Calculate the standard heat of reaction at 250C (298K) and 1 atm pressure of
3FeO(s) + 2Al(s) → Al2O3(s) + 3Fe(s)
in terms of per mole of Al2O3 formed, per mole of Fe formed, per mole of FeO reacted, per mole of Al reacted and per g of Fe formed.
ΔH298K,FeO(s) = -63.3 kcal/mol (-264.84 kJ/mol)
ΔH298K,Al2O3(s) = -400 kcal/mol (-1673.6kJ/mol)
Aw,Fe(s) = 56 g/mol
dHrxn = (n*dHf products) - (n*dHf reatants).
That gives you dH/mol Al2O3, per 3 mols Fe, per 3 mols FeO
, or per 2 mols Al.
And note that 250 C is not 298K. I assume that 0 is a typo and you meant for it to be a degree symbol.
To calculate the standard heat of reaction (∆H°) at 298K, we need to use the given enthalpy values and apply the Hess's law.
First, let's determine the molar masses of Al2O3 and Fe:
Molar mass Al2O3 = 2(27 g/mol) + 3(16 g/mol) = 102 g/mol
Molar mass Fe = 56 g/mol
Now, let's calculate the standard heat of reaction per mole of Al2O3 formed:
∆H° (per mole of Al2O3 formed) = (∆H° Al2O3(s)) - [∆H° FeO(s) + ∆H° Fe(s)]
= (-400 kcal/mol) - [-63.3 kcal/mol + 0]
= -400 + 63.3 kcal/mol
= -336.7 kcal/mol
Next, let's calculate the standard heat of reaction per mole of Fe formed:
∆H° (per mole of Fe formed) = (∆H° Fe(s)) - [∆H° Al2O3(s) + ∆H° FeO(s)]
= 0 - [-400 kcal/mol + (-63.3 kcal/mol)]
= -400 + 63.3 kcal/mol
= -336.7 kcal/mol
Now, let's calculate the standard heat of reaction per mole of FeO reacted:
∆H° (per mole of FeO reacted) = [∆H° FeO(s)] - [∆H° Fe(s) + ∆H° Al2O3(s)]
= (-63.3 kcal/mol) - [0 + (-400 kcal/mol)]
= -63.3 + 400 kcal/mol
= 336.7 kcal/mol
Next, let's calculate the standard heat of reaction per mole of Al reacted:
∆H° (per mole of Al reacted) = 2[∆H° Al2O3(s)] - [3∆H° FeO(s) + 2∆H° Fe(s)]
= 2(-400 kcal/mol) - [3(-63.3 kcal/mol) + 0]
= 800 + 189.9 kcal/mol
= 989.9 kcal/mol
Finally, let's calculate the standard heat of reaction per gram of Fe formed:
∆H° (per gram of Fe formed) = (∆H° Fe(s)) / (molar mass Fe)
= 0 kcal/mol / 56 g/mol
= 0 kcal/g
Therefore, the values for ∆H° are as follows:
Per mole of Al2O3 formed: -336.7 kcal/mol
Per mole of Fe formed: -336.7 kcal/mol
Per mole of FeO reacted: 336.7 kcal/mol
Per mole of Al reacted: 989.9 kcal/mol
Per gram of Fe formed: 0 kcal/g
To calculate the standard heat of reaction, we need to use the standard enthalpy of formation values for each compound involved in the reaction.
The standard enthalpy of formation (ΔHf) is the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states. The standard state means the most stable form of an element or compound at a given temperature and pressure.
We can use the given standard enthalpy of formation values and stoichiometry of the reaction to calculate the standard heat of reaction.
Given data:
ΔH298K,FeO(s) = -63.3 kcal/mol (-264.84 kJ/mol)
ΔH298K,Al2O3(s) = -400 kcal/mol (-1673.6 kJ/mol)
Aw,Fe(s) = 56 g/mol
First, let's balance the equation by ensuring the number of atoms on both sides is the same:
3FeO(s) + 2Al(s) → Al2O3(s) + 3Fe(s)
Now, let's calculate the standard heat of reaction per mole of Al2O3 formed:
ΔH298K,Al2O3(s) = -400 kcal/mol
The coefficient of Al2O3 in the balanced equation is 1. Therefore, the standard enthalpy change per mole of Al2O3 formed is -400 kcal/mol.
Next, let's calculate the standard heat of reaction per mole of Fe formed:
ΔH298K,Fe(s) = 0 kcal/mol (Fe is in its standard state)
The coefficient of Fe in the balanced equation is 3. Therefore, the standard enthalpy change per mole of Fe formed is 3 * 0 kcal/mol = 0 kcal/mol.
Now, let's calculate the standard heat of reaction per mole of FeO reacted:
ΔH298K,FeO(s) = -63.3 kcal/mol
The coefficient of FeO in the balanced equation is -3. Therefore, the standard enthalpy change per mole of FeO reacted is -3 * -63.3 kcal/mol = +189.9 kcal/mol.
Next, let's calculate the standard heat of reaction per mole of Al reacted:
The coefficient of Al in the balanced equation is -2. Since Al is in its standard state, the enthalpy change is 0 kcal/mol. Therefore, the standard heat of reaction per mole of Al reacted is 0 kcal/mol.
Finally, let's calculate the standard heat of reaction per gram of Fe formed:
Aw,Fe(s) = 56 g/mol
The coefficient of Fe in the balanced equation is 3, so the mass of Fe formed is 3 * 56 g/mol = 168 g.
To convert the per mole values to per gram values, we divide by the molar mass of Fe:
Standard heat of reaction per g of Fe formed = (Standard heat of reaction per mole of Fe formed) / (Mass of Fe formed)
= 0 kcal/mol / 168 g
= 0 kcal/g
Therefore, the standard heat of reaction per g of Fe formed is 0 kcal/g.
To summarize:
- The standard heat of reaction per mole of Al2O3 formed: -400 kcal/mol
- The standard heat of reaction per mole of Fe formed: 0 kcal/mol
- The standard heat of reaction per mole of FeO reacted: 189.9 kcal/mol
- The standard heat of reaction per mole of Al reacted: 0 kcal/mol
- The standard heat of reaction per g of Fe formed: 0 kcal/g