Question

For which one of the following system $\Delta E<\Delta H$?

• A. $2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\to 2S{O}_{3\left(g\right)}$
• B. $N_{2\left(g\right)}+O_{2\left(g\right)}\to 2N{O}_{\left(g\right)}$
• C. $2N{H}_{3\left(g\right)}\to N_{2\left(g\right)}+3H_{2\left(g\right)}$
• D. $H_{2\left(g\right)}+I_{2\left(g\right)}\to 2H{I}_{\left(g\right)}$

Answer 1

Answer: (C)

$2N{H}_{3\left(g\right)}\to N_{2\left(g\right)}+3H_{2\left(g\right)}$

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

As we know that,

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

For $\Delta n_g>0$

$\Delta H>\Delta E$

Now for the reaction-

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

$\Delta n_g=(3+1)-2=+2>0$

$\therefore \Delta H>\Delta E$

Hence for the reaction $2{N{H}_3}_{\left(g\right)}⟶{N_2}_{\left(g\right)}+3{H_2}_{\left(g\right)};\Delta E<\Delta H$.