Chapter 17, Problem 17.109QP

Consider the gas-phase reaction between A₂ (green) and B₂ (red) to form AB at 298 K:

${\text{A}}_2(g)+{\text{B}}_2(g)\rightleftharpoons 2{\text{AB}}_2(g)\Delta G°=-3.4\text{ kJ/mol}$

(1)    Which of the following reaction mixtures is at equilibrium?

(2)    Which of the following reaction mixtures has a negative ΔG value?

(3)    Which of the following reaction mixtures has a positive ΔG value?

The partial pressures of the gases in each frame are equal to the number of A₂, B₂, and AB molecules times 0.10 atm. Round your answers to two significant figures.

Interpretation Introduction

Interpretation:

The equilibrium process should be determined from the given equilibrium reaction.

Concept Introduction:

Free energy$(\text{ΔG})$: In thermodynamics free energy or Gibbs free energy is the energy that is used to express the total energy content of a system.  According to second law of thermodynamics, in all spontaneous process is associated with the decrease in free energy of the system.  That is the change in free energy will be negative.

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}\Delta \text{G=}\Delta \text{G}°\text{+}\text{RTlnQ}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(8.314\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{Q=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

We calculate the equilibrium constant $(\text{Kp})$ from the $\Delta \text{G}°$ value

$\begin{array}{l}{\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}\\ \text{ΔG}°\text{=-RTln}\text{K}\\ \text{-3}\text{.4}\times {\text{10}}^{\text{3}}\text{J/mol=-(8}\text{.314J/mol}\cdot \text{K)(298K)lnKp}\\ \text{lnKp=}\frac{\text{-3}\text{.4}\times {\text{10}}^{\text{3}}\text{J/mol}}{\text{-(8}\text{.314J/K}\text{.mol)(298K)}}\\ \text{lnK=}1.37\text{rounded}\text{value}\text{1}\text{.40}\\ {\text{K=e}}^{\text{-1}\text{.40}}\\ \text{K=3}\text{.9}\text{(or)}\text{K=4}\end{array}$

$\begin{array}{l}\text{Reaction-c}\\ {\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}\\ \text{Q=}\frac{{\text{[AB]}}^{\text{2}}}{{\text{[A}}_{\text{2}}{\text{][B}}_{\text{2}}\text{]}}\\ \text{Given the partial pressure values 0}\text{.10}\text{atm}\\ \text{No}\text{. reactants}\text{(A=2)\&}\text{(B=2) and (C=4)}\text{products}\\ \text{Q}=\frac{{[0.40]}^2}{[0.20][0.20]}=4.0\end{array}$

$\Delta \text{G}\text{=}\Delta \text{G}°+\text{RTlnQ}$

The reaction mixture (c) is at equilibrium $(\text{Q=K})$ and $\Delta \text{G=0}$

Interpretation Introduction

Interpretation:

The negative entropy value has to be calculated and identified for the given equilibrium reaction.

Concept Introduction:

Free energy$(\text{ΔG})$: In thermodynamics free energy or Gibbs free energy is the energy that is used to express the total energy content of a system.  According to second law of thermodynamics, in all spontaneous process is associated with the decrease in free energy of the system.  That is the change in free energy will be negative.

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}\Delta \text{G=}\Delta \text{G}°\text{+}\text{RTlnQ}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(8.314\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{Q=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Reaction quotient: This type of chemical equilibrium reaction proceeds likely to produced, given either the pressure (or) the concentration of the reactants and the products. The value can be compared to the equilibrium constant, to determine the direction of the reaction that is take place. Then reaction quotient (Qc) the indication of Q can be used to determine which direction will shift to reach of chemical equilibrium process.

Explanation of Solution

$\begin{array}{l}\text{Reaction-a}\\ {\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}\\ {\text{Q}}_{\text{P}}\text{=}\frac{{\text{[AB]}}^{\text{2}}}{{\text{[A}}_{\text{2}}{\text{][B}}_{\text{2}}\text{]}}\\ \text{Given the partial pressure values 0}\text{.10}\text{atm}\\ {\text{Q}}_{\text{P}}=\frac{{[0.30]}^2}{[0.30][0.20]}=1.5\end{array}$

The value of $\text{ΔG}$ could also be calculated using the equation,

$\begin{array}{l}\Delta \text{G}\text{=}\Delta \text{G}°+\text{RTlnQ}\\ \text{The (Q) values is substituted equation}\\ \text{ΔG}=-\text{3400}\text{J/mol}\text{+}\text{(8}\text{.314}\text{J/K}\text{.mol)}\text{(298K)}\text{In(1}\text{.5)}\\ \text{=}\text{-3400}\text{J/mol}+(8.314J/K.mol)(298K)(0.4054)\\ \text{ΔG}=-2.4\times {10}^3\text{J/mol}\end{array}$

Reaction mixture (a) has a negative $\text{ΔG}$ value, because $\text{Q<K}$ the system of equilibrium will shift into right towards products, to reach the equilibrium.

Interpretation Introduction

Interpretation:

The positive entropy value has to be calculate and identified given equilibrium reaction. 

Concept Information:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}\Delta \text{G=}\Delta \text{G}°\text{+}\text{RTlnQ}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(8.314\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{Q=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\\ \ln =(-ve(\log )\text{State Function})\end{array}$

Reaction quotient: This type of chemical equilibrium reaction proceeds likely to produced, given either the pressure (or) the concentration of the reactants and the products. The value can be compared to the equilibrium constant, to determine the direction of the reaction that is take place. Then reaction quotient (Qc) the indication of Q can be used to determine which direction will shift to reach of chemical equilibrium process.

Explanation of Solution

$\begin{array}{l}\text{Reaction-c}\\ {\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}\\ {\text{Q}}_{\text{P}}\text{=}\frac{{[AB]}^2}{[A_2][B_2]}\\ {\text{Q}}_{\text{P}}=\frac{{[0.60]}^2}{[0.20][0.30]}=6.0\end{array}$

$\begin{array}{l}\Delta \text{G}\text{+}\Delta \text{G}°\text{+}\text{RTlnQ}\\ \text{The respative (Q) value is substituted equation}\\ \text{ΔG}=-\text{3400}\text{J/mol}\text{+}\text{(8}\text{.314}\text{J/K}\text{.mol)}\text{(298K)}\text{In(6)}\\ \text{=}\text{-3400}\text{J/mol}+(8.314J/K.mol)(298K)(1.7917)\\ \text{ΔG}=1.0\times {10}^3\text{kJ/mol}\end{array}$

The reaction mixture (b) has positive $\Delta \text{G}$ value, because $\text{Q>K}$ the system of equilibrium will shift left, toward reactants, to reach the equilibrium. This shift corresponds to a positive $\Delta \text{G}$ value.