Question

Define standard enthalpy of formation?

Answer 1

Hint: “The enthalpy change accompanying the formation of one mole of any compound from its elements, all the substances being in their standard states is called standard enthalpy of formation”.
standard enthalpy of formation $ {\Delta _r}{{\rm H}^0} = \sum {{\Delta _f}{{\rm H}_{\left( {products} \right)}}} - {\Delta _f}{{\rm H}_{\left( {reac\tan ts} \right)}} $

Complete Step By Step Answer:
Standard enthalpy of formation depends on various factors like temperature, pressure and physical state of the reactant and product i.e. solid, liquid or gas. Enthalpy of formation is termed as standard enthalpy of formation when all of its constituents are present in their standard state. Standard state of any constituent is a stable form of co pound at one bar atmospheric pressure. Standard enthalpy of formation is expressed as $ {\Delta _f}{{\rm H}^0} $ . Standard molar enthalpy of formation is a special condition where one mole of product is formed from its elements.
For example- reaction between hydrogen and oxygen to form one mole of water has Standard enthalpy of formation equal to $ {\Delta _f}{{\rm H}^0} = - 285.8kJ/mol $
 $ {H_2}_{\left( g \right)} + \dfrac{1}{2}{O_{2\left( g \right)}} \to {H_2}{O_{\left( l \right)}} $
Reaction between calcium oxide and carbon dioxide to form calcium carbonate is not considered as Standard enthalpy of formation because calcium carbonate is synthesized from other compounds and not from its constituent elements like carbon, oxygen and calcium.
 $ Ca{O_{\left( s \right)}} + C{O_{3\left( g \right)}} \to CaC{O_{3\left( s \right)}} $
Enthalpy of this reaction is $ {\Delta _r}{{\rm H}^0} = - 178.3kJ/mol $
Similarly, reaction between hydrogen and bromine atoms to form hydrogen bromide is not considered as Standard enthalpy of formation because this reaction leads to formation of two moles of hydrogen bromide.
 $ {H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}} $
Enthalpy of this reaction is $ {\Delta _r}{{\rm H}^0} = - 72.8kJ/mol $ .

Note:
Enthalpy of any chemical reaction is calculated experimentally by calorimetry. It is recommended that always write the chemical equation for enthalpy of formation for one mole of compound. Standard enthalpy of formation of any element that exists in its standard state is considered as zero.