Question

127. Which of the following reactions (or processes) are expected to have a negative value for $\mathbf{∆}{\mathbf{S}}^{\mathbf{o}}$?
a. ${\mathbf{SiF}}_{\mathbf{6}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{HF}\left(g\right)\mathbf{+}{\mathbf{SiF}}_{\mathbf{4}}\left(g\right)$
b.$\mathbf{4}\mathbf{Al}\left(s\right)\mathbf{+}\mathbf{3}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\left(s\right)$
c.$\mathbf{CO}\left(g\right)\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{COCl}}_{\mathbf{2}}\left(g\right)$
d.${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\left(g\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{\to }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\left(l\right)$
e.${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(s\right)\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)$

Short answer

Determination ofnegative value for$\mathbf{∆}{\mathbf{S}}^{\mathbf{o}}$,

1. $\mathbf{∆}{\mathbf{S}}^{\mathbf{o}}$is positive value
2. $\mathbf{∆}{\mathbf{S}}^{\mathbf{o}}$is negative value
3. $\mathbf{∆}{\mathbf{S}}^{\mathbf{o}}$is negative value
4. $\mathbf{∆}{\mathbf{S}}^{\mathbf{o}}$is negative value
5. $\mathbf{∆}{\mathbf{S}}^{\mathbf{o}}$is positive value

Step-by-step solution

Step 1: Introduction to the Concept

The entropy of a molecule is a measure of its disorder or unpredictability. The higher the unpredictability, the higher will be the entropy of the system. Atoms and molecules are confined to a fixed place in solids. Atoms and molecules are free to move about in the liquid phase, although they remain relatively close together. Atoms and molecules are free to move in a significantly larger volume in the gas phase than they are in the liquid phase. As a result, the entropy of substances varies as follows:

${\mathbf{S}}_{\mathbf{gas}}\mathbf{>}{\mathbf{S}}_{\mathbf{liquid}}\mathbf{>}{\mathbf{S}}_{\mathbf{solid}}$

Step 2: Determination of negative value of reaction

a)

The given reaction is,

${\mathrm{SiF}}_6\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2\mathrm{HF}\left(\mathrm{g}\right)+{\mathrm{SiF}}_4\left(\mathrm{g}\right)$

The reaction starts with one mole of liquid substance and one mole of gaseous substance and ends with three moles of gaseous substance. The entropy of the system grows as the number of moles of substance in the gas phase increases. As a result,entropy change is positive.

Step 3: Determination of negative value of reaction

b)

The given reaction is,

$4\mathrm{Al}\left(\mathrm{s}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$

Starting with four moles of solid substance and three moles of gas phase substance, the reaction produces two moles of solid phase substance. Because only solid-phase substances are created, the system's entropy reduces from its original value. As a result,entropy change is negative.

Step 4: Determination of negative value of reaction

c)

The given reaction is,

$\mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to {\mathrm{COCl}}_2\left(\mathrm{g}\right)$

In the gas phase, the reaction starts with two moles of material and ends with only one mole of substance in the gas phase. The entropy of the system has decreased since the quantity of moles of material in the gas phase has dropped. As a result,entropy change is negative.

Step 5: Determination of negative value of reaction

d)

The given reaction is,

${\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)$

The reaction begins with one mole of gaseous substance and one mole of liquid substance, and ends with one mole of liquid substance. Because the entropy of a substance in the liquid phase is lower than that of a substance in the gas phase, the system's entropy falls below its original value. As a result, entropy change is a negative number.

Step 6: Determination of negative value of reaction

e)

The given reaction is,

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

The reaction begins with one mole of a solid substance and ends with one mole of a liquid substance. Because the entropy of a substance in the liquid phase is higher than that of a substance in the solid phase, the system's entropy rises above its original value. As a result, entropy change is positive.