Question

For which reaction from the following, $\Delta S$ will be maximum?

• A. $Ca\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to CaO\left(s\right)$
• B. $Ca{CO}_3\left(s\right)\to CaO\left(s\right)+{CO}_2\left(g\right)$
• C. $C\left(s\right)+O_2\left(g\right)\to {CO}_2\left(g\right)$
• D. $N_2\left(g\right)+O_2\left(g\right)\to 2NO\left(g\right)$

Answer 1

The correct option is B $Ca{CO}_3\left(s\right)\to CaO\left(s\right)+{CO}_2\left(g\right)$

Change in entropy i.e, $\Delta S$ is proportional to the change in randomness of the system.
Hence, the one in which the randomness increases maximum will have maximum $\Delta S$.
Since, when the number of gaseous molecules increases in any reaction then its randomness i.e movement increases as gases have more randomness compared to solid and liquid.
So, the reaction with maximum value of $\Delta n_g$ will be having maximum $\Delta S$.
$\Delta n_g=\text{change in gaseous moles in any reaction}$
Since, for option B we have maximum $\Delta n_g=1$
Hence, answer is option B.