Question

The entropy change will be positive for which reaction?

• A. $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
• B. $HCl\left(g\right)+N{H}_3\left(g\right)\rightleftharpoons N{H}_{4}Cl\left(s\right)$
• C. $N{H}_{4}N{O}_3\left(s\right)\rightleftharpoons N_{2}O\left(g\right)+2H_{2}O\left(g\right)$
• D. $MgO\left(s\right)+H_2\left(g\right)\rightleftharpoons Mg\left(s\right)+H_{2}O\left(l\right)$

Answer 1

The correct option is C $N{H}_{4}N{O}_3\left(s\right)\rightleftharpoons N_{2}O\left(g\right)+2H_{2}O\left(g\right)$

Entropy change will be positive in the cases where the number of gaseous moles of the product is more than the number of gaseous moles of the reactants.
Which means,
$△n_g>0$
where, $△n_g=n_p-n_R$
$n_P=$ total number of gaseous moles of the products
$n_R=$ total number of gaseous moles of the reactants

In the cases,
A. $△n_g=2-2=0$
B. $△n_g=0-2=-2$
C. ​​​$△n_g=3-0=3$
D. $△n_g=0-1=-1$
So, we can clearly see only in case (c), the $△n_g$ is positive.
So, Entropy change will be +ve in case (C).