Chapter 10, Problem 69E

(a)

Interpretation Introduction

Interpretation:

The direction in which the reaction will shift to reach equilibrium should be predicted for $P_{N{O}_2}=P_{N_2{O}_4}=1.0\text{ atm}$.

Reaction: ${\text{2NO}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$

Concept Introduction:

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{\text{f}}^{\text{o}}$.

If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$ or,

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The expression for $\text{ΔG}$ is:

  $\Delta G=\Delta G^o+RT\ln Q$

Where, $\text{ΔG}$ is Gibbs free energy

  $\text{ΔG}°$ is standard Gibbs free energy

  $Q$ is reaction quotient

  $T$ is temperature

  $R$ is universal gas constant

Answer to Problem 69E

Reaction will shift to right to reach equilibrium.

Explanation of Solution

The given reaction is:

  ${\text{2NO}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{N}}_2{\text{O}}_4\text{(g)}$ is $\text{98 kJ/mol}$

The value of standard free energy for ${\text{NO}}_{\text{2}}\text{(g)}$ is $52\text{ kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(G{\text{°}}_{\text{298}}{\text{(N}}_2{\text{O}}_4\text{(g)}))-(2\times G{\text{°}}_{\text{298}}{\text{(NO}}_{\text{2}}\text{(g)}))$

  $\Delta G{°}_{298}=[(98)-(2\times (52)]\text{ kJ/mol}$

  $\Delta G{°}_{298}=-6\text{ kJ/mol}$

Convert temperature given in degree Celsius to Kelvin:

Temperature in K = $25+\text{ 273 K}$

  = 298 K

The expression of reaction quotient is:

Reaction quotient ( $Q$ ) = $\frac{P_{N_2{O}_4}}{{\left(P_{N{O}_2}\right)}^2}$

  = $\frac{1.0\text{ atm}}{{(1.0\text{ atm)}}^2}$ = 1

The value of $\Delta G$ is calculated as:

  $\Delta G=\Delta G^o+RT\ln Q$

Put the values,

  $\Delta G=-6\text{ kJ/mol}+8.3145\text{ J/Kmol}\times \text{298}\text{.15 K}\times \ln (1)$

  = $-6\text{ kJ/mol+0}$

  = $-6\text{ kJ/mol}$

The negative sign of $\Delta G$ implies that the Gibbs free energy of products is less than the Gibbs free energy of reactants. Thus, reaction shifts to right to reach equilibrium.

(b)

Interpretation Introduction

Interpretation:

The direction in which the reaction will shift to reach equilibrium should be predicted for $P_{N{O}_2}=0.21\text{ atm }{P}_{N_2{O}_4}=0.50\text{ atm}$.

Reaction: ${\text{2NO}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$

Concept Introduction:

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{\text{f}}^{\text{o}}$.

If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$ or,

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The expression for $\text{ΔG}$ is:

  $\Delta G=\Delta G^o+RT\ln Q$

Where, $\text{ΔG}$ is Gibbs free energy

  $\text{ΔG}°$ is standard Gibbs free energy

  $Q$ is reaction quotient

  $T$ is temperature

  $R$ is universal gas constant

Answer to Problem 69E

No, shift in direction of the reaction as it is present in equilibrium.

Explanation of Solution

The given reaction is:

  ${\text{2NO}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{N}}_2{\text{O}}_4\text{(g)}$ is $\text{98 kJ/mol}$

The value of standard free energy for ${\text{NO}}_{\text{2}}\text{(g)}$ is $52\text{ kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(G{\text{°}}_{\text{298}}{\text{(N}}_2{\text{O}}_4\text{(g)}))-(2\times G{\text{°}}_{\text{298}}{\text{(NO}}_{\text{2}}\text{(g)}))$

  $\Delta G{°}_{298}=[(98)-(2\times (52)]\text{ kJ/mol}$

  $\Delta G{°}_{298}=-6\text{ kJ/mol}$

Convert temperature given in degree Celsius to Kelvin:

Temperature in K = $25+\text{ 273 K}$

  = 298 K

The expression of reaction quotient is:

Reaction quotient ( $Q$ ) = $\frac{P_{N_2{O}_4}}{{\left(P_{N{O}_2}\right)}^2}$

  = $\frac{0.5\text{ atm}}{{(0.21\text{ atm)}}^2}$ = 11.3

The value of $\Delta G$ is calculated as:

  $\Delta G=\Delta G^o+RT\ln Q$

Put the values,

  $\Delta G=-6\text{ kJ/mol}+\left(8.3145\text{ J/Kmol}\times \text{298}\text{.15 K}\times \ln (11.3)\right)\times \left(\frac{1\text{ kJ}}{1000\text{ kJ}}\right)$

  = $0\text{ kJ/mol}$

The zero value of $\Delta G$ implies that the Gibbs free energy of products is equal toGibbs free energy of reactants. Thus, reaction doesn’t shifts because reaction is present at equilibrium.

(c)

Interpretation Introduction

Interpretation:

The direction in which the reaction will shift to reach equilibrium should be predicted for $P_{N{O}_2}=0.29\text{ atm }{P}_{N_2{O}_4}=1.6\text{ atm}$.

Reaction: ${\text{2NO}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$

Concept Introduction:

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{\text{f}}^{\text{o}}$.

If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$ or,

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The expression for $\text{ΔG}$ is:

  $\Delta G=\Delta G^o+RT\ln Q$

Where, $\text{ΔG}$ is Gibbs free energy

  $\text{ΔG}°$ is standard Gibbs free energy

  $Q$ is reaction quotient

  $T$ is temperature

  $R$ is universal gas constant

Answer to Problem 69E

Reaction will shift to left to reach equilibrium.

Explanation of Solution

The given reaction is:

  ${\text{2NO}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}$

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{N}}_2{\text{O}}_4\text{(g)}$ is $\text{98 kJ/mol}$

The value of standard free energy for ${\text{NO}}_{\text{2}}\text{(g)}$ is $52\text{ kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(G{\text{°}}_{\text{298}}{\text{(N}}_2{\text{O}}_4\text{(g)}))-(2\times G{\text{°}}_{\text{298}}{\text{(NO}}_{\text{2}}\text{(g)}))$

  $\Delta G{°}_{298}=[(98)-(2\times (52)]\text{ kJ/mol}$

  $\Delta G{°}_{298}=-6\text{ kJ/mol}$

Convert temperature given in degree Celsius to Kelvin:

Temperature in K = $25+\text{ 273 K}$

  = 298 K

The expression of reaction quotient is:

Reaction quotient ( $Q$ ) = $\frac{P_{N_2{O}_4}}{{\left(P_{N{O}_2}\right)}^2}$

  = $\frac{1.6\text{ atm}}{{(0.29\text{ atm)}}^2}$ = 19.0

The value of $\Delta G$ is calculated as:

  $\Delta G=\Delta G^o+RT\ln Q$

Put the values,

  $\Delta G=-6\text{ kJ/mol}+\left(8.3145\text{ J/Kmol}\times \text{298}\text{.15 K}\times \ln (19.0)\right)\times \left(\frac{1\text{ J}}{1000\text{ kJ}}\right)$

  = $1.3\text{ kJ/mol}$

The positive value of $\Delta G$ implies that the Gibbs free energy of products is more than the Gibbs free energy of reactants. Thus, reaction shifts to left to reach equilibrium.