Chapter 15, Problem 15.104QP

Consider the gas-phase reaction between A₂ (green) and B₂ (red) to form AB at 298 K: $\begin{array}{cc}{\text{A}}_{\text{2}}\left(g\right)+{\text{B}}_{\text{2}}\left(g\right)\rightleftarrows \text{2AB}\left(g\right)& \Delta G^{\circ }\end{array}=–\text{3}\text{.4kJ/mol}$

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

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

(c) 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 results to two significant figures.

Interpretation Introduction

Interpretation:

Calculate the free energy values $(\text{ΔG})$ and equilibrium constant (Kp) values for given the gas phase equilibrium reaction and its images with respective partial pressure at $0.10\text{atm}$.

Concept Introduction:

Chemical equilibrium: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Entropy $({\text{ΔG}}^0)$: The entropy is second law of thermodynamics it indicated for $({\text{ΔG}}^0)$ symbol, the many of chemical reactions cause changes in entropy and it plays on important role in determining in which direction (forward and backward) a chemical reaction spontaneously proceeds.

Gibbs free energy (G): The thermodynamic quantity to the ( $\text{ΔH}$) enthalpy of a system process and minutes the product of entropy and the absolute temperature. The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system.

To find: Calculate the pressure values $(Kp)$ for given the statement.

Answer to Problem 15.104QP

The system equilibrium reaction on given respective images (a-c) are shown below.

${\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}{\text{ΔG}}^{\text{0}}=\text{-3}\text{.4}\text{KJ/mol}$

The given equilibrium reaction (Image (A) as positive and Images (ii and iii) has negative $(\text{ΔG})$ vales.

Explanation of Solution

Calculate the chemical equilibrium process $(Kp)$.

Consider the following equation (1)

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K---------[1]}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.08314 (or) 8}\text{.314}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\\ \ln =(-ve(\log )\text{State Function})\end{array}$

The equilibrium constant is related to the to the standard free energy change by the followed above equation (1). Than the given statement of values (K_P, R and T) are substituted in same equation.

$\begin{array}{l}{\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}{\text{ΔG}}^{\text{0}}=\text{-3}\text{.4}\text{KJ/mol}\\ \text{ΔG}\text{=-RTln}\text{K-------------[1]}\\ \text{Respactive vaalues are substitued for this equation }\\ \text{-3400J/mol=-(8}\text{.314J/K}\text{.mol)(298K)lnK}\\ \text{lnK=}\frac{\text{3400}\text{J/mol}}{\text{-(8}\text{.314J/K}\text{.mol)(298K)}}\\ \text{lnK=}1.37\\ \text{Taking the anti-log of both side}\\ K=e^{\text{-1}\text{.37}}\\ K=3.9(or)K=4\end{array}$

The standard free energy values and partial pressure values are derived given the equilibrium reactions.

Interpretation Introduction

Interpretation:

Calculate the free energy values $(\text{ΔG})$ and equilibrium constant (Kp) values for given the gas phase equilibrium reaction and its images with respective partial pressure at $0.10\text{atm}$.

Concept Introduction:

Chemical equilibrium: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Entropy $({\text{ΔG}}^0)$: The entropy is second law of thermodynamics it indicated for $({\text{ΔG}}^0)$ symbol, the many of chemical reactions cause changes in entropy and it plays on important role in determining in which direction (forward and backward) a chemical reaction spontaneously proceeds.

Gibbs free energy (G): The thermodynamic quantity to the ( $\text{ΔH}$) enthalpy of a system process and minutes the product of entropy and the absolute temperature. The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system.

To find: Calculate the reaction quotient (Qr) values for given the equilibrium reaction.

Answer to Problem 15.104QP

The system equilibrium reaction on given respective images (a-c) are shown below.

${\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}{\text{ΔG}}^{\text{0}}=\text{-3}\text{.4}\text{KJ/mol}$

The given equilibrium reaction (Image (A) as positive and Images (ii and iii) has negative $(\text{ΔG})$ vales.

Explanation of Solution

Calculate and analyze the (Qr) values for equilibrium reactions.

$\begin{array}{l}{\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}{\text{ΔG}}^{\text{0}}=\text{-3}\text{.4}\text{KJ/mol}\\ Q=\frac{{[AB]}^2}{[A_2][B_2]}\\ \text{Given the partial pressure values 0}\text{.10}\text{atm}\\ \text{No}\text{. reactants and products are 3}\\ \text{3}\times \text{0}\text{.10}\text{atm}\text{=0}\text{.30}\text{atm}\\ {\text{Q}}_{\text{i}}=\frac{{[0.30]}^2}{[0.30][0.20]}=1.5\\ {\text{Q}}_{\text{ii}}=\frac{{[0.60]}^2}{[0.20][0.30]}=6.0\\ {\text{Q}}_{\text{iii}}=\frac{{[0.40]}^2}{[0.20][0.20]}=4.0\end{array}$

Analysis for image (1): The equilibrium process (a) also smallest equilibrium constant, here three products are present in the diagrams.

Analyzing image (2): Further we consider the equilibrium images (2), this process has the smallest equilibrium constant (Kc), because low amount of reactant present in this equilibrium.

Analyzing image (3): Given the equilibrium reaction (3) has largest equilibrium constant, because the three diagrams, there is the most products present at equilibrium process, and this products indicated the respective image.

Interpretation Introduction

Interpretation:

Calculate the free energy values $(\text{ΔG})$ and equilibrium constant (Kp) values for given the gas phase equilibrium reaction and its images with respective partial pressure at $0.10\text{atm}$.

Concept Introduction:

Chemical equilibrium: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Entropy $({\text{ΔG}}^0)$: The entropy is second law of thermodynamics it indicated for $({\text{ΔG}}^0)$ symbol, the many of chemical reactions cause changes in entropy and it plays on important role in determining in which direction (forward and backward) a chemical reaction spontaneously proceeds.

Gibbs free energy (G): The thermodynamic quantity to the ( $\text{ΔH}$) enthalpy of a system process and minutes the product of entropy and the absolute temperature. The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system.

To find: Calculate the entropy values $({\text{ΔG}}^0)$ for given the statement.

Calculate the chemical equilibrium process $({\text{ΔG}}^0)$.

Answer to Problem 15.104QP

The system equilibrium reaction on given respective images (a-c) are shown below.

${\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}{\text{ΔG}}^{\text{0}}=\text{-3}\text{.4}\text{KJ/mol}$

The given equilibrium reaction (Image (A) as positive and Images (ii and iii) has negative $(\text{ΔG})$ vales.

Explanation of Solution

Let us consider the following equation (1)

$\begin{array}{l}{\text{A}}_{\text{2}}\text{(g)}\text{+}{\text{B}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2AB(g)}{\text{ΔG}}^{\text{0}}=\text{-3}\text{.4}\text{KJ/mol}\\ {\text{ΔG}}^{\text{0}}\text{=-RTln}Q\text{---------[1]}\\ {\text{The respative (Q}}_{\text{i}}{\text{, Q}}_{\text{ii}}{\text{, Q}}_{\text{iii}}\text{) values are substituted equation (1)}\\ {\text{ΔG}}_{\text{i}}=-\text{3400}\text{J/mol}\text{+}\text{(8}\text{.314}\text{J/K}\text{.mol)}\text{(298K)}\text{In(1}\text{.5)}\\ \text{=}\text{-3400}\text{J/mol}+(8.314J/K.mol)(298K)(0.4054)\\ {\text{ΔG}}_{\text{i}}=-2.4\times {10}^3\text{J/mol}\\ {\text{ΔG}}_{\text{ii}}=-\text{3400}\text{J/mol}\text{+}\text{(8}\text{.314}\text{J/K}\text{.mol)}\text{(298K)}\text{In(6)}\\ \text{=}\text{-3400}\text{J/mol}+(8.314J/K.mol)(298K)(1.7917)\\ {\text{ΔG}}_{\text{ii}}=1.0\times {10}^3\text{J/mol}\\ {\text{ΔG}}_{\text{iii}}=-\text{3400}\text{J/mol}\text{+}\text{(8}\text{.314}\text{J/K}\text{.mol)}\text{(298K)}\text{In(4)}\\ \text{=}\text{-3400}\text{J/mol}+(8.314J/K.mol)(298K)(1.3862)\\ {\text{ΔG}}_{\text{ii}}=35\text{J/mol}\end{array}$

The entropy values of given equilibrium reaction (1) has negative $(\text{ΔG})$ values, and reaction (ii and iii) are positive $(\text{ΔG})$ values.