Question

Which reaction from the following, will have maximum entropy change?

• A. Ca(s) + $\frac{1}{2}$O$_2$(g) → CaO(s)
• B. CaCO$_3$(s) → CaO(s) + CO$_2$(g)
• C. C(s) + O$_2$(g) → CO$_2$(g)
• D. N$_2$(g) + O$_2$(g) → 2NO(g)

Answer 1

Answer: (B)

CaCO$_3$(s) → CaO(s) + CO$_2$(g)

Answer 2

The entropy change ($\Delta S$) of a reaction is primarily determined by the change in the number of moles of gaseous species ($\Delta n_g$). An increase in the number of gas moles leads to a positive entropy change (increase in disorder), while a decrease leads to a negative entropy change (decrease in disorder).

Let's analyze each reaction:

1. Ca(s) + $\frac{1}{2}$O$_2$(g) → CaO(s)

   • Moles of gaseous reactants = 0.5
   • Moles of gaseous products = 0
   • $\Delta n_g = 0 - 0.5 = -0.5$
   • Entropy decreases ($\Delta S < 0$).

2. CaCO$_3$(s) → CaO(s) + CO$_2$(g)

   • Moles of gaseous reactants = 0
   • Moles of gaseous products = 1
   • $\Delta n_g = 1 - 0 = +1$
   • Entropy increases significantly ($\Delta S > 0$).

3. C(s) + O$_2$(g) → CO$_2$(g)

   • Moles of gaseous reactants = 1
   • Moles of gaseous products = 1
   • $\Delta n_g = 1 - 1 = 0$
   • Entropy change is relatively small, as the number of gas moles remains constant.

4. N$_2$(g) + O$_2$(g) → 2NO(g)

   • Moles of gaseous reactants = 1 + 1 = 2
   • Moles of gaseous products = 2
   • $\Delta n_g = 2 - 2 = 0$
   • Entropy change is relatively small, as the number of gas moles remains constant.

Comparing the $\Delta n_g$ values:

• Option 1: -0.5
• Option 2: +1
• Option 3: 0
• Option 4: 0

The reaction with the largest positive value of $\Delta n_g$ will have the maximum positive entropy change. In this case, Option 2 has $\Delta n_g = +1$, which is the largest positive change.