Question

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

Answer 1

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

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,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]$

Putting the values we get :

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

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

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

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