Chapter 13, Problem 61E

Interpretation Introduction

(a)

Interpretation:

For the given reaction, the value of $\text{ΔrS}°$ at 298 K needs to be calculated.

$2NaHC{O}_3\left(s\right)\to N{a}_{2}C{O}_3\left(s\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$

Concept introduction:

The Gibb’s equation of thermodynamic purposed a relation between $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$ with temperature. The mathematical expression of Gibb’s equation can be written as:

$\text{ΔrG}°\text{ = ΔrH}°\text{ - TΔrS}°$

Here, $\text{ ΔrH}°$ and $\text{ΔrS}°$ can be calculated with help of given equation:

$\begin{array}{l}\text{ΔrH° = ΔrH°(product) - ΔrH°(reactant) }\\ \text{ΔrS° = ΔrS°(product) - ΔrS°(reactant) }\end{array}$

For 298 K, the value of $\text{ΔrG}°$ can also be calculated using the following relation:

$\text{ΔrG° = }\sum \text{ΔrG°(product)- }\sum \text{ΔrG°(reactant)}$

Interpretation Introduction

(b)

Interpretation:

For the given reaction, the value of $\text{ΔrH}°$ at 298 K needs to be calculated.

$2NaHC{O}_3\left(s\right)\to N{a}_{2}C{O}_3\left(s\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$

Concept introduction:

The Gibb’s equation of thermodynamic purposed a relation between $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$ with temperature. The mathematical expression of Gibb’s equation can be written as:

$\text{ΔrG}°\text{ = ΔrH}°\text{ - TΔrS}°$

Here, $\text{ ΔrH}°$ and $\text{ΔrS}°$ can be calculated with help of given equation:

$\begin{array}{l}\text{ΔrH° = ΔrH°(product) - ΔrH°(reactant) }\\ \text{ΔrS° = ΔrS°(product) - ΔrS°(reactant) }\end{array}$

For 298 K, the value of $\text{ΔrG}°$ can also be calculated using the following relation:

$\text{ΔrG° = }\sum \text{ΔrG°(product)- }\sum \text{ΔrG°(reactant)}$

${\text{K}}_{\text{p}}$ or $\text{Kc}$ are the equilibrium constants for the reaction which are ratio of gaseous and aqueous products to the reactant molecules. The relation between equilibrium constant and $\Delta \text{rG}°$ can be written as:

$\text{ΔrG° = - 2}\text{.303 RT log K}$

Here:

• R = 8.314 J / mol K
• T = temperature in Kelvin

Interpretation Introduction

(c)

Interpretation:

For the given reaction, the value of $\text{ΔrG}°$ at 298 K needs to be calculated.

$2NaHC{O}_3\left(s\right)\to N{a}_{2}C{O}_3\left(s\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$

Concept introduction:

The Gibb’s equation of thermodynamic purposed a relation between $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$ with temperature. The mathematical expression of Gibb’s equation can be written as:

$\text{ΔrG}°\text{ = ΔrH}°\text{ - TΔrS}°$

Here, $\text{ ΔrH}°$ and $\text{ΔrS}°$ can be calculated with help of given equation:

$\begin{array}{l}\text{ΔrH° = ΔrH°(product) - ΔrH°(reactant) }\\ \text{ΔrS° = ΔrS°(product) - ΔrS°(reactant) }\end{array}$

For 298 K, the value of $\text{ΔrG}°$ can also be calculated using the following relation:

$\text{ΔrG° = }\sum \text{ΔrG°(product)- }\sum \text{ΔrG°(reactant)}$

${\text{K}}_{\text{p}}$ or $\text{Kc}$ are the equilibrium constants for the reaction which are ratio of gaseous and aqueous products to the reactant molecules. The relation between equilibrium constant and $\Delta \text{rG}°$ can be written as:

$\text{ΔrG° = - 2}\text{.303 RT log K}$

Here:

• R = 8.314 J / mol K
• T = temperature in Kelvin

Interpretation Introduction

(d)

Interpretation:

For the given reaction, the value of equilibrium constant $K$ at 298 K needs to be calculated.

$2NaHC{O}_3\left(s\right)\to N{a}_{2}C{O}_3\left(s\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$

Concept introduction:

The Gibb’s equation of thermodynamic purposed a relation between $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$ with temperature. The mathematical expression of Gibb’s equation can be written as:

$\text{ΔrG}°\text{ = ΔrH}°\text{ - TΔrS}°$

Here, $\text{ ΔrH}°$ and $\text{ΔrS}°$ can be calculated with help of given equation:

$\begin{array}{l}\text{ΔrH° = ΔrH°(product) - ΔrH°(reactant) }\\ \text{ΔrS° = ΔrS°(product) - ΔrS°(reactant) }\end{array}$

For 298 K, the value of $\text{ΔrG}°$ can also be calculated using the following relation:

$\text{ΔrG° = }\sum \text{ΔrG°(product)- }\sum \text{ΔrG°(reactant)}$

With the help of this equation one can predict the change in $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$.

The ${\text{K}}_{\text{p}}$ or $\text{Kc}$ are the equilibrium constants for the reaction which are ratio of gaseous and aqueous products to the reactant molecules. The relation between equilibrium constant and $\Delta \text{rG}°$ can be written as:

$\text{ΔrG° = - 2}\text{.303 RT log K}$

Here:

• R = 8.314 J / mol K
• T = temperature in Kelvin