Chapter 10, Problem 45E

(a)

Interpretation Introduction

Interpretation:

The sign of the entropy change for the following process should be predicted and value of $\Delta S{°}_{298}$ should be calculated.

  ${\text{H}}_2\text{(g)}+\frac{1}{2}{\text{O}}_2\text{(g)}\to {\text{H}}_{\text{2}}\text{O(l)}$

Concept Introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged. In different physical states, entropy depends upon the phase of the substance; the order of the entropy is given by: $S_{solid}<S_{liquid}<S_{gas}$

The sign of the entropy should be predicted with the help of above information.

Answer to Problem 45E

Sign of entropy is negative as the randomness decreases.

  $\Delta S{°}_{298}=-163.5\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{H}}_2\text{(g)}+\frac{1}{2}{\text{O}}_2\text{(g)}\to {\text{H}}_{\text{2}}\text{O(l)}$

In this case, sign of entropy is negative as the number of moles of reactant in gaseous state is more in comparison to number of moles of product in gaseous state. Due to decreases in randomness, sign of entropy is negative.

The mathematical expression for the standard entropy value at room temperature is:

  $\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for ${\text{H}}_{\text{2}}\text{O(l)}$ is $70\text{ J/Kmol}$

The value of standard entropy for ${\text{O}}_2(g)$ is $205.2\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}_2(g)$ is $131\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}_2\text{O(g)}$ is $189\text{ J/Kmol}$

Put the values, we get:

  $\Delta S{°}_{298}=\left(1\times {\text{S°}}_{\text{298}}{\text{(H}}_{\text{2}}\text{O(g}))\right)-\left(1\times {\text{S°}}_{\text{298}}{\text{(H}}_{\text{2}}(g))\frac{1}{2}\times {\text{S°}}_{\text{298}}{\text{(O}}_{\text{2}}(g))\right)$

  $\Delta S{°}_{298}=[(1\times 70)-(1\times 131+\frac{1}{2}\times 205)]\text{ J/Kmol}$

  $\Delta S{°}_{298}=-163.5\text{ J/Kmol}$

(b)

Interpretation Introduction

Interpretation:

The sign of the entropy change for the following process should be predicted and value of $\Delta S{°}_{298}$ should be calculated.

  ${\text{2CH}}_{\text{3}}{\text{OH (g)+3O}}_2(g)\to {\text{2CO}}_{\text{2}}{\text{(g)+4H}}_{\text{2}}\text{O(g)}$

Concept Introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged. In different physical states, entropy depends upon the phase of the substance; the order of the entropy is given by: $S_{solid}<S_{liquid}<S_{gas}$

The sign of the entropy should be predicted with the help of above information.

Answer to Problem 45E

Sign of entropy is positive.

  $\Delta S{°}_{298}=+89\text{ J/Kmol}$

Explanation of Solution

Given process is: ${\text{2CH}}_{\text{3}}{\text{OH (g)+3O}}_2(g)\to {\text{2CO}}_{\text{2}}{\text{(g)+4H}}_{\text{2}}\text{O(g)}$

In this case, sign of entropy is positive as in this process, number of moles of products in gaseous state is more than the number of moles of reactants in gaseous state.

The mathematical expression for the standard entropy value at room temperature is:

  $\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for ${\text{CH}}_{\text{3}}\text{OH(g)}$ is $240\text{ J/Kmol}$

The value of standard entropy for ${\text{O}}_{\text{2}}\text{(g)}$ is $205\text{ J/Kmol}$

The value of standard entropy for ${\text{CO}}_2\text{(g)}$ is $214\text{ J/Kmol}$

The value of standard entropy for ${\text{H}}_2\text{O(g)}$ is $189\text{ J/Kmol}$

Put the values,

  $\Delta S{°}_{298}=\left(4\times {\text{S°}}_{\text{298}}{\text{(H}}_2\text{O}(g))+2\times {\text{S°}}_{\text{298}}{\text{(CO}}_{\text{2}}\text{(g)})\right)-\left(2\times {\text{S°}}_{\text{298}}{\text{(CH}}_3\text{OH(g))+3}\times {\text{S°}}_{\text{298}}{\text{(O}}_{\text{2}}\text{(g)}\right)$

  $\Delta S{°}_{298}=[(4\times 189+2\times 214)-(2\times 240+3\times 205)]\text{ J/Kmol}$

  $\Delta S{°}_{298}=+89\text{ J/Kmol}$

(c)

Interpretation Introduction

Interpretation:

The sign of the entropy change for the following process should be predicted and value of $\Delta S{°}_{298}$ should be calculated.

  $\text{HCl}(\text{g})\to {\text{H}}^{\text{+}}\text{(aq})+{\text{Cl}}^{-}\text{(aq)}$

Concept Introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged. In different physical states, entropy depends upon the phase of the substance; the order of the entropy is given by: $S_{solid}<S_{liquid}<S_{gas}$

The sign of the entropy should be predicted with the help of above information.

Answer to Problem 45E

Sign of entropy is negative.

  $\Delta S{°}_{298}=-130\text{ J/Kmol}$

Explanation of Solution

Given process is: $\text{HCl}(\text{g})\to {\text{H}}^{\text{+}}\text{(aq})+{\text{Cl}}^{-}\text{(aq)}$

In this case, number of moles of products in gaseous state is less than number of moles of reactants in gaseous state. Thus, sign of entropy is negative.

The mathematical expression for the standard entropy value at room temperature is:

  $\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for $\text{HCl(g)}$ is $\text{187 J/Kmol}$

The value of standard entropy for ${\text{H}}^{+}\text{(aq)}$ is $\text{0 J/Kmol}$

The value of standard entropy for ${\text{Cl}}^{-}\text{(aq)}$ is $57\text{ J/Kmol}$

Put the values,

  $\Delta S{°}_{298}=\left(1\times {\text{S°}}_{\text{298}}{\text{(H}}^{+}(aq))+1\times {\text{S°}}_{\text{298}}{\text{(Cl}}^{-}\text{(aq)})\right)-\left(1\times {\text{S°}}_{\text{298}}\text{(HCl(g))}\right)$

  $\Delta S{°}_{298}=[(1\times 0+1\times 57)-(1\times 187)]\text{ J/Kmol}$

  $\Delta S{°}_{298}=-130\text{ J/Kmol}$