Question

A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:

Fe2O3(s) + 3 H2(
92Fe(s) + 3 H2O(9)
Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation
of H2O(g) is
kJ/mol.

Answer 1

Answer: The standard enthalpy of formation of
`H_2O(g)` is -252.1 kJ/mol.

Step-by-step explanation:

The balanced chemical reaction is,

`Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)`

The expression for enthalpy change is,

`\Delta H=[n* H_f_(products)]-[n* H_f_(reactantss)]`

Putting the values we get :

`\Delta H=[2* H_f{Fe}+3* H_f{H_2O}]-[1* H_f{Fe_2O_3}+3* H_f{H_2}]`

`67.9kJ=[(2* 0)+(3* H_f{H_2O})]-[(1* -824.2kJ/mol)+3* 0kJ/mol)]`

`H_f{H_2O}=-252.1kJ/mol`

Thus standard enthalpy of formation of
`H_2O(g)` is -252.1 kJ/mol.