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Entropy

Δ S for the r...

Question

$Δ S$ for the reaction, $M g C O_{3} (s) \to M g O (s) + C O_{2} (g)$ will be:

A

zero

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B

- ve

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C

+ ve

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D

\infty

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Solution

The correct option is C $+ ve$
In the reaction,
$M g C O_{3} (s) \to M g O (s) + C O_{2} (g)$
$M g C O_{3} (s)$ decomposes to give $M g O (s)$ and $C O_{2} (g)$ .

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 $Δ n_{g} = 1 - 0 = 1$
Thus, entropy increases.

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