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Hint :Thermodynamics is the study of heat, work and temperature and their relation with energy and other physical properties of matter. Enthalpy (H) is a state function that measures energy of a system. Enthalpy is the sum of internal energy (E) and product of pressure and volume. Internal energy (E) is the total energy of the system- sum of kinetic and potential energy.
Complete Step By Step Answer:
The first law of thermodynamics adopts the law of conservation of energy and can be formulated as:
$ \Delta H = \Delta E + P\Delta V $
where, $ \Delta H $ is the change in enthalpy, $ \Delta E $ is the change in internal energy, $ P $ is the pressure and $ \Delta V $ is the change in volume.
From ideal gas equation, $ PV = nRT $ , the above equation can be rewritten as:
$ \Delta H = \Delta E + \Delta {n_g}RT $
$ \Delta {n_g} $ $ = $ change in number of moles in gaseous form $ = $ $ {n_p} - {n_r} $
R $ = $ universal gas constant
T $ = $ Temperature in Kelvin
$ \Delta H \ne \Delta E\,\, $ when $ \Delta {n_g}\, \ne \,0 $
Now, let us examine the given reactions :
(A) $ {H_2}(g)\, + \,{I_2}(g)\, \to \,2HI(g) $
Here, $ \Delta {n_g}\, = $ $ 2 - 2\, = \,0 $
Therefore, $ \Delta H\, = \,\Delta E $ .
(B) $ HCl(aq)\, + \,NaOH(aq)\, \to \,NaCl(aq) + {H_2}O(l) $
This is an acid base neutralisation reaction. There are neither gaseous products or reactants. $ \Delta H\, = \,\Delta E $ .
(C) $ C(s)\, + \,{O_2}(g)\, \to \,C{O_2}(g) $
$ \Delta {n_g}\, = \,1 - 1 = 0 $
Therefore, $ \Delta H\, = \,\Delta E $ .
(D) $ {N_2}(g)\, + \,3{H_2}(g)\, \to \,2N{H_3}(g) $
Here, $ \Delta {n_g}\, = \,2 - 4 = - 2 $
The equation becomes $ \Delta H = \Delta E - 2RT $ .
$ \Delta H \ne \Delta E $ in the above chemical reaction.
The correct option is (D) $ {N_2}(g)\, + \,3{H_2}(g)\, \to \,2N{H_3}(g) $ .
Note :
The thermodynamic state of a system is a function of temperature, pressure, and quantity of a substance. State functions depend only on these parameters and do not depend on the path taken. Enthalpy and internal energy are state functions. Certain functions like heat and work depend on the path followed. They are called path functions.
Complete Step By Step Answer:
The first law of thermodynamics adopts the law of conservation of energy and can be formulated as:
$ \Delta H = \Delta E + P\Delta V $
where, $ \Delta H $ is the change in enthalpy, $ \Delta E $ is the change in internal energy, $ P $ is the pressure and $ \Delta V $ is the change in volume.
From ideal gas equation, $ PV = nRT $ , the above equation can be rewritten as:
$ \Delta H = \Delta E + \Delta {n_g}RT $
$ \Delta {n_g} $ $ = $ change in number of moles in gaseous form $ = $ $ {n_p} - {n_r} $
R $ = $ universal gas constant
T $ = $ Temperature in Kelvin
$ \Delta H \ne \Delta E\,\, $ when $ \Delta {n_g}\, \ne \,0 $
Now, let us examine the given reactions :
(A) $ {H_2}(g)\, + \,{I_2}(g)\, \to \,2HI(g) $
Here, $ \Delta {n_g}\, = $ $ 2 - 2\, = \,0 $
Therefore, $ \Delta H\, = \,\Delta E $ .
(B) $ HCl(aq)\, + \,NaOH(aq)\, \to \,NaCl(aq) + {H_2}O(l) $
This is an acid base neutralisation reaction. There are neither gaseous products or reactants. $ \Delta H\, = \,\Delta E $ .
(C) $ C(s)\, + \,{O_2}(g)\, \to \,C{O_2}(g) $
$ \Delta {n_g}\, = \,1 - 1 = 0 $
Therefore, $ \Delta H\, = \,\Delta E $ .
(D) $ {N_2}(g)\, + \,3{H_2}(g)\, \to \,2N{H_3}(g) $
Here, $ \Delta {n_g}\, = \,2 - 4 = - 2 $
The equation becomes $ \Delta H = \Delta E - 2RT $ .
$ \Delta H \ne \Delta E $ in the above chemical reaction.
The correct option is (D) $ {N_2}(g)\, + \,3{H_2}(g)\, \to \,2N{H_3}(g) $ .
Note :
The thermodynamic state of a system is a function of temperature, pressure, and quantity of a substance. State functions depend only on these parameters and do not depend on the path taken. Enthalpy and internal energy are state functions. Certain functions like heat and work depend on the path followed. They are called path functions.
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