The value of equilibrium constant for the given reaction at 298 K . Solution: The value of equilibrium constant for the given reaction is 7.856 × 10 15 . Explanation: Given The balanced chemical equation for the given reaction is, SO 2 ( g ) + 2 H 2 S ( g ) ⇌ 3 S ( s ) + 2 H 2 O ( g ) The temperature of the reaction is 298 K . The standard free energy change of the given reaction is calculated by the formula, Δ G 0 = ∑ n Δ G 0 ( products ) − ∑ m Δ G 0 ( reactants ) = ( 2 Δ G H 2 O ( g ) 0 + 3 Δ G S ( s ) 0 ) − ( Δ G SO 2 ( g ) 0 + 2 Δ G H 2 S ( g ) 0 ) Substitute the values of Δ G H 2 O ( g ) 0 , Δ G S ( s ) 0 , Δ G SO 2 ( g ) 0 and Δ G H 2 S ( g ) 0 from appendix C in equation. Δ H 0 = ( 2 × − 228.57 kJ/mol + 3 × 0 kJ/mol ) − ( 1 × − 300.4 kJ/mol + 2 × − 33.01 kJ/mol ) = − 90.72 kJ/mol The conversion of free energy from kJ/mol into J/mol is done as, 1 kJ/mol = 10 3 J/mol Hence, the conversion of free energy from − 90.72 kJ/mol into J/mol is, − 90.72 kJ/mol = − 90.72 × 10 3 J/mol The equilibrium constant for the given reaction is calculated by the formula, Δ G 0 = − R T ln K ln K = − Δ G 0 R T Where, Δ G 0 is the standard free energy of the reaction. R is the gas constant ( 8.314 J/mol ⋅ K ) . T is the temperature of the reaction. K is the equilibrium constant of the reaction. Substitute the values of Δ G 0 , T and R in above equation. ln K = − ( − 90.72 × 10 3 J/mol ) 8.314 J/mol ⋅ K × 298 K = 36.6 The equilibrium constant of the reaction is calculated by taking exponential of the above equation. K = e 36.6 = 7.856 × 10 15 Conclusion: The value of equilibrium constant for the given reaction is 7.856 × 10 15 .

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Answer 1

Question

Chapter 5, Problem 34E

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at $298 K$ .

Solution: The value of equilibrium constant for the given reaction is $7.856×1015$ .

Explanation:

Given

The balanced chemical equation for the given reaction is,

$SO2(g)+2H2S(g)⇌3S(s)+2H2O(g)$

The temperature of the reaction is $298 K$ .

The standard free energy change of the given reaction is calculated by the formula,

$ΔG0=∑nΔG0( products)−∑mΔG0( reactants)=(2ΔG H 2 O( g )0+3ΔG S( s )0)−(ΔG SO 2 ( g )0+2ΔG H 2 S( g )0)$

Substitute the values of $ΔGH2O(g)0$ , $ΔGS(s)0$ , $ΔGSO2(g)0$ and $ΔGH2S(g)0$ from appendix C in equation.

$ΔH0=(2×−228.57 kJ/mol+3×0 kJ/mol)−(1×−300.4 kJ/mol+2×−33.01 kJ/mol)=−90.72 kJ/mol$

The conversion of free energy from $kJ/mol$ into $J/mol$ is done as,

$1 kJ/mol=103 J/mol$

Hence, the conversion of free energy from $−90.72 kJ/mol$ into $J/mol$ is,

$−90.72 kJ/mol=−90.72×103 J/mol$

The equilibrium constant for the given reaction is calculated by the formula,

$ΔG0=−RTlnKlnK=−ΔG0RT$

Where,
$ΔG0$ is the standard free energy of the reaction.
$R$ is the gas constant $(8.314 J/mol⋅K)$ .
$T$ is the temperature of the reaction.
$K$ is the equilibrium constant of the reaction.

Substitute the values of $ΔG0$ , $T$ and $R$ in above equation.

$lnK=−( −90.72× 10 3 J/mol)8.314 J/mol⋅K×298 K=36.6$

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

$K=e36.6=7.856×1015$

Conclusion:

The value of equilibrium constant for the given reaction is $7.856×1015$ .

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove $SO2$ or not.

(c)

Interpretation Introduction

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.

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Answer 2

Question

Chapter 5, Problem 34E

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at $298 K$ .

Solution: The value of equilibrium constant for the given reaction is $7.856×1015$ .

Explanation:

Given

The balanced chemical equation for the given reaction is,

$SO2(g)+2H2S(g)⇌3S(s)+2H2O(g)$

The temperature of the reaction is $298 K$ .

The standard free energy change of the given reaction is calculated by the formula,

$ΔG0=∑nΔG0( products)−∑mΔG0( reactants)=(2ΔG H 2 O( g )0+3ΔG S( s )0)−(ΔG SO 2 ( g )0+2ΔG H 2 S( g )0)$

Substitute the values of $ΔGH2O(g)0$ , $ΔGS(s)0$ , $ΔGSO2(g)0$ and $ΔGH2S(g)0$ from appendix C in equation.

$ΔH0=(2×−228.57 kJ/mol+3×0 kJ/mol)−(1×−300.4 kJ/mol+2×−33.01 kJ/mol)=−90.72 kJ/mol$

The conversion of free energy from $kJ/mol$ into $J/mol$ is done as,

$1 kJ/mol=103 J/mol$

Hence, the conversion of free energy from $−90.72 kJ/mol$ into $J/mol$ is,

$−90.72 kJ/mol=−90.72×103 J/mol$

The equilibrium constant for the given reaction is calculated by the formula,

$ΔG0=−RTlnKlnK=−ΔG0RT$

Where,
$ΔG0$ is the standard free energy of the reaction.
$R$ is the gas constant $(8.314 J/mol⋅K)$ .
$T$ is the temperature of the reaction.
$K$ is the equilibrium constant of the reaction.

Substitute the values of $ΔG0$ , $T$ and $R$ in above equation.

$lnK=−( −90.72× 10 3 J/mol)8.314 J/mol⋅K×298 K=36.6$

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

$K=e36.6=7.856×1015$

Conclusion:

The value of equilibrium constant for the given reaction is $7.856×1015$ .

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove $SO2$ or not.

(c)

Interpretation Introduction

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.

Expert Solution & Answer

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Answer 3

Question

Chapter 5, Problem 34E

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at $298 K$ .

Solution: The value of equilibrium constant for the given reaction is $7.856×1015$ .

Explanation:

Given

The balanced chemical equation for the given reaction is,

$SO2(g)+2H2S(g)⇌3S(s)+2H2O(g)$

The temperature of the reaction is $298 K$ .

The standard free energy change of the given reaction is calculated by the formula,

$ΔG0=∑nΔG0( products)−∑mΔG0( reactants)=(2ΔG H 2 O( g )0+3ΔG S( s )0)−(ΔG SO 2 ( g )0+2ΔG H 2 S( g )0)$

Substitute the values of $ΔGH2O(g)0$ , $ΔGS(s)0$ , $ΔGSO2(g)0$ and $ΔGH2S(g)0$ from appendix C in equation.

$ΔH0=(2×−228.57 kJ/mol+3×0 kJ/mol)−(1×−300.4 kJ/mol+2×−33.01 kJ/mol)=−90.72 kJ/mol$

The conversion of free energy from $kJ/mol$ into $J/mol$ is done as,

$1 kJ/mol=103 J/mol$

Hence, the conversion of free energy from $−90.72 kJ/mol$ into $J/mol$ is,

$−90.72 kJ/mol=−90.72×103 J/mol$

The equilibrium constant for the given reaction is calculated by the formula,

$ΔG0=−RTlnKlnK=−ΔG0RT$

Where,
$ΔG0$ is the standard free energy of the reaction.
$R$ is the gas constant $(8.314 J/mol⋅K)$ .
$T$ is the temperature of the reaction.
$K$ is the equilibrium constant of the reaction.

Substitute the values of $ΔG0$ , $T$ and $R$ in above equation.

$lnK=−( −90.72× 10 3 J/mol)8.314 J/mol⋅K×298 K=36.6$

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

$K=e36.6=7.856×1015$

Conclusion:

The value of equilibrium constant for the given reaction is $7.856×1015$ .

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove $SO2$ or not.

(c)

Interpretation Introduction

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 5, Problem 34E

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at $298 K$ .

Solution: The value of equilibrium constant for the given reaction is $7.856×1015$ .

Explanation:

Given

The balanced chemical equation for the given reaction is,

$SO2(g)+2H2S(g)⇌3S(s)+2H2O(g)$

The temperature of the reaction is $298 K$ .

The standard free energy change of the given reaction is calculated by the formula,

$ΔG0=∑nΔG0( products)−∑mΔG0( reactants)=(2ΔG H 2 O( g )0+3ΔG S( s )0)−(ΔG SO 2 ( g )0+2ΔG H 2 S( g )0)$

Substitute the values of $ΔGH2O(g)0$ , $ΔGS(s)0$ , $ΔGSO2(g)0$ and $ΔGH2S(g)0$ from appendix C in equation.

$ΔH0=(2×−228.57 kJ/mol+3×0 kJ/mol)−(1×−300.4 kJ/mol+2×−33.01 kJ/mol)=−90.72 kJ/mol$

The conversion of free energy from $kJ/mol$ into $J/mol$ is done as,

$1 kJ/mol=103 J/mol$

Hence, the conversion of free energy from $−90.72 kJ/mol$ into $J/mol$ is,

$−90.72 kJ/mol=−90.72×103 J/mol$

The equilibrium constant for the given reaction is calculated by the formula,

$ΔG0=−RTlnKlnK=−ΔG0RT$

Where,
$ΔG0$ is the standard free energy of the reaction.
$R$ is the gas constant $(8.314 J/mol⋅K)$ .
$T$ is the temperature of the reaction.
$K$ is the equilibrium constant of the reaction.

Substitute the values of $ΔG0$ , $T$ and $R$ in above equation.

$lnK=−( −90.72× 10 3 J/mol)8.314 J/mol⋅K×298 K=36.6$

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

$K=e36.6=7.856×1015$

Conclusion:

The value of equilibrium constant for the given reaction is $7.856×1015$ .

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove $SO2$ or not.

(c)

Interpretation Introduction

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 5, Problem 34E

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at $298 K$ .

Solution: The value of equilibrium constant for the given reaction is $7.856×1015$ .

Explanation:

Given

The balanced chemical equation for the given reaction is,

$SO2(g)+2H2S(g)⇌3S(s)+2H2O(g)$

The temperature of the reaction is $298 K$ .

The standard free energy change of the given reaction is calculated by the formula,

$ΔG0=∑nΔG0( products)−∑mΔG0( reactants)=(2ΔG H 2 O( g )0+3ΔG S( s )0)−(ΔG SO 2 ( g )0+2ΔG H 2 S( g )0)$

Substitute the values of $ΔGH2O(g)0$ , $ΔGS(s)0$ , $ΔGSO2(g)0$ and $ΔGH2S(g)0$ from appendix C in equation.

$ΔH0=(2×−228.57 kJ/mol+3×0 kJ/mol)−(1×−300.4 kJ/mol+2×−33.01 kJ/mol)=−90.72 kJ/mol$

The conversion of free energy from $kJ/mol$ into $J/mol$ is done as,

$1 kJ/mol=103 J/mol$

Hence, the conversion of free energy from $−90.72 kJ/mol$ into $J/mol$ is,

$−90.72 kJ/mol=−90.72×103 J/mol$

The equilibrium constant for the given reaction is calculated by the formula,

$ΔG0=−RTlnKlnK=−ΔG0RT$

Where,
$ΔG0$ is the standard free energy of the reaction.
$R$ is the gas constant $(8.314 J/mol⋅K)$ .
$T$ is the temperature of the reaction.
$K$ is the equilibrium constant of the reaction.

Substitute the values of $ΔG0$ , $T$ and $R$ in above equation.

$lnK=−( −90.72× 10 3 J/mol)8.314 J/mol⋅K×298 K=36.6$

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

$K=e36.6=7.856×1015$

Conclusion:

The value of equilibrium constant for the given reaction is $7.856×1015$ .

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove $SO2$ or not.

(c)

Interpretation Introduction

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.

Answer 6

Chapter 5, Problem 34E

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at $298 K$ .

Solution: The value of equilibrium constant for the given reaction is $7.856×1015$ .

Explanation:

Given

The balanced chemical equation for the given reaction is,

$SO2(g)+2H2S(g)⇌3S(s)+2H2O(g)$

The temperature of the reaction is $298 K$ .

The standard free energy change of the given reaction is calculated by the formula,

$ΔG0=∑nΔG0( products)−∑mΔG0( reactants)=(2ΔG H 2 O( g )0+3ΔG S( s )0)−(ΔG SO 2 ( g )0+2ΔG H 2 S( g )0)$

Substitute the values of $ΔGH2O(g)0$ , $ΔGS(s)0$ , $ΔGSO2(g)0$ and $ΔGH2S(g)0$ from appendix C in equation.

$ΔH0=(2×−228.57 kJ/mol+3×0 kJ/mol)−(1×−300.4 kJ/mol+2×−33.01 kJ/mol)=−90.72 kJ/mol$

The conversion of free energy from $kJ/mol$ into $J/mol$ is done as,

$1 kJ/mol=103 J/mol$

Hence, the conversion of free energy from $−90.72 kJ/mol$ into $J/mol$ is,

$−90.72 kJ/mol=−90.72×103 J/mol$

The equilibrium constant for the given reaction is calculated by the formula,

$ΔG0=−RTlnKlnK=−ΔG0RT$

Where,
$ΔG0$ is the standard free energy of the reaction.
$R$ is the gas constant $(8.314 J/mol⋅K)$ .
$T$ is the temperature of the reaction.
$K$ is the equilibrium constant of the reaction.

Substitute the values of $ΔG0$ , $T$ and $R$ in above equation.

$lnK=−( −90.72× 10 3 J/mol)8.314 J/mol⋅K×298 K=36.6$

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

$K=e36.6=7.856×1015$

Conclusion:

The value of equilibrium constant for the given reaction is $7.856×1015$ .

Answer 7

Chapter 5, Problem 34E

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at $298 K$ .

Solution: The value of equilibrium constant for the given reaction is $7.856×1015$ .

Explanation:

Given

The balanced chemical equation for the given reaction is,

$SO2(g)+2H2S(g)⇌3S(s)+2H2O(g)$

The temperature of the reaction is $298 K$ .

The standard free energy change of the given reaction is calculated by the formula,

$ΔG0=∑nΔG0( products)−∑mΔG0( reactants)=(2ΔG H 2 O( g )0+3ΔG S( s )0)−(ΔG SO 2 ( g )0+2ΔG H 2 S( g )0)$

Substitute the values of $ΔGH2O(g)0$ , $ΔGS(s)0$ , $ΔGSO2(g)0$ and $ΔGH2S(g)0$ from appendix C in equation.

$ΔH0=(2×−228.57 kJ/mol+3×0 kJ/mol)−(1×−300.4 kJ/mol+2×−33.01 kJ/mol)=−90.72 kJ/mol$

The conversion of free energy from $kJ/mol$ into $J/mol$ is done as,

$1 kJ/mol=103 J/mol$

Hence, the conversion of free energy from $−90.72 kJ/mol$ into $J/mol$ is,

$−90.72 kJ/mol=−90.72×103 J/mol$

The equilibrium constant for the given reaction is calculated by the formula,

$ΔG0=−RTlnKlnK=−ΔG0RT$

Where,
$ΔG0$ is the standard free energy of the reaction.
$R$ is the gas constant $(8.314 J/mol⋅K)$ .
$T$ is the temperature of the reaction.
$K$ is the equilibrium constant of the reaction.

Substitute the values of $ΔG0$ , $T$ and $R$ in above equation.

$lnK=−( −90.72× 10 3 J/mol)8.314 J/mol⋅K×298 K=36.6$

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

$K=e36.6=7.856×1015$

Conclusion:

The value of equilibrium constant for the given reaction is $7.856×1015$ .

Answer 8

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove $SO2$ or not.

Answer 9

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove $SO2$ or not.

Answer 10

(c)

Interpretation Introduction

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .

Answer 11

(c)

Interpretation Introduction

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .

Answer 12

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.

Answer 13

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Answer 17

Ch. 5.2 - Prob. 5.1.1PE Ch. 5.2 - Prob. 5.1.2PE Ch. 5.3 - Prob. 5.2.1PE Ch. 5.3 - Prob. 5.2.2PE Ch. 5.3 - Prob. 5.3.1PE Ch. 5.3 - Prob. 5.3.2PE Ch. 5.4 - Prob. 5.4.1PE Ch. 5.4 - Prob. 5.4.2PE Ch. 5.5 - Prob. 5.5.1PE Ch. 5.5 - Prob. 5.5.2PE

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To determine: If the given reaction is feasible to remove $SO2$ or not.

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .

To determine: If the given reaction is more or less effective at higher temperature.

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Chapter 5 Solutions

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To determine: If the given reaction is feasible to remove SO2<m:math><m:mrow><m:msub><m:mrow><m:mtext>SO</m:mtext></m:mrow><m:mn>2</m:mn></m:msub></m:mrow></m:math> or not.

To determine: The equilibrium pressure of SO2<m:math><m:mrow><m:msub><m:mrow><m:mtext>SO</m:mtext></m:mrow><m:mn>2</m:mn></m:msub></m:mrow></m:math> gas at 298 K<m:math><m:mrow><m:mn>298</m:mn><m:mtext> </m:mtext><m:mtext>K</m:mtext></m:mrow></m:math> .

To determine: If the given reaction is more or less effective at higher temperature.

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Chapter 5 Solutions

To determine: If the given reaction is feasible to remove $SO2$ or not.

To determine: The equilibrium pressure of $SO2$ gas at $298 K$ .