Question

a) Calculate the standard reaction entropy for the combustion of methane to carbon dioxide and liquid water at 298 K. b) Calculate the standard reaction entropy for the hydrolysis of liquid water to form oxygen and hydrogen gas. c) Do your results in (a) and (b) agree with the general rule that reaction entropies are positive if there is a net formation of gas in a reaction, and negative if the is a net reduction of gas?

Answer 1

Answer: (a) $\Delta S^{0}$ = - 242.2J/K.mol

              (b) $\Delta S^{0}$ = 163.34J/K.mol

              (c) Yes

Explanation: Entropy is the measure of randomness or disorder of a system. Standard Entropy has the symbol $S^{0}$.

Mathematically, Entropy is defined as

$\Delta S^{0}=\Sigma S^{0}(products)-\Sigma S^{0}(reagents)$

Each substance has its own entropy value.

(a) Combustion of Methane:

$CH_{4}+2O_{2}$ ⇒ $CO_{2}+2H_{2}O$

Using standard entropy table, entropy will be:

$\Sigma S^{0}(products)=214+2(69.91)$ = 353.8

$\Sigma S^{0}=186.264+2(205.138)$ = 596.54

$\Delta S^{0}=$ 353.8 - 596.54

$\Delta S^{0}=$ - 242.2

Standard Entropy for combustion of methane is $\Delta S^{0}=$ - 242.2 J/K.mol.

(b) Hydrolysis of Water:

$H_{2}O_{(l)}$ ⇒ $H_{2}_{(g)}+1/2O_{2}_{(g)}$

$\Sigma S^{0}(products)=130.684+0.5(205.138)$ = 233.253

$\Sigma S^{0}(reagents)=69.91$

$\Delta S^{0}=$ 233.253 - 69.91

$\Delta S^{0}=$ +163.34

Standard Entropy for hydrolysis of water is $\Delta S^{0}=$ + 163.34 J/K.mol

(c) Yes, the results prove the general rule: Reactions Entropies are positive for net formation of gas and negative for net reduction of gas. This happens because, as entropy is the measure of disorder, the entropy for gases are greater than for liquids and solids. So, when there is formation of a gas, Entropy increases and when there isn't, entropy decreases.