Chapter 8, Problem 8.49AP

Interpretation Introduction

(a)

Interpretation:

The change in the concentration of the complex as the concentration of ethanol is increased is to be stated.

Concept introduction:

Equilibrium constant represented by $K$ gives the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature. It is expressed as the ratio of the concentrations of the products over the reactants raised to the power equal to their coefficients in the equation.

Answer to Problem 8.49AP

The concentration of the complex increases if the concentration of ethanol is increased.

Explanation of Solution

The equilibrium constant for the reaction is given as shown below.

$K_{\text{eq}}=\frac{{\left[\text{Reactants}\right]}^{\text{r}}}{{\left[\text{Products}\right]}^{\text{p}}}$ …(1)

Where,

• $K_{\text{eq}}$ is the equilibrium constant.
• ${\left[\text{Reactants}\right]}^{\text{r}}$ is the concentration of reactant raise to the power of its coefficient.
• ${\left[\text{Products}\right]}^{\text{p}}$ is the concentration of product raise to the power of its coefficient.

Figure 1

The value of equilibrium constant is $K_{\text{eq}}=11$ so, the reaction is favored in forward direction.

The equilibrium constant for the above reaction is given as shown below.

$K_{\text{eq}}=\frac{\left[\text{Complex}\right]}{{\left[\text{Ethanol}\right]}^2}$

If the concentration of ethanol is increased in order to reach a constant value of equilibrium constant $K_{\text{eq}}=11$, the concentration of the complex is also increased.

Conclusion

If the concentration of ethanol is increased the concentration of the complex also increases.

Interpretation Introduction

(b)

Interpretation:

The standard free-energy change for this reaction at $25°\text{C}$ is to be calculated.

Concept introduction:

The standard Gibbs free energy of a compound is the change in the formation of 1 mol of a substance from its component elements, at standard states like temperature and pressure. Its symbol is $\text{Δ}G°$.

Answer to Problem 8.49AP

The standard free-energy change for the reaction at $25°\text{C}$ is $-5.9\text{kJ}{\text{mol}}^{-1}$.

Explanation of Solution

The expression to calculate the standard free-energy change is given as shown below.

$\text{Δ}G°=-2.3\text{RT}\log K_{\text{eq}}$ …(2)

Where,

• $K_{\text{eq}}$ is the equilibrium constant.
• $\text{T}$ is the temperature of the reaction.
• $\text{R}$ is the universal gas constant.
• $\text{Δ}G°$ is the standard free energy change.

Substitute the value of $2.3\text{RT}=\text{5}\text{.7}\text{kJ}{\text{mol}}^{-1}$ at $25°\text{C}$ and $K_{\text{eq}}=11$ in equation (2) as shown below.

$\begin{array}{rll}\text{Δ}G°& =& -\text{5}\text{.7}\left(\log 11\right)\text{kJ}{\text{mol}}^{-1}\\ & =& -\text{5}\text{.7}\times 1.0413\text{kJ}{\text{mol}}^{-1}\\ & =& -5.9\text{kJ}{\text{mol}}^{-1}\end{array}$

Conclusion

The standard free-energy change for this reaction at $25°\text{C}$ is found to be $-5.9\text{kJ}{\text{mol}}^{-\text{1}}$.

Interpretation Introduction

(c)

Interpretation:

The concentrations of free ethanol and complex if one mole of ethanol is dissolved in one liter of ${\text{CCl}}_4$ are to be stated.

Concept introduction:

Equilibrium constant represented by $K$ gives the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature. It is expressed as the ratio of the concentrations of the products over the reactants raised to the power equal to their coefficients in the equation.

Answer to Problem 8.49AP

The concentrations of free ethanol and complex are $1\text{M}$ and $\text{11}\text{M}$ respectively.

Explanation of Solution

Molarity is the number of moles per unit volume so according to question one mole of ethanol is dissolved in one liter of ${\text{CCl}}_4$ so the concentration of ethanol is 1M.

The equilibrium constant for the reaction is given as shown below.

$K_{\text{eq}}=\frac{{\left[\text{Reactants}\right]}^{\text{r}}}{{\left[\text{Products}\right]}^{\text{p}}}$ …(1)

Where,

• $K_{\text{eq}}$ is the equilibrium constant.
• ${\left[\text{Reactants}\right]}^{\text{r}}$ is the concentration of reactant raise to the power its coefficient.
• ${\left[\text{Products}\right]}^{\text{p}}$ is the concentration of product raise to the power its coefficient.

Substitute the value of concentration of ethanol is $1\text{M}$ and $K_{\text{eq}}=11$ in equation (1) $\begin{array}{rll}11{\text{M}}^{-1}& =& \frac{\left[\text{Complex}\right]}{{\left(1\text{M}\right)}^2}\\ \left[\text{Complex}\right]& =& 11{\text{M}}^{-1}\times {\left(1\text{M}\right)}^2\\ & =& 11\text{M}\end{array}$

Conclusion

The concentration of $1\text{M}$ and $11\text{M}$ are the concentrations of free ethanol and complex respectively.

Interpretation Introduction

(d)

Interpretation:

Among thiols and alcohols which form stronger hydrogen bonds if the equilibrium constant for the analogous reaction of ethanethiol is $0.004$ is to be stated.

Concept introduction:

Equilibrium constant represented by $K$ gives the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature. It is expressed as the ratio of the concentrations of the products over the reactants raised to the power equal to their coefficients in the equation.

Answer to Problem 8.49AP

The hydrogen bond is much stronger in ethanol than in ethanethiol.

Explanation of Solution

The equilibrium constant for the above reaction is given as shown below.

$K_{\text{eq}}=\frac{\left[\text{Complex}\right]}{{\left[\text{Ethanol}\right]}^2}$ …(3)

This shows that the concentration of complex is directly proportional to the equilibrium constant; more the concentration of complex stronger will be the hydrogen bond.

The equilibrium constant for the analogous reaction of ethanethiol is $0.004$ and for ethanol is $K_{\text{eq}}=11$ so from equation (3) the concentration of ethanol is higher than that of ethanethiol. Therefore, hydrogen bonding is much stronger in ethanol as compared to ethanethiol.

Conclusion

Hydrogen bonding in ethanol is stronger than that of ethanethiol.

Interpretation Introduction

(e)

Interpretation:

The more water soluble compound among ${\text{CH}}_3{\text{OCH}}_{\text{2}}{\text{CH}}_2\text{SH}$ and its isomer ${\text{CH}}_3{\text{SCH}}_{\text{2}}{\text{CH}}_2\text{OH}$ is to be stated.

Concept introduction:

The solubility of a compound depends upon the strength of hydrogen bond; stronger the hydrogen bond more will be its solubility. The compounds having weak hydrogen bonding results in lower solubility.

Answer to Problem 8.49AP

The compound ${\text{CH}}_3{\text{SCH}}_{\text{2}}{\text{CH}}_2\text{OH}$ is more soluble in water than ${\text{CH}}_3{\text{OCH}}_{\text{2}}{\text{CH}}_2\text{SH}$.

Explanation of Solution

The hydrogen bonding in alcohol is much stronger than that in thiol so ${\text{CH}}_3{\text{SCH}}_{\text{2}}{\text{CH}}_2\text{OH}$ will make stronger bond in water than ${\text{CH}}_3{\text{OCH}}_{\text{2}}{\text{CH}}_2\text{SH}$. The solubility of a compound depends upon the strength of hydrogen bond, stronger the hydrogen bond more will be its solubility. Therefore, solubility of ${\text{CH}}_3{\text{SCH}}_{\text{2}}{\text{CH}}_2\text{OH}$ is higher than the solubility of ${\text{CH}}_3{\text{OCH}}_{\text{2}}{\text{CH}}_2\text{SH}$.

Conclusion

The compound ${\text{CH}}_3{\text{SCH}}_{\text{2}}{\text{CH}}_2\text{OH}$ is found to be more soluble in water than ${\text{CH}}_3{\text{OCH}}_{\text{2}}{\text{CH}}_2\text{SH}$.