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Which among the following represents the reaction of formation of the product?
A.${{{C}}_{{{diamond}}}}{{ + }}{{{O}}_{{{2}}\left( {{g}} \right)}}\xrightarrow{{}}{{C}}{{{O}}_{{{2}}\left( {{g}} \right)}}$
B. ${{{S}}_{{{monoclinic}}}}{{ + }}{{{O}}_{{2}}}\xrightarrow{{}}{{S}}{{{O}}_{{{2}}\left( {{g}} \right)}}$
C. ${{2}}{{{N}}_{{2}}}{{ + }}{{{O}}_{{{2}}\left( {{g}} \right)}}\xrightarrow{{}}{{2}}{{{N}}_{{2}}}{{{O}}_{\left( {{g}} \right)}}$
D. None of the above

Last updated date: 24th Feb 2024
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IVSAT 2024
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Hint: Reaction of formation or enthalpy of formation involves the formation of${{1 mole}}$product from its constituent elements in its standard state. The reactants should be in standard state in order to get the ${{\Delta }}{{{H}}_{{f}}}$ of products.

Complete step by step answer:
Change in enthalpy required for the formation of ${{1mol}}$ of a compound from its elements in their standard states at ${{25}}{{{ }}^{{o}}}{{C}}$ is called as Standard enthalpy of formation ${{\Delta }}{{{H}}^{{o}}}_{{f}}$
For a free element in its standard state ${{\Delta }}{{{H}}^{{o}}}_{{f}}$ will be zero.
Carbon has a stable state of graphite but not diamond, thus if the elements are not in their standard states than there will be no reaction of formation, thus first reaction doesn’t involve the reaction of formation
Sulphur has a stable state of rhombic when compared to monoclinic thus ${{S}}{{{O}}_{{2}}}$ cannot be formed
In the third reaction both ${{{N}}_{{2}}}$ and ${{{O}}_{{2}}}$ are in standard state but there is a formation of ${{2mol}}$ of product instead of ${{1mol}}$ thus the correct answer is none of the above.

So, the correct answer is Option D.

Additional information:
According to Hess’s law if a reaction occurs in multiple steps, the sum of the enthalpy changes $\left( {{{\Delta H}}} \right)$ for the individual reactions is equal to ${{\Delta H}}$ for the overall reaction.
Eg: ${{{N}}_{{{2}}\left( {{g}} \right)}}{{ + }}{{{O}}_{{{2}}\left( {{g}} \right)}}\xrightarrow{{}}{{2N}}{{{O}}_{\left( {{g}} \right)}}{{ \Delta H = 180 kJ}}$
${{2N}}{{{O}}_{\left( {{g}} \right)}}{{ + }}{{{O}}_{{{2}}\left( {{g}} \right)}}\xrightarrow{{}}{{2N}}{{{O}}_{{{2}}\left( {{g}} \right)}}{{ \Delta H = - 112 kJ}}$

${{{N}}_{{{2}}\left( {{g}} \right)}}{{ + 2}}{{{O}}_{{{2}}\left( {{g}} \right)}}\xrightarrow{{}}{{2N}}{{{O}}_{{{2}}\left( {{g}} \right)}}{{ \Delta H = 68 kJ}}$

Note: Rhombic sulphur is considered as the stable form of sulphur because of its stable structure; here 8 sulphur atoms are arranged in an octahedral form. Among diamond and graphite, graphite has got Van der Waals forces whereas diamond has covalent bonds within but the melting point of graphite is higher than that of diamond making them thermodynamically more stable.
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