Question

$\Delta S$ for the reaction, $MgC{O}_3\left(s\right)\to MgO\left(s\right)+C{O}_2\left(g\right)$ will be:

• A. $\text{zero}$
• B. $-\text{ve}$
• C. $+\text{ve}$
• D. $\infty$

Answer 1

The correct option is C $+\text{ve}$

In the reaction,
$MgC{O}_3\left(s\right)\to MgO\left(s\right)+C{O}_2\left(g\right)$
$MgC{O}_3\left(s\right)$ decomposes to give $MgO\left(s\right)$ and $C{O}_2\left(g\right)$.

Entropy of gases is more than that of solids and liquids.
Entropy increases when there is an increase in the number of moles (gaseous) on the product side i.e., when $△n_g>0$
$△n_g=n_p-n_R$
$n_P=$ total number of gaseous moles of products
$n_R=$ total number of gaseous moles of reactants
So, the value of $\Delta n_g=1-0=1$
Thus, entropy increases.