Question

$N_2\left(g\right)+{3H}_2\left(g\right)\to {2NH}_3\left(g\right);\Delta H^O=-44Kcal$
For the above reaction, the correct statement is:

• A. $\Delta H<\Delta E$
• B. $\Delta H=\Delta E$
• C. $\Delta H>\Delta E$
• D. none of these

Answer 1

Answer: (C)

$\Delta H>\Delta E$

According to first law of thermodynamics-

$\Delta H=\Delta E+\Delta n_{g}RT$

where as,

$\Delta E$ = Internal energy

$\Delta H$ = Enthalpy

$\Delta n_g$ = difference between the final and initial numbers of moles of gas.

For all the reaction,

Case I:-

$\Delta n_g<0$

$\Rightarrow \Delta E>\Delta H$

Case II:-

$\Delta n_g>0$

$\Rightarrow \Delta H>\Delta E$

$\therefore$ For the given reaction,

${N_2}_{\left(g\right)}+3{H_2}_{\left(g\right)}⟶2{N{H}_3}_{\left(g\right)}$

$\Delta n_g=n_P-n_R$

$\Rightarrow \Delta n_g=2-(1+3)$

$\Rightarrow \Delta n_g=2-4=-2=-ve$

$\Rightarrow \Delta E<\Delta H$

Hence, the correct answer is $\Delta E<\Delta H$.