The reaction of iron with oxygen is very familiar. You can see the resulting rust on buildings, vehicles and bridges. You may be surprised, however at the large amount of heat that is produced by this reaction.
4Fe(g) + 3O2(g) 2Fe2O3(g) + 165 x 103 kJ
(a) What is the enthalpy change for this reaction?
(b) Draw the enthalpy diagram that corresponds to the thermochemical equation.
(c) What is the enthalpy change for the formation of 23.6 g iron (III) oxide?
A. The reaction produces 165E3 kJ for 2 mols; therefore, delta H = -165E3 kJ for 2 mols. My thermodynamic tables don't give this value. Look to see that you have posted the right number. I'm thinking you meant 1650. I will use your value to answer A and C.
B. This site does not support us drawing diagrams. Sorry.
C. So you have 165E3 kJ for 159.7 x 2 or 319.4 g Fe2O3. For 23.6 you will have 165E3 kJ x (23.6/319.4) = ? kJ.
(a) To find the enthalpy change for the reaction, you can use the stoichiometric coefficients and the given enthalpy value.
From the balanced equation:
4Fe(g) + 3O2(g) -> 2Fe2O3(g)
The coefficient in front of Fe2O3 is 2, indicating that 2 moles of Fe2O3 are formed.
Given that the enthalpy change is 165 x 10^3 kJ, this corresponds to the enthalpy change for the formation of 2 moles of Fe2O3.
Thus, the enthalpy change for the reaction is:
(165 x 10^3 kJ) / 2 = 82.5 x 10^3 kJ
Therefore, the enthalpy change for this reaction is 82.5 x 10^3 kJ.
(b) To draw the enthalpy diagram that corresponds to the thermochemical equation, you will need to include the reactants, the products, and the enthalpy change.
On the reactant side, you can draw 4Fe(g) and 3O2(g) as the initial state. On the product side, you can draw 2Fe2O3(g) as the final state.
To represent the enthalpy change, you can draw a delta H (ΔH) arrow pointing from the reactant side to the product side, and label it with the value of 82.5 x 10^3 kJ.
(c) To find the enthalpy change for the formation of 23.6 g of iron (III) oxide, you can use the molar mass of Fe2O3 and the enthalpy change calculated in part (a).
First, calculate the number of moles of Fe2O3:
Number of moles = mass / molar mass
Molar mass of Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O)
Using the atomic masses of Fe and O from the periodic table:
Molar mass of Fe2O3 = (2 x 55.845 g/mol) + (3 x 16.00 g/mol) = 159.69 g/mol
Number of moles = 23.6 g / 159.69 g/mol = 0.1479 mol
Next, use the enthalpy change calculated in part (a) to find the enthalpy change for the formation of 0.1479 mol of Fe2O3:
Enthalpy change = number of moles x enthalpy change
Enthalpy change = 0.1479 mol x 82.5 x 10^3 kJ/mol = 12,184.175 kJ
Therefore, the enthalpy change for the formation of 23.6 g of iron (III) oxide is approximately 12,184.175 kJ.
(a) To find the enthalpy change for this reaction, we need to look at the coefficients of the balanced equation. The coefficient in front of Fe2O3 tells us that the enthalpy change for the formation of 2 moles of Fe2O3 is 165 x 103 kJ.
We can set up a proportion to find the enthalpy change for the formation of 1 mole of Fe2O3:
(165 x 103 kJ) / (2 moles Fe2O3) = ΔH / (1 mole Fe2O3)
Rearranging the equation, we find:
ΔH = (165 x 103 kJ) / (2 moles Fe2O3)
Calculating this expression will give us the enthalpy change for the reaction.