Question

Select the correct relation:
${{C}}\,\left( {{{diamond}}} \right)\, + \,{{2}}\,{{{H}}_2}\,\left( g \right)\, \to \,{{C}}{{{H}}_4}\,\left( g \right)\,;\,\,\,\,\vartriangle {{{H}}_1} \\
  {{C}}\,\left( g \right)\, + \,4\,{{H}}\, \to \,{{C}}{{{H}}_4}\,\left( g \right)\,\,;\,\,\,\vartriangle {{{H}}_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
$
A.$\Delta {{{H}}_1}\, = \,\Delta {{{H}}_2}$
B.\[\Delta {{{H}}_1}\,{{ > }}\,\Delta {{{H}}_2}\]
C.$\Delta {{{H}}_{1\,}}\, < \,\Delta {{{H}}_2}$
D.$\Delta {{{H}}_1}\, = \,\Delta {{{H}}_2}\, + \,{\Delta _{{{vap}}}}{{{H}}_{{C}}}\, + \,{\Delta _{{{diss}}}}{{{H}}_{{{{H}}_{{2}}}}}$

Answer 1

Hint: The enthalpy of a reaction which is also known as the heat of a reaction is the change in the enthalpy of a chemical reaction that occurs at constant pressure. It is a thermodynamic measurement which is useful for calculating the amount of energy per mole released or produced during a reaction. Also enthalpy is a state function, whose value doesn’t depend upon the path followed in the reaction but only depends on the initial and final state of the system.

Complete step by step solution: For a chemical reaction, enthalpy of reaction is defined as the difference in enthalpy between products and reactants. The units of enthalpy of a reaction is measured in kilojoules per mole. Reversing the chemical reaction reverses the sign of enthalpy.
Hence enthalpy of a reaction is given by
\[\Delta {{H}}\, = \,{{{H}}_{{f}}}\, - \,{{{H}}_{{i}}}\]
For a chemical reaction it can written as $\Delta {{H}}\, = \,{{{H}}_{{P}}} - {{{H}}_{{R}}}$
A positive $\Delta {{H}}$ value indicates endothermic reaction whereas a negative $\Delta {{H}}$ value indicates exothermic reaction.
Hess’s law is the most important law used for calculating enthalpy of a given chemical reaction, it states that if reactions take place by more than one route and the initial and that of final conditions are same, then total enthalpy change is the same. That means enthalpy change is independent of the route of the reaction
If the values of any two of the three enthalpy changes are known then the third one can easily workout.
Here given that, the formation of methane by reaction
${{C}}\,\left( {{s}} \right)\, + \,2{{{H}}_2}\left( {{g}} \right) \to \,{{C}}{{{H}}_4}\left( {{g}} \right)\, \Leftrightarrow \Delta {{{H}}_1}$
Here the carbon is in solid state. We have to find the enthalpy change for the above reaction.
Here we are looking for the formation of methane which is also possible by other chemical reaction given by
${{C}}\,\left( {{g}} \right)\,\, + \,4\,{{H}}\,\left( {{g}} \right)\, \to \,{{C}}{{{H}}_4}\,\left( {{g}} \right)\, \Leftrightarrow \Delta {{{H}}_2}\, \to {{1}}$
Here the solid carbon convert to vapor state, vaporisation reaction take place as
${{C}}\,\left( {{s}} \right)\, \to \,{{C}}\,\left( {{g}} \right)$ $ \Leftrightarrow \Delta {{{H}}_{{{vap}}}}\, \to \,{{2}}$
where $\Delta {{{H}}_{{{Vap}}}}$ be the enthalpy of vaporization.
Next, two ${{{H}}_2}$ molecules dissociate into four hydrogen atoms. The reaction is as follows
$2{{{H}}_2}\left( {{g}} \right)\, \to \,4{{H}}\,\left( {{g}} \right) \Leftrightarrow \Delta {{{H}}_{{{diss}}}} \to {{3}}$
$\vartriangle {{{H}}_{{{diss}}}}$ be the dissociation enthalpy
Now we can calculate the enthalpy change of the reaction by using Hess’s law.
By adding above three reactions (please note that if same atoms/molecules comes on reactants and product side the it has to be cancelled)
  $ {{{{C}}\,{{(g)}}}}\, + \, {{4\,{{H}}\left( g \right)}} \to \,{{C}}{{{H}}_{{4}}}\left( {{g}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {{{H}}_2}\, \\
  {{C}}\left( {{s}} \right) \to {{{{C}}\left( {{g}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {{{H}}_{{{vap}}}} \\
  {{2}}{{{H}}_{{2}}}\left( {{g}} \right) \to {{{{4H}}\left( {{g}} \right)\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {{{H}}_{{{diss}}}} \\
{{{{C(g}}) + {{2}}{{{H}}_{{2}}}{{(g)}} \to {{C}}{{{H}}_{{4}}}{{(g)}}}}$
From this the enthalpy of formation for the reaction is given by
$\Delta {{{H}}_2} + \Delta {{{H}}_{{{vap}}}} + \Delta {{{H}}_{{{diss}}}}$
Here ${{C}}\left( {{g}} \right)$ comes in both product and reactant side so cancelled also $4{{H}}\left( {{g}} \right)$ comes on both side so it also cancelled so we get the reaction for which the enthalpy of formation given by $\Delta {{{H}}_1}$.
So it is clear that $\Delta {{{H}}_1} = \,\Delta {{{H}}_2} + \Delta {{{H}}_{{{vap}}}} + \Delta {{{H}}_{{{diss}}}}$.

Hence option D is correct.

Note: Three factors can affect enthalpy of a reaction that are
1.The concentrations of reactants and products.
2.The temperature of the system
3.The partial pressure of gases involved.