Question

What is enthalpy change in the formation of MgO?
\[(i)Mg(s)\to Mg(g)\text{ };\Delta H=S\]
\[(ii)Mg(g)\to M{{g}^{+}}(g)+{{e}^{-}}\text{ ;}\Delta H={{l}_{1}}\]
\[(iii)M{{g}^{+}}(g)\to M{{g}^{2+}}(g)+{{e}^{-}}\text{ ;}\Delta H={{l}_{2}}\]
\[(iv){{O}_{2}}(g)\to 2O(g)\text{ };\Delta H=D\]
\[(v)O(g)+{{e}^{-}}\to {{O}^{-}}(g)\text{ };\Delta H={{E}_{1}}\]
\[(vi){{O}^{-}}(g)+{{e}^{-}}\to {{O}^{2-}}(g)\text{ };\Delta H={{E}_{2}}\]
\[(vii)M{{g}^{2+}}(g)+{{O}^{2-}}(g)\to MgO(s)\text{ };\Delta H=-U\]
(A) \[\Delta H=S-{{l}_{1}}+{{l}_{2}}-\dfrac{D}{2}-{{E}_{1}}+{{E}_{2}}-U\]
(B) \[\Delta H=S+{{l}_{1}}+{{l}_{2}}+\dfrac{D}{2}-{{E}_{1}}+{{E}_{2}}-U\]
(C) \[\Delta H=S+{{l}_{1}}+{{l}_{2}}+\dfrac{D}{2}-{{E}_{1}}-{{E}_{2}}+U\]
(D) None of the above

Answer 1

Hint: The above reactions are based on Hess's law. This law is based on the first law of thermodynamics. This law in simple terms means the enthalpy of reaction depends only on the initial reactant and final products and not on the intermediate products.

Complete step by step solution:
Before going into enthalpy of formation, let’s first understand the basic definition for enthalpy of solution. Enthalpy of solution is defined as the amount of heat absorbed or evolved when one mole of solute is dissolved in some amount of the solvent. When the heat is absorbed from the solution, then the enthalpy will be positive. When the heat is given to the solution, then enthalpy will be negative.
Enthalpy of formation can be calculated by the application of Hess’s law. Hess’s law states that the amount of heat absorbed or evolved in a process is the same, whether the process takes place in single or multi-step.
\[\Delta H=\Delta {{H}_{1}}+\Delta {{H}_{2}}+\Delta {{H}_{3}}\]
From the above Equation
\[Mg(g)\to M{{g}^{+}}(g)+{{e}^{-}}\text{ ;}\Delta H={{l}_{1}}\]
\[M{{g}^{+}}(g)\to M{{g}^{2+}}(g)+{{e}^{-}}\text{ ;}\Delta H={{l}_{2}}\]
\[{{O}_{2}}(g)\to 2O(g)\text{ };\Delta H=D\text{ }\Rightarrow \text{ }\dfrac{1}{2}{{O}_{2}}(g)\to O(g)\text{ };\Delta H=\dfrac{D}{2}\]
\[O(g)+{{e}^{-}}\to {{O}^{-}}(g)\text{ };\Delta H={{E}_{1}}\]
\[{{O}^{-}}(g)+{{e}^{-}}\to {{O}^{2-}}(g)\text{ };\Delta H={{E}_{2}}\]
\[M{{g}^{2+}}(g)+{{O}^{2-}}(g)\to MgO(s)\text{ };\Delta H=-U\]
The overall reaction of formation of \[MgO\]is given by
\[Mg(s)+\dfrac{1}{2}{{O}_{2}}\to MgO(s)\text{ }\]
The enthalpy formation of \[MgO\]
\[\Delta H=\Delta {{H}_{1}}+\Delta {{H}_{2}}+\Delta {{H}_{3}}+\Delta {{H}_{4}}+\Delta {{H}_{5}}+\Delta {{H}_{6}}+\Delta {{H}_{7}}\]
\[\Delta H=S+{{l}_{1}}+{{l}_{2}}+\dfrac{D}{2}+{{E}_{1}}+{{E}_{2}}-U\]
The enthalpy of formation of\[MgO\] is \[\Delta H=S+{{l}_{1}}+{{l}_{2}}+\dfrac{D}{2}+{{E}_{1}}+{{E}_{2}}-U\]

Therefore, correct answer is option(D) None of the above

Additional information:
- First law of thermodynamics states that energy can neither be created nor destroyed, but it can be transformed from one form to another.
- This is also known as the law of conservation of energy.

Note: Hess’s law can be used for the calculation of the following:
- Calculation of enthalpies of reaction.
- Determination of slow reaction.
- Calculation of Enthalpy of formation.
- Calculation of enthalpy of combustion.