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Question: Given, \({{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(l)\) \[{{H}_{2}}O(l)\to {{H}_{2}}O(g...

Given, H2(g)+12O2(g)→H2O(l){{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(l)
H2O(l)→H2O(g);ΔH=x4{{H}_{2}}O(l)\to {{H}_{2}}O(g);\Delta H={{x}_{4}}
Given,
EH−H=x1{{E}_{H-H}}={{x}_{1}}
EO−O=x2{{E}_{O-O}}={{x}_{2}}
EO−H=x3{{E}_{O-H}}={{x}_{3}}
ΔHf\Delta {{H}_{f}}ofH2O{{H}_{2}}O vapour is:
A. x1+x22−x3+x4{{x}_{1}}+\dfrac{{{x}_{2}}}{2}-{{x}_{3}}+{{x}_{4}}
B.2x3−x1−x22−x3−x42{{x}_{3}}-{{x}_{1}}-\dfrac{{{x}_{2}}}{2}-{{x}_{3}}-{{x}_{4}}
C.x1+x22−2x3−x4{{x}_{1}}+\dfrac{{{x}_{2}}}{2}-2{{x}_{3}}-{{x}_{4}}
D.x1+x22−2x3+x4{{x}_{1}}+\dfrac{{{x}_{2}}}{2}-2{{x}_{3}}+{{x}_{4}}

Explanation

Solution

The basic concept of physical chemistry that deals with the formula of the enthalpy of formation of the gaseous product of water by the production of liquid water molecule that is given by,ΔH\Delta H= (Bond energy of reactant) - (bond energy of product).

Complete answer:
We have studied in the previous classes of physical chemistry about the enthalpy of formation of a product and also several parameters like, entropy, internal energy calculation and also some other parameters that are related to the thermodynamics chapter of the physical chemistry part.
We shall now see in brief about the meaning of enthalpy and its calculation.
- Enthalpy is basically a property of the thermodynamic system which is defined as the sum of the internal energy of the system and the product of its pressure and volume which is given by,
H = E + PV
In terms of bond energies of the molecules, the enthalpy is calculated using the formula,
ΔH\Delta H= (Bond energy of reactant) - (bond energy of product).
ΔH=B.Ereactan⁡t−B.Eproduct\Delta H=B.{{E}_{reac\tan t}}-B.{{E}_{product}}
Since, the reaction given is,
H2(g)+12O2(g)→H2O(l){{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(l)
where hydrogen is having two moles and oxygen is present in half moles and the product formed is one mole, the ΔH1\Delta {{H}_{1}} formula for this above reaction in terms of the data given in the form of ‘x’ can be written as shown below,
ΔH1=(x1+x22)−2x3\Delta {{H}_{1}}=\left( {{x}_{1}}+\dfrac{{{x}_{2}}}{2} \right)-2{{x}_{3}}
The other reaction is,
H2O(l)→H2O(g){{H}_{2}}O(l)\to {{H}_{2}}O(g) and theΔH2=x4\Delta {{H}_{2}}={{x}_{4}}for this reaction
Thus, the formation of water vapour that is,
ΔHf=ΔH1+ΔH2\Delta {{H}_{f}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}
⇒ΔHf=x1+x22−2x3+x4\Rightarrow \Delta {{H}_{f}}={{x}_{1}}+\dfrac{{{x}_{2}}}{2}-2{{x}_{3}}+{{x}_{4}}

Therefore, the correct answer is option D).

Note:
Do not be confused with the parameters entropy and enthalpy because these two are different terms where enthalpy is the measure of total heat present in the thermodynamic system with constant pressure whereas entropy is the measure of disorder in the thermodynamic system.