3O4(s) + 3/2Cl2(g) + 6HCl(g) + 3FeCl3(s) → Fe3O4(s) + 6HCl(g) + H2(g) + 3FeCl2(s) + 3H2O(g) + 12O2(g) + 3FeCl2(s) + 3/2Cl2(g)

By canceling out the common species on both sides, we can simplify the equation:

4H2O(g) + 3/2Cl2(g) + 3FeCl3(s) → H2(g) + 3H2O(g) + 12O2(g)

As you can see, water (H2O) is decomposed into hydrogen gas (H2) and oxygen gas (O2) in the net reaction.

(b) To determine the enthalpy change for the decomposition of one mole of water, we can use Hess's law, which states that the enthalpy change for a reaction is the same regardless of the pathway taken.

Let's assign a letter to each equation:

1. Equation 1: A: 3FeCl2(s) + 4H2O(g) → Fe3O4(s) + 6HCl(g) + H2(g)
2. Equation 2: B: Fe3O4(s) + 3/2Cl2(g) + 6HCl(g) → 3FeCl3(s) + 3H2O(g) + 12O2(g)
3. Equation 3: C: 3FeCl3(s) → 3FeCl2(s) + 3/2Cl2(g)

Now, let's manipulate these equations to get the desired equation: H2O(g) → H2(g) + 1/2O2(g)

To do this, we can reverse Equation A and multiply by a factor of 3/4 to match the number of moles of water on both sides:

3/4A: 9/4FeCl2(s) + 3H2O(g) → 3/4Fe3O4(s) + 9/2HCl(g) + 3/4H2(g)

We also need to reverse Equation B and multiply by a factor of 1/3 to match the number of moles of water on both sides:

1/3B: 1/3Fe3O4(s) + 1/2Cl2(g) + 2HCl(g) → 1FeCl3(s) + 1/3H2O(g) + 4/3O2(g)

Lastly, we need to reverse Equation C and multiply by a factor of 1/3 to match the number of moles of water on both sides:

1/3C: 1/3FeCl3(s) → 1/3FeCl2(s) + 1/2Cl2(g)

Now let's add these three equations together:

3/4A + 1/3B + 1/3C: 9/4FeCl2(s) + 3H2O(g) + 1/3Fe3O4(s) + 1/2Cl2(g) + 2HCl(g) + 1/3FeCl3(s) → 3/4Fe3O4(s) + 9/2HCl(g) + 3/4H2(g) + 1/3FeCl2(s) + 1/3H2O(g) + 4/3O2(g)

Simplifying this equation by canceling out common species gives us:

7/2H2O(g) + 1/2Cl2(g) + 1/3FeCl3(s) → 3/4H2(g) + 4/3O2(g)

The equation above represents the desired equation for the decomposition of water. From this equation, we can see that the molar ratio of H2O(g) to H2(g) is 7/2 and the molar ratio of H2O(g) to O2(g) is 7/6.

Now, using the given enthalpy changes for the reactions:

ΔH°(A) = 318 kJ
ΔH°(B) = -249 kJ
ΔH°(C) = 173 kJ

We can sum up the enthalpy changes for the three equations:

ΔH°(7/2H2O(g) + 1/2Cl2(g) + 1/3FeCl3(s) → 3/4H2(g) + 4/3O2(g)) = 7/2ΔH°(A) + 1/2ΔH°(B) + 1/3ΔH°(C)

Substituting the given values:

ΔH°(7/2H2O(g) + 1/2Cl2(g) + 1/3FeCl3(s) → 3/4H2(g) + 4/3O2(g)) = 7/2 × 318 kJ + 1/2 × (-249 kJ) + 1/3 × 173 kJ

Calculating this expression will give you the enthalpy change for the decomposition of one mole of water using the given reactions.