Chapter 6, Problem 6.35P

Interpretation Introduction

(a)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.35P

The formation of starting material is favored at the given value of $K_{\text{eq}}$.

Explanation of Solution

Given

The value of $K_{\text{eq}}$ is $0.5$.

The given value of $K_{\text{eq}}$ is smaller than $1$. It implies that the concentration of the starting material is comparatively higher than that of the product at equilibrium. Therefore, the formation of starting material is favored at equilibrium.

Conclusion

(a) The formation of starting material is favored at the given value of $K_{\text{eq}}$.

Interpretation Introduction

(b)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ respectively. They are state functions. The relation of $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ is shown as,

$\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}$

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Answer to Problem 6.35P

The formation of the productis favored at the given value of $\Delta G°$.

Explanation of Solution

Given:

The value of $\Delta G°$ is $-100\text{kJ}/\text{mol}$.

The given value of $\Delta G°$ is negative. The reaction becomes spontaneous if $\Delta G°$ has negative value. Therefore, for the given value of $\Delta G°$, the formation of the product is favored at equilibrium.

Conclusion

The formation of the product is favored at the given value of $\Delta G°$.

Interpretation Introduction

(c)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ respectively. They are state functions. The relation of $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ is shown as,

$\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}$

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Answer to Problem 6.35P

The formation of the starting material is favoredat the given values of $\text{Δ}H\text{°}$.

Explanation of Solution

Given

The values of $\text{Δ}H\text{°}$ are $\text{8}\text{.0}\text{kJ}/\text{mol}$ and $200\text{kJ}/\text{mol}$.

The given values of $\text{Δ}H\text{°}$ are positive. It indicates that bond strength in starting material is stronger than bond strength in the product. Therefore, the breaking of bonds in the starting material becomes difficult. Therefore, the formation of the starting material is favored at equilibrium.

Conclusion

The formation of the starting material is favouredat the given values of $\text{Δ}H\text{°}$.

Interpretation Introduction

(d)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy is represented by $\Delta G°$. It is a state function. The value of $\Delta G°$ depends on $K_{\text{eq}}$. The free energy change is calculated as,

$\text{Δ}G\text{°}=-2.303\text{R}T\log K_{\text{eq}}$

If the $\Delta G°$ is greater than zero and $K_{\text{eq}}$ is smaller than one, then the formation of starting material is favored at equilibrium. However, if the $\Delta G°$ is smaller than zero and $K_{\text{eq}}$ is greater than one, the formation of product is favored at equilibrium.

Answer to Problem 6.35P

The formation of the productis favored at the given value of $K_{\text{eq}}$.

Explanation of Solution

Given

The value of $K_{\text{eq}}$ is $16$.

The given value of $K_{\text{eq}}$ is greater than $1$. It implies that the concentration of the starting material is comparatively lower than that of the product. Therefore, the formation of product is favored at equilibrium.

Conclusion

The formation of the product is favored at the given value of $K_{\text{eq}}$.

Interpretation Introduction

(e)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ respectively. They are state functions. The relation of $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ is shown as,

$\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}$

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Answer to Problem 6.35P

The formation of the starting materialis favored at the given value of $\Delta G°$.

Explanation of Solution

Given

The value of $\Delta G°$ is $2.0\text{kJ}/\text{mol}$.

The given value of $\Delta G°$ is positive. The reaction becomes spontaneous if $\Delta G°$ has negative Value. Therefore, at the given value of $\Delta G°$ the formation of the starting material is favored at equilibrium.

Conclusion

The formation of the starting material is favored at the given value of $\Delta G°$.

Interpretation Introduction

(f)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ respectively. They are state functions. The relation of $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ is shown as,

$\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}$

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Answer to Problem 6.35P

The formation of the productis favored at the given value of $\text{Δ}S\text{°}$.

Explanation of Solution

Given

The value of $\text{Δ}S\text{°}$ is $8\text{J}/\left(\text{K}\cdot \text{mol}\right)$.

For the spontaneous reaction, the value of $\Delta G°$ and $\text{Δ}S\text{°}$ must be negative and positive respectively. The given value of $\text{Δ}S\text{°}$ is positive. Therefore, the formation of product is favored at equilibrium.

Conclusion

The formation of the product is favored at the given value of $\text{Δ}S\text{°}$.

Interpretation Introduction

(g)

Interpretation: The starting material or product which isfavored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ respectively. They are state functions. The relation of $\Delta G°$, $\text{Δ}H\text{°}$ and $\text{Δ}S\text{°}$ is shown as,

$\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}$

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Answer to Problem 6.35P

The formation of the starting material is favored at the given value of $\text{Δ}S\text{°}$.

Explanation of Solution

Given

The value of $\text{Δ}S\text{°}$ is $-8\text{J}/\left(\text{K}\cdot \text{mol}\right)$.

The value of $\Delta G°$ and $\text{Δ}S\text{°}$ must be negative and positive respectively for the spontaneous reaction. The given value of $\text{Δ}S\text{°}$ is negative. Hence, the reaction is non spontaneous and therefore, the formation of starting material is favored at equilibrium

Conclusion

The formation of the starting material is favored at the given value of $\text{Δ}S\text{°}$.