For the following reaction free energy value ΔG o has to be calculated at 298 K , using the given K value. 2 H C l ( g ) + B r 2 ( l ) ⇌ 2 H B r ( g ) + C l 2 ( g ) K = 2.22 × 10 − 15 a t 298 K Concept introduction: 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 G . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG ° rxn ) is the difference in free energy of the reactants and products in their standard state. ΔG ° rxn = ∑ nΔG f ° (Products)- ∑ nΔG f ° (Reactants) Free energy change ΔG : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant K ΔG = ΔG 0 + RT ln ( K ) ΔG 0 = ΔH 0 − TΔS 0 Where, T is the temperature ΔG is the free energy ΔG 0 , ΔH 0 and ΔS 0 is standard free energy, enthalpy and entropy values. The formula of ΔH in terms of ΔG and TΔS is, ΔH ∘ = ΔG ∘ + T Δ S The formula of ΔG in terms of equilibrium constant, ΔG = ΔG ∘ − RT ln(K)

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Answer 1

Question

Chapter 20.4, Problem 20.9BFP

Interpretation Introduction

Interpretation:

For the following reaction free energy value $ΔGo$ has to be calculated at $298 K$ , using the given $K$ value.

     $2HCl (g) + Br2(l) ⇌ 2HBr(g) + Cl2(g) K= 2.22×10−15 at 298 K$

Concept introduction:

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔG0+ RT ln (K)ΔG0 = ΔH0 − TΔS0$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ , $ΔH0$ and $ΔS0$ is standard free energy, enthalpy and entropy values.

The formula of $ΔH$ in terms of $ΔG$ and $TΔS$ is,

         $ΔH∘ = ΔG∘ +T ΔS$

The formula of $ΔG$ in terms of equilibrium constant,

         $ΔG = ΔG∘ − RT ln(K)$

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Answer 2

Question

Chapter 20.4, Problem 20.9BFP

Interpretation Introduction

Interpretation:

For the following reaction free energy value $ΔGo$ has to be calculated at $298 K$ , using the given $K$ value.

     $2HCl (g) + Br2(l) ⇌ 2HBr(g) + Cl2(g) K= 2.22×10−15 at 298 K$

Concept introduction:

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔG0+ RT ln (K)ΔG0 = ΔH0 − TΔS0$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ , $ΔH0$ and $ΔS0$ is standard free energy, enthalpy and entropy values.

The formula of $ΔH$ in terms of $ΔG$ and $TΔS$ is,

         $ΔH∘ = ΔG∘ +T ΔS$

The formula of $ΔG$ in terms of equilibrium constant,

         $ΔG = ΔG∘ − RT ln(K)$

Expert Solution & Answer

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Answer 3

Question

Chapter 20.4, Problem 20.9BFP

Interpretation Introduction

Interpretation:

For the following reaction free energy value $ΔGo$ has to be calculated at $298 K$ , using the given $K$ value.

     $2HCl (g) + Br2(l) ⇌ 2HBr(g) + Cl2(g) K= 2.22×10−15 at 298 K$

Concept introduction:

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔG0+ RT ln (K)ΔG0 = ΔH0 − TΔS0$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ , $ΔH0$ and $ΔS0$ is standard free energy, enthalpy and entropy values.

The formula of $ΔH$ in terms of $ΔG$ and $TΔS$ is,

         $ΔH∘ = ΔG∘ +T ΔS$

The formula of $ΔG$ in terms of equilibrium constant,

         $ΔG = ΔG∘ − RT ln(K)$

Expert Solution & Answer

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Answer 4

Question

Chapter 20.4, Problem 20.9BFP

Interpretation Introduction

Interpretation:

For the following reaction free energy value $ΔGo$ has to be calculated at $298 K$ , using the given $K$ value.

     $2HCl (g) + Br2(l) ⇌ 2HBr(g) + Cl2(g) K= 2.22×10−15 at 298 K$

Concept introduction:

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔG0+ RT ln (K)ΔG0 = ΔH0 − TΔS0$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ , $ΔH0$ and $ΔS0$ is standard free energy, enthalpy and entropy values.

The formula of $ΔH$ in terms of $ΔG$ and $TΔS$ is,

         $ΔH∘ = ΔG∘ +T ΔS$

The formula of $ΔG$ in terms of equilibrium constant,

         $ΔG = ΔG∘ − RT ln(K)$

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 20.4, Problem 20.9BFP

Interpretation Introduction

Interpretation:

For the following reaction free energy value $ΔGo$ has to be calculated at $298 K$ , using the given $K$ value.

     $2HCl (g) + Br2(l) ⇌ 2HBr(g) + Cl2(g) K= 2.22×10−15 at 298 K$

Concept introduction:

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔG0+ RT ln (K)ΔG0 = ΔH0 − TΔS0$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ , $ΔH0$ and $ΔS0$ is standard free energy, enthalpy and entropy values.

The formula of $ΔH$ in terms of $ΔG$ and $TΔS$ is,

         $ΔH∘ = ΔG∘ +T ΔS$

The formula of $ΔG$ in terms of equilibrium constant,

         $ΔG = ΔG∘ − RT ln(K)$

Answer 6

Chapter 20.4, Problem 20.9BFP

Interpretation Introduction

Interpretation:

For the following reaction free energy value $ΔGo$ has to be calculated at $298 K$ , using the given $K$ value.

     $2HCl (g) + Br2(l) ⇌ 2HBr(g) + Cl2(g) K= 2.22×10−15 at 298 K$

Concept introduction:

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔG0+ RT ln (K)ΔG0 = ΔH0 − TΔS0$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ , $ΔH0$ and $ΔS0$ is standard free energy, enthalpy and entropy values.

The formula of $ΔH$ in terms of $ΔG$ and $TΔS$ is,

         $ΔH∘ = ΔG∘ +T ΔS$

The formula of $ΔG$ in terms of equilibrium constant,

         $ΔG = ΔG∘ − RT ln(K)$

Answer 7

Chapter 20.4, Problem 20.9BFP

Interpretation Introduction

Interpretation:

For the following reaction free energy value $ΔGo$ has to be calculated at $298 K$ , using the given $K$ value.

     $2HCl (g) + Br2(l) ⇌ 2HBr(g) + Cl2(g) K= 2.22×10−15 at 298 K$

Concept introduction:

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Free energy change $ΔG$ : change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $K$

$ΔG = ΔG0+ RT ln (K)ΔG0 = ΔH0 − TΔS0$

Where,

$T$ is the temperature

$ΔG$ is the free energy

$ΔG0$ , $ΔH0$ and $ΔS0$ is standard free energy, enthalpy and entropy values.

The formula of $ΔH$ in terms of $ΔG$ and $TΔS$ is,

         $ΔH∘ = ΔG∘ +T ΔS$

The formula of $ΔG$ in terms of equilibrium constant,

         $ΔG = ΔG∘ − RT ln(K)$

Answer 8

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Answer 9

Expert Solution & Answer

Answer 10

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Answer 11

Ch. 20.1 - Select the substance with the higher entropy in...Ch. 20.1 - Select the substance with the lower entropy in...Ch. 20.2 - Balance each equation, predict the sign of if...Ch. 20.2 - Balance each equation, predict the sign of if...Ch. 20.2 - Gaseous phosphorus trichloride forms from the...Ch. 20.2 - Prob. 20.3BFP Ch. 20.3 - Use and to calculate at 298 K for this...Ch. 20.3 - Prob. 20.4BFP Ch. 20.3 - Use values to calculate at 298 K: 2NO(g) +...Ch. 20.3 - Prob. 20.5BFP

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