Question

For a reaction: $X\to Y+Z$
The absolute entropies for $X,Y$ and $Z$ are $120J{K}^{-1}mo{l}^{-1},213.8J{K}^{-1}mo{l}^{-1}and197.9J{K}^{-1}mo{l}^{-1}$ respectively.
What will be the entropy change of the reaction at 298 K and 1 atm?

• A. $291.7J{K}^{-1}$
• B. $255J{K}^{-1}$
• C. $213.8J{K}^{-1}$
• D. $257.3J{K}^{-1}$

Answer 1

The correct option is A $291.7J{K}^{-1}$

Change in entropy for a reaction is given as:
$\Delta S=$ Entropy of the product - Entropy of the reactant
So, $\Delta S$ = $S_{product}-S_{reactant}$
$\Delta S$ = $(S_Y+S_Z)-S_X$
= $(213.8+197.9)-120$
$\Delta S$ = $291.7J{K}^{-1}mo{l}^{-1}$

Theory:
$\Delta S=S_{final}-S_{initial}$

$\Delta S=\sum S_{products}-\sum S_{reactants}$