Question

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

Answer 1

Answer : The standard enthalpy of the reaction is, -849.5 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2Al(s)+Fe_2O_3(s)\rightleftharpoons 2Fe(s)+Al_2O_3(s)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(Al_2O_3)}\times \Delta H^o_f_{(Al_2O_3)})+(n_{(Fe)}\times \Delta H^o_f_{(Fe)})]-[(n_{(Fe_2O_3)}\times \Delta H^o_f_{(Fe_2O_3)})+(n_{(Al)}\times \Delta H^o_f_{(Al)})]$

We are given:

$\Delta H^o_f_{(Al_2O_3(s))}=-1675kJ/mol\\\Delta H^o_f_{(Al(s))}=0kJ/mol\\\Delta H^o_f_{(Fe_2O_3(s))}=-825.5kJ/mol\\\Delta H^o_f_{(Fe(s))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times -1675)+(2\times 0)]-[(1\times -825.5)+(2\times 0)]=-849.5kJ/mol$

Therefore, the standard enthalpy of the reaction is, -849.5 kJ/mol

Answer 2

Answer:

The answer to your question is:  ΔH = -849.5 KJ/mol

Explanation:

Data

ΔH = ?

H Fe2O3 = -825.5 KJ/mol

H Al2O3 = -1675 KJ/mol

H Al = 0 KJ/mol

H Fe = 0 KJ/ mol

Reaction

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

ΔH = H products - H reactants

ΔH = H Al2O3 - H Fe2O3

ΔH = -1675 - (- 825.5)

ΔH = -1675 + 825.5

ΔH = -849.5 KJ/mol