Question

For the reaction,
${\text{3}}{{\text{N}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right){\text{ + 2N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right) \to 4{{\text{N}}_{\text{2}}}\left( {\text{g}} \right) + {\text{3}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right)$ ; $\Delta {{\text{H}}^0} = - 879.6{\text{kJmo}}{{\text{l}}^{ - 1}}$
If $\Delta {{\text{H}}^0}_{\text{f}}\left[ {{\text{N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right)} \right] = - 45.9{\text{kJmo}}{{\text{l}}^{ - 1}}$ ; $\Delta {{\text{H}}^0}_{\text{f}}\left[ {{{\text{H}}_2}{\text{O}}\left( {\text{g}} \right)} \right] = - 241.8{\text{kJmo}}{{\text{l}}^{ - 1}}$
Then $\Delta {{\text{H}}^0}_{\text{f}}\left[ {{{\text{N}}_2}{\text{O}}\left( {\text{g}} \right)} \right]$ will be:
A.$ + 246{\text{kJ}}$
B.$ + 82{\text{kJ}}$
C.${\text{ - 82kJ}}$
D.$ - 246{\text{kJ}}$

Answer 1

Hint: The standard enthalpy or heat of formation refers to the change in enthalpy or heat when one mole of a substrate is formed from its elements in their standard state.
The standard enthalpy of any reaction is considered to be equal to the difference between the standard enthalpy of formation values of all the products and the standard enthalpy of formation values of all the reactants.

Complete step by step answer:
The symbol $\Delta {{\text{H}}^0}_{\text{f}}$ is generally used to denote the standard enthalpy or heat of formation. For elements, it is considered to be equal to zero.
If $\Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{products}}} \right)$ represents the enthalpy of formation of all the products and $\Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{reactants}}} \right)$ represents the enthalpy of formation of all the reactants, then the standard enthalpy of a reaction is given by the mathematical expression:
$\Delta {{\text{H}}^{\text{0}}} = \Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{products}}} \right) - \Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{reactants}}} \right)$
Thus, we can determine the enthalpy of reactions by using the enthalpy of formation values.
Now, according to the given question, the enthalpy of formation of ammonia is $ - 45.9{\text{kJmo}}{{\text{l}}^{ - 1}}$ and the enthalpy of formation of water is $ - 241.8{\text{kJmo}}{{\text{l}}^{ - 1}}$ . The enthalpy change of the reaction is also given to be $ - 879.6{\text{kJmo}}{{\text{l}}^{ - 1}}$ . We need to find out the enthalpy of formation of nitrous oxide.
The given reaction is ${\text{3}}{{\text{N}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right){\text{ + 2N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right) \to 4{{\text{N}}_{\text{2}}}\left( {\text{g}} \right) + {\text{3}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right)$ .
In this reaction, there are 3 moles of nitrous oxide, 2 moles of ammonia, 4 moles of nitrogen and 3 moles of water.
So, we have:
$\Delta {{\text{H}}^{\text{0}}} = 4\Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left[ {{{\text{N}}_{\text{2}}}\left( {\text{g}} \right)} \right] + 3\Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left[ {{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right)} \right] - 2\Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left[ {{\text{N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right)} \right] - 3\Delta {{\text{H}}^0}_{\text{f}}\left[ {{{\text{N}}_2}{\text{O}}\left( {\text{g}} \right)} \right]$
$\Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left[ {{{\text{N}}_{\text{2}}}\left( {\text{g}} \right)} \right]$ is equal to zero as nitrogen gas is an element in standard state.
Substitute all the values and we will have
$
   - 879.6 = 4 \times 0 + 3 \times \left( { - 241.8} \right) - 2 \times \left( { - 45.9} \right) - 3\Delta {{\text{H}}^0}_{\text{f}}\left[ {{{\text{N}}_2}{\text{O}}\left( {\text{g}} \right)} \right] \\
   \Rightarrow \Delta {{\text{H}}^0}_{\text{f}}\left[ {{{\text{N}}_2}{\text{O}}\left( {\text{g}} \right)} \right] = \dfrac{{879.6 - 725.4 + 91.8}}{3} \\
   \Rightarrow \Delta {{\text{H}}^0}_{\text{f}}\left[ {{{\text{N}}_2}{\text{O}}\left( {\text{g}} \right)} \right] = 82{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}} \\
 $

Hence, the correct option is B.

Note:
Some other common kinds of enthalpy of reactions are enthalpy of combustion and enthalpy of hydration.
The term ‘enthalpy of combustion’ refers to the enthalpy change when the complete combustion of one mole of a substance. For example, the enthalpy of combustion of carbon is found to be \[ - 393.5{\text{kJ/mol}}\] .
${\text{C}}\left( {\text{s}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right);\Delta {\text{H}} = - 393.5{\text{kJ/mol}}$
The enthalpy of hydration refers to the enthalpy change when one mole of a partially hydrated or anhydrous salt combines with water to give its hydrate. For example:
${\text{CuS}}{{\text{O}}_4}\left( {\text{s}} \right){\text{ + 5}}{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) \to {\text{CuS}}{{\text{O}}_4}{\text{.5}}{{\text{H}}_2}{\text{O}}\left( {\text{s}} \right);\Delta {\text{H}} = - 78.2{\text{kJ/mol}}$