Question

The standard enthalpy change for the following reaction is $436.4$kJ
${{\text{H}}_{\text{2}}}{\text{(g)}}\,\, \to \,{\text{H(g)}}\,{\text{ + }}\,{\text{H(g)}}$
What is the ${{\text{\Delta }}_{\text{f}}}{{\text{H}}^0}$of atomic hydrogen (H)
A. $872.8$ kJ/mol
B. $218.2$ kJ/mol
C. $- 218.2$ kJ/mol
D. $- 436.9$ kJ/mol

Answer 1

Standard enthalpy change of a reaction is determined by subtracting the standard enthalpy change for the formation of reactants from the standard enthalpy change for the formation of products. If the standard enthalpy change for the reaction is known we can determine the standard enthalpy for reactant or product.

Formula used: $\Delta {\text{H}} = \,\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{products)}}} \, - \,\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{reactants)}}}$

Complete step-by-step answer:
The formula to determine the standard enthalpy change for a reaction as follows:

$\Delta {\text{H}} = \,\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{products)}}} \, - \,\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{reactants)}}}$

Where,
$\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{products)}}} \,$is the summation of enthalpy of products

$\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{reactants)}}}$is the summation of enthalpy of products

We will use the given standard enthalpy change for reaction to determine the standard enthalpy of the product as follows:

The standard enthalpy of formation of an element in its natural state is zero.

For the given reaction$\Delta {\text{H}}$formula can be written as follows:

$\Delta {\text{H}} = \,\left[ {2\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{H)}}} \right] - \left[ {1\,\, \times {\Delta _{\text{f}}}{\text{H}}\,({{\text{H}}_{\text{2}}}{\text{)}}} \right]$

On substituting $436.4$for $\Delta {\text{H}}$and 0 for${\Delta _{\text{f}}}{\text{H}}$of ${{\text{H}}_{\text{2}}}$.

$\Rightarrow 436.4 = \,\left[ {2\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{H)}}} \right] - \left[ {1\,\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{0)}}} \right]$

$\Rightarrow 436.4 = \,2\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{H}})$
$\Rightarrow {\Delta _{\text{f}}}{\text{H}}\,({\text{H}}) = \,\dfrac{{436.4}}{2}$
$\Rightarrow {\Delta _{\text{f}}}{\text{H}}\,({\text{H}}) = \,218.2$
So, the ${{{\Delta }}_{\text{f}}}{{\text{H}}^0}$of atomic hydrogen (H) is$218.2$ KJ/mole.

Therefore, option (B) $218.2$ KJ/mole is correct.

Note: To determine the change in enthalpy of reaction a balanced chemical equation is necessary. Standard enthalpy of formation is measured in kJ/mol. Oxygen is found in a gaseous state, carbon is in a solid-state. So, there is no change in enthalpy during the formation of solid carbon or gaseous oxygen. Standard enthalpy of formation is also different in different phases. The natural form of carbon is graphite so, its standard enthalpy of formation is zero but the standard enthalpy of formation of diamond is not zero.