Question

Calculate the enthalpy of reaction for the following reaction:

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

ΔHfo(Fe2O3(s)) = -824.2 kJ/mol
ΔHfo(CO(g)) = -110.5
ΔHfo(Fe(s)) = ?
ΔHfo(CO2(g)) = -393.5 kJ/mol

Use SF.

Answer 1

To calculate the enthalpy of the reaction, use Hess's Law and the enthalpy of formation values. Fe(s) has an enthalpy of formation of -400.5 kJ/mol.

Step-by-step explanation:

To calculate the enthalpy of reaction for the given reaction, we can use Hess's Law and the enthalpy of formation values provided.

The enthalpy of formation of Fe2O3 is -824.2 kJ/mol, the enthalpy of formation of CO is -110.5 kJ/mol, and the enthalpy of formation of CO2 is -393.5 kJ/mol.

By applying Hess's Law, we can add the enthalpy of formation values for the products and reactants to determine the enthalpy change for the reaction.

The enthalpy of formation for Fe(s) can be calculated by subtracting the enthalpy of formation of Fe2O3, 3 times the enthalpy of formation of CO, and 3 times the enthalpy of formation of CO2, from the enthalpy of formation of the reactant Fe2O3. After performing the calculation, we find that the enthalpy of the formation of Fe(s) is -400.5 kJ/mol.

Answer 2

The enthalpy of reaction for Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) is calculated using standard enthalpies of formation and equals -24.8 kJ, which indicates the reaction is endothermic.

Step-by-step explanation:

To calculate the enthalpy of reaction for the reaction Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g), you need to use the standard enthalpies of formation for each substance involved in the reaction. The standard enthalpy of formation (ΔHf°) is the heat change that results when one mole of a compound is formed from its elements in their standard states.

The enthalpy of reaction (ΔHrxn) can be found using the equation:

ΔHrxn = ∑ΔHf°(products) - ∑ΔHf°(reactants)

For the reactants:

• 
  
• Fe2O3: -824.2 kJ/mol
• 
  
• CO: -110.5 kJ/mol

For the products:

• 
  
• Fe: The standard enthalpy of formation for elemental substances in their standard states is zero.
• 
  
• CO2: -393.5 kJ/mol

Putting these values into the equation gives:

ΔHrxn = [2(0) + 3(-393.5)] - [-824.2 + 3(-110.5)]
= [0 - 1180.5] - [-824.2 - 331.5]
= -1180.5 - (-1155.7)
= —24.8 kJ

Therefore, the enthalpy of reaction for the given chemical reaction is ——25.1 kJ, indicating that the reaction is endothermic.