Chapter 21, Problem 77GQ

(a)

Interpretation Introduction

Interpretation:

The value for ${\Delta }_f{\text{G}}^{\text{°}}$ for ${\text{MX}}_{\text{n}}$ for the reaction to be product favored at equilibrium is to be stated.

Concept introduction:

The change of Gibbs free energy to form one mole of a substance from its constituent elements when all the substances in the standard form is known as standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$.

The standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ in terms of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ is written as,

${\Delta }_r{\text{G}}^{\circ }={\displaystyle \sum {\Delta }_f{\text{G}}^{\circ }{}_{\text{products}}-{\displaystyle \sum {\Delta }_f{\text{G}}^{\circ }{}_{\text{reactants}}}}$

If the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ is positive then the reaction is reactant favored. If the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ is negative then the reaction is product favoured and spontaneous.

Answer to Problem 77GQ

The value for ${\Delta }_f{\text{G}}^{\text{°}}$ for ${\text{MCl}}_{\text{n}}$ for the reaction to be product favored at equilibrium should be negative and a bigger value.

Explanation of Solution

Any metal M reacts with hydrogen chloride to give metal halide and hydrogen. The chemical equation of any metal M with hydrogen chloride is written as,

    $\text{M}\left(\text{s}\right)+\text{nHCl}\left(\text{g}\right)\to {\text{MCl}}_{\text{n}}\left(\text{s}\right)+\frac{1}{2}{\text{nH}}_{\text{2}}\left(\text{g}\right)$ (1)

The standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ in terms of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ is written as,

${\Delta }_r{\text{G}}^{\circ }={\Delta }_f{\text{G}}^{\circ }\left({\text{MCl}}_{\text{n}}\right)-\text{n}{\Delta }_f{\text{G}}^{\circ }\left(\text{HCl}\right)$ (2)

The standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for $\text{HCl}$ is $-95.1\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for $\text{HCl}$ in equation (2),

$\begin{array}{l}{\Delta }_r{\text{G}}^{\circ }={\Delta }_f{\text{G}}^{\circ }\left({\text{MCl}}_{\text{n}}\right)-\text{n}\left(-95.1\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ {\Delta }_r{\text{G}}^{\circ }={\Delta }_f{\text{G}}^{\circ }\left({\text{MCl}}_{\text{n}}\right)+\text{n}\left(95.1\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}$

Therefore, the value for ${\Delta }_f{\text{G}}^{\text{°}}$ for ${\text{MCl}}_{\text{n}}$ for the reaction to be product favored at equilibrium should be negative and a bigger value.

(b)

Interpretation Introduction

Interpretation:

Among the metals, $\text{Ba}$, $\text{Pb}$, $\text{Hg}$ and $\text{Ti}$ which is/are predicted to have a product-favored reaction with $\text{HCl}$ is to be stated.

Concept introduction:

The change of Gibbs free energy to form one mole of a substance from its constituent elements when all the substances in the standard form is known as standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$.

The standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ in terms of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ is written as,

${\Delta }_r{\text{G}}^{\circ }={\displaystyle \sum {\Delta }_f{\text{G}}^{\circ }{}_{\text{products}}-{\displaystyle \sum {\Delta }_f{\text{G}}^{\circ }{}_{\text{reactants}}}}$

If the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ is positive then the reaction is reactant favoured. If the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ is negative then the reaction is product favoured and spontaneous.

Answer to Problem 77GQ

The metals $\text{Ba}$, $\text{Pb}$ and $\text{Ti}$ gives a product-favored reaction with hydrogen chloride.

Explanation of Solution

The standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for the reaction of the metals, $\text{Ba}$, $\text{Pb}$, $\text{Hg}$ and $\text{Ti}$ with $\text{HCl}$ is calculated below. 

Given:

Refer to the appendix L for the value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$.

The chemical reaction of barium with hydrogen chloride is written as,

    $\text{Ba}\left(\text{s}\right)+2\text{HCl}\left(\text{g}\right)\to {\text{BaCl}}_2\left(\text{s}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

The value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for ${\text{BaCl}}_2$ is

$-810.4\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for $\text{HCl}$ is

$-95.1\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute these values in equation (2) to calculate the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for ${\text{BaCl}}_2$

$\begin{array}{c}{\Delta }_r{\text{G}}^{\circ }=\left(-810.4\text{kJ}\cdot {\text{mol}}^{-1}\right)-\left(2\right)\left(-95.1\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-620.2\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for ${\text{BaCl}}_2$ is

 $-620.2\text{kJ}\cdot {\text{mol}}^{-1}$.

Hence, the negative value of standard Gibbs free energy of reaction indicates that the barium metal gives a product-favoured reaction with $\text{HCl}$.

The chemical reaction of lead with hydrogen chloride is written as,

    $\text{Pb}\left(\text{s}\right)+2\text{HCl}\left(\text{g}\right)\to {\text{PbCl}}_2\left(\text{s}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

The value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for ${\text{PbCl}}_2$ is

$-314\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for $\text{HCl}$ is

$-95.1\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute these values in equation (2) to calculate the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for ${\text{PbCl}}_2$

$\begin{array}{c}{\Delta }_r{\text{G}}^{\circ }=\left(-314\text{kJ}\cdot {\text{mol}}^{-1}\right)-\left(2\right)\left(-95.1\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-123.8\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for ${\text{PbCl}}_2$ is

 $-123.8\text{kJ}\cdot {\text{mol}}^{-1}$.

Hence, the negative value of standard Gibbs free energy of reaction indicates that the lead metal gives a product-favoured reaction with $\text{HCl}$.

The chemical reaction of mercury with hydrogen chloride is written as,

    $\text{Hg}\left(\text{s}\right)+2\text{HCl}\left(\text{g}\right)\to {\text{HgCl}}_2\left(\text{s}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

The value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for ${\text{HgCl}}_2$ is

$-178.6\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for $\text{HCl}$ is

$-95.1\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute these values in equation (2) to calculate the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for ${\text{HgCl}}_2$

$\begin{array}{c}{\Delta }_r{\text{G}}^{\circ }=\left(-178.6\text{kJ}\cdot {\text{mol}}^{-1}\right)-\left(2\right)\left(-95.1\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =11.6\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for ${\text{HgCl}}_2$ is

 $11.6\text{kJ}\cdot {\text{mol}}^{-1}$.

Hence, the positive value of standard Gibbs free energy of reaction indicates that the mercury metal does not form a product-favored reaction with $\text{HCl}$.

The chemical reaction of titanium with hydrogen chloride is written as,

    $\text{Ti}\left(\text{s}\right)+4\text{HCl}\left(\text{g}\right)\to {\text{TiCl}}_4\left(\text{g}\right)+2{\text{H}}_{\text{2}}\left(\text{g}\right)$

The value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for ${\text{TiCl}}_4$ is

$-726.7\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of standard Gibbs free energy of formation $\left({\Delta }_f{\text{G}}^{\text{°}}\right)$ for $\text{HCl}$ is

$-95.1\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute these values in equation (2) to calculate the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for ${\text{TiCl}}_4$

$\begin{array}{c}{\Delta }_r{\text{G}}^{\circ }=\left(-726.7\text{kJ}\cdot {\text{mol}}^{-1}\right)-\left(4\right)\left(-95.1\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-346.3\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, the value of standard Gibbs free energy of reaction $\left({\Delta }_r{\text{G}}^{\circ }\right)$ for ${\text{TiCl}}_4$ is

 $-346.3\text{kJ}\cdot {\text{mol}}^{-1}$.

Hence, the negative value of standard Gibbs free energy of reaction indicates that the titanium metal gives a product-favoured reaction with $\text{HCl}$.

Therefore, the metals $\text{Ba}$, $\text{Pb}$ and $\text{Ti}$ gives a product-favored reaction with hydrogen chloride.