Calculate the standard enthalpy of the reaction, δh∘rxn, for the thermite reaction:
2al(s)+fe2o3(s)→2fe(s)+al2o3(s)
elements in their standard state have an enthalpy of formation value of zero. the standard enthalpies of formation of fe2o3 and al2o3 are
δh∘f of fe2o3(s)=−825.5 kj/molδh∘f of al2o3(s)=−1675 kj/mol
express your answer to three significant figures and include the appropriate units.

The standard enthalpy of reaction can be defined as the reaction change that takes place when there has been transfer of energy in the reactions.

The standard enthalpy of reaction () can be given by:

= Enthalpy of formation of products - Enthalpy of formation of reactants

The enthalpy of formation of reactants :

= 2 + 1

Since Al is in a standard state, the enthalpy of formation is 0.

= 0 + (-825.5 kJ/mol)

= -825.5 kJ/mol

The enthalpy of formation of products:

=  2 + 1

Since Fe is in a standard state, the enthalpy of formation is 0.

= 0 + (-1675 kJ/mol.)

= -1675 kJ/mol

= -1675 kJ/mol - (-825.5 kJ/mol)

= -1675 kJ/mol + 825.5 kJ/mol

= -849.5 kJ/mol.

The standard enthalpy for the given reaction has been -849.5 kJ/mol.

For more information about the standard enthalpy, refer to the link:

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The answer to your question is ΔH rxn = - 849.5 kJ/mol

Explanation:

Reaction

                   2Al(s)  + Fe₂O₃ (s)   ⇒   2Fe (s)   +  Al₂O₃ (s)

Enthalpy of reaction

ΔHrxn =  ∑ H products - ∑ H reactants

ΔHrxn = ∑( H (Al₂O₃) + H (Fe)  - ∑ ( H Fe₂O₃ + H Al)

ΔHrxn = [-1675 + 0] - [ -825.5 + 0]

ΔHrxn = -1675 + 825.5

ΔHrxn = - 849.5 kJ/mol

answer: h+ its less than oh

explanation: