Answer :
Answer : Only option A shows [tex]\Delta H=\Delta E[/tex]
Explanation :
Formula used :
[tex]\Delta H=\Delta E+\Delta n_gRT[/tex]
[tex]\Delta H[/tex] = change in enthalpy
[tex]\Delta E[/tex] = change in internal energy
[tex]\Delta n_g[/tex] = change in moles
R = gas constant = 8.314 J/mol.K
T = temperature
According to the question, [tex]\Delta H=\Delta E[/tex] when the value of [tex]\Delta n_gRT[/tex] will be zero.
Now we have to determine the value of [tex]\Delta n_gRT[/tex] for the following cases.
(A) [tex]2HI(g)\rightarrow H_2(g)+I_2(g)[/tex] at atmospheric pressure.
In this case:
[tex]\Delta n_g[/tex] = change in moles
Change in moles = Number of moles of product side - Number of moles of reactant side
According to the reaction:
Change in moles = (1+1) - 2 = 2 - 2 = 0 mole
That means, value of [tex]\Delta n_gRT[/tex] = 0
So, in this process [tex]\Delta H=\Delta E[/tex]
(B) Two moles of ammonia gas are cooled from 325 °C to 300 °C at 1.2 atm.
In this case:
[tex]\Delta n_g[/tex] = change in moles = 2
That means, value of [tex]\Delta n_gRT[/tex] ≠ 0
So, in this process [tex]\Delta H\neq \Delta E[/tex]
(C) [tex]H_2O(l)\rightarrow H_2O(g)[/tex] at 100 °C at atmospheric pressure.
In this case:
Change in moles = Number of moles of product side - Number of moles of reactant side
According to the reaction:
Change in moles = 1 - 0 = 1 mole
[tex]\Delta n_g[/tex] = change in moles = 1
That means, value of [tex]\Delta n_gRT[/tex] ≠ 0
So, in this process [tex]\Delta H\neq \Delta E[/tex]
(D) [tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex] at 800 °C at atmospheric pressure.
In this case:
Change in moles = Number of moles of product side - Number of moles of reactant side
According to the reaction:
Change in moles = 1 - 0 = 1 mole
[tex]\Delta n_g[/tex] = change in moles = 1
That means, value of [tex]\Delta n_gRT[/tex] ≠ 0
So, in this process [tex]\Delta H\neq \Delta E[/tex]
(E) [tex]CO_2(s)\rightarrow CO_2(g)[/tex] at atmospheric pressure.
In this case:
Change in moles = Number of moles of product side - Number of moles of reactant side
According to the reaction:
Change in moles = 1 - 0 = 1 mole
[tex]\Delta n_g[/tex] = change in moles = 1
That means, value of [tex]\Delta n_gRT[/tex] ≠ 0
So, in this process [tex]\Delta H\neq \Delta E[/tex]
Hence, from this we conclude that, only option A shows [tex]\Delta H=\Delta E[/tex]