Chapter 5, Problem 87AP

Interpretation Introduction

Interpretation:

Whether the standard enthalpy of the reaction is equal to the standard enthalpy of formation or not is to be determined.

Concept introduction:

The standard enthalpy of a reaction is the amount of enthalpy to occur under standard conditions.

The standard enthalpy of a reaction can be determined using the equation given below:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum n}\Delta H_{\text{f}}^{\text{°}}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_{\text{f}}^{\text{°}}\left(\text{reactants}\right)$

Here, the stoichiometric coefficients are represented by m for the reactants and by n for the products, while the enthalpy of formation under standard conditions is represented by $\Delta H_{\text{f}}^{\text{°}}$.

The value of enthalpy of the formation of an element is zero at its most stable state.

Answer to Problem 87AP

Solution:

(a) $\Delta H{°}_{\text{rxn}}=\Delta H{°}_{\text{f}}$

(b) $\Delta H{°}_{\text{rxn}}\cancel{=}\Delta H{°}_{\text{f}}$

(c) $\Delta H{°}_{\text{rxn}}\cancel{=}\Delta H{°}_{\text{f}}$

(d) $\Delta H{°}_{\text{rxn}}\cancel{=}\Delta H{°}_{\text{f}}$

Explanation of Solution

a) ${\text{H}}_{\text{2}}\left(g\right)+\text{S}\left(\text{rhombic}\right)\to {\text{H}}_{\text{2}}\text{S}\left(g\right)$

The reaction is given as follows:

${\text{H}}_{\text{2}}\left(g\right)+\text{S}\left(\text{rhombic}\right)\to {\text{H}}_{\text{2}}\text{S}\left(g\right)$

Hydrogen in its ${\text{H}}_{\text{2}}\left(g\right)$ form and sulfur in its rhombic form are the standard or most stable states of hydrogen (diatomic) and sulfur (transition state). So, their enthalpy of formation is zero, due to which the enthalpy of the reaction becomes the enthalpy of formation for ${\text{H}}_{\text{2}}\text{S}\left(g\right)$.

For the given equation, the enthalpy of the reaction is as follows:

$\Delta H{°}_{\text{rxn}}=\left[\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{S}\left(g\right)\right)\right]-\left[\Delta H{°}_{\text{f}}\left({\text{H}}_2\left(g\right)\right)+\Delta H{°}_{\text{f}}\left(\text{S}\left(\text{rhombic}\right)\right)\right]$

The enthalpy of formation is zero for ${\text{H}}_{\text{2}}\left(g\right)$ and $\text{S}\left(\text{rhombic}\right)$.

Hence, for the reaction ${\text{H}}_{\text{2}}\left(g\right)+\text{S}\left(\text{rhombic}\right)\to {\text{H}}_{\text{2}}\text{S}\left(g\right)$, the value of $\Delta H{°}_{\text{rxn}}=\Delta H{°}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{S}\left(g\right)\right]$.

b) $\text{C}\left(\text{diamond}\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)$

The reaction is given as follows:

$\text{C}\left(\text{diamond}\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)$

Oxygen in its ${\text{O}}_{\text{2}}\left(g\right)$ form is the standard (diatomic) or most stable state of oxygen. So, its enthalpy of formation is zero. However, carbon is in its diamond form, and, for carbon, the most stable form is graphite, due to which its enthalpy of formation is not zero. Thus, the enthalpy of the reaction is not equal to the enthalpy of formation of ${\text{CO}}_{\text{2}}\left(g\right)$.

For the given equation, the enthalpy of the reaction is as follows:

$\Delta H{°}_{\text{rxn}}=\left[\Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\left(g\right)\right)\right]-\left[\Delta H{°}_{\text{f}}\left({\text{O}}_{\text{2}}\left(g\right)\right)+\Delta H{°}_{\text{f}}\left(C\left(\text{diamond}\right)\right)\right]$

The enthalpy of formation is zero for ${\text{O}}_2\left(g\right)$, but nonzero for $\text{C}\left(\text{diamond}\right)$.

Hence, for the reaction $\text{C}\left(\text{diamond}\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)$, the value of $\Delta H{°}_{\text{rxn}}\cancel{=}\Delta H{°}_{\text{f}}\left[{\text{CO}}_{\text{2}}\left(g\right)\right]$.

c) ${\text{H}}_{\text{2}}\left(g\right)+\text{CuO}\left(g\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+\text{Cu}\left(s\right)$

The reaction is given as follows:

${\text{H}}_{\text{2}}\left(g\right)+\text{CuO}\left(g\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+\text{Cu}\left(s\right)$

Hydrogen in its ${\text{H}}_{\text{2}}\left(g\right)$ form is the standard (diatomic) or most stable state of hydrogen. So, its enthalpy of formation is zero. However, we have the formation of liquid as well in the by-products of the reaction, which has a nonzero enthalpy of formation. Thus, the enthalpy of the reaction is not equal to the enthalpy of formation of $\text{Cu}\left(s\right)$.

For the given equation, the enthalpy of the reaction is as follows:

$\Delta H{°}_{\text{rxn}}=\left[\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\left(l\right)\right)+\Delta H{°}_{\text{f}}\left(\text{Cu}\left(s\right)\right)\right]-\left[\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\left(g\right)\right)+\Delta H{°}_{\text{f}}\left(\text{CuO}\left(g\right)\right)\right]$

The enthalpy of formation is zero for ${\text{H}}_{\text{2}}\left(g\right)$, but nonzero for ${\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Hence, for the reaction ${\text{H}}_{\text{2}}\left(g\right)+\text{CuO}\left(g\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+\text{Cu}\left(s\right)$, the value of $\Delta H{°}_{\text{rxn}}\cancel{=}\Delta H{°}_{\text{f}}\left[\text{Cu}\left(s\right)\right]$.

d) $\text{O}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{O}}_{\text{3}}\left(g\right)$

The reaction is given as follows:

$\text{O}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{O}}_{\text{3}}\left(g\right)$

Oxygen is in its diatomic form, that is, ${\text{O}}_{\text{2}}\left(g\right)$ form, which is the standard or most stable state of oxygen. So, its enthalpy of formation is zero. However, monoatomic oxygen is not the most stable form, due to which its enthalpy of formation is nonzero. Thus, the enthalpy of the reaction is not equal to the enthalpy of formation of ${\text{O}}_{\text{3}}\left(g\right)$.

For the given equation, the enthalpy of the reaction is as follows:

$\Delta H{°}_{\text{rxn}}=\left[\Delta H{°}_{\text{f}}\left({\text{O}}_3\left(g\right)\right)\right]-\left[\Delta H{°}_{\text{f}}\left({\text{O}}_{\text{2}}\left(g\right)\right)+\Delta H{°}_{\text{f}}\left(\text{O}\left(g\right)\right)\right]$

The enthalpy of formation is zero for ${\text{O}}_2\left(g\right)$, but nonzero for $\text{O}\left(g\right)$.

Hence, for the reaction $\text{O}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{O}}_{\text{3}}\left(g\right)$, the value of $\Delta H{°}_{\text{rxn}}\cancel{=}\Delta H{°}_{\text{f}}\left[{\text{O}}_3\left(g\right)\right]$.