The values of Δ r H ο and Δ r S ο for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated. Concept Introduction: The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

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Answer 1

Question

Chapter 16, Problem 50QRT

(a)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

(a)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ and the value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]$

The value of $Hfο(CO2(g))$ is $−393.509 kJ mol−1$ .

The value of $Hfο(H2O(l))$ is $−285.83 kJ mol−1$ .

The value of $Hfο(C6H12O6(s))$ is $−1274.4 kJ mol−1$ .

The value of $Hfο(O2(g))$ is $0 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]=[6×(−393.509 kJ mol−1)+6×(−285.83 kJ mol−1)]−[1×(−1274.4 kJ mol−1)+6×(0 kJ mol−1)]=−4075.98 kJ mol−1+1274.4 kJ mol−1=-2801.58 kJ mol-1_$

The value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]$

The value of $Sfο(CO2(g))$ is $213.74 J k−1 mol−1$ .

The value of $Sfο(H2O(l))$ is $69.91 J k−1 mol−1$ .

The value of $Sfο(C6H12O6(s))$ is $235.9 J k−1 mol−1$ .

The value of $Sfο(O2(g))$ is $205.138 J k−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]=[6×(213.74 J k−1 mol−1)+6×(69.91 J k−1 mol−1)]−[1×(235.9 J k−1 mol−1)+6×(205.138 J k−1 mol−1)]=1701.9 J K−1 mol−1−1466.728 J K−1 mol−1=235.172 J K-1 mol-1_$

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is also negative. Therefore, the Gibbs free energy will be negative. Hence, the reaction will be product-favored.

(b)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

Refer to part (a)

(b)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $-121.316 J K-1 mol-1_$ and the value of $ΔrHο$ is $166.36 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$MgO(s)+C(s,graphite)→Mg(s)+CO(g)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]$

The value of $Hfο(MgO(s))$ is $−601.7 kJ mol−1$ .

The value of $Hfο(C(s))$ is $0 kJ mol−1$ .

The value of $Hfο(Mg(s))$ is $0 kJ mol−1$ .

The value of $Hfο(CO(g))$ is $−110.525 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]=[1×(0 kJ mol−1)+1×(−110.525 kJ mol−1)]−[1×(−601.7 kJ mol−1)+1×(0 kJ mol−1)]=491.175 kJ mol-1_$

The value of $ΔrHο$ is $491.175 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]$

The value of $Sfο(MgO(s))$ is $26.94 J K−1 mol−1$ .

The value of $Sfο(C(s))$ is $5.74 J K−1 mol−1$ .

The value of $Sfο(Mg(s))$ is $32.68 J K−1 mol−1$ .

The value of $Sfο(CO(g))$ is $197.674 J K−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]=[1×(32.68 J K−1 mol−1)+1×(197.674 J K−1 mol−1)]−[1×(26.94 J K−1 mol−1)+1×(5.74 J K−1 mol−1)]=230.354 J K−1 mol−1−32.68 J K−1 mol−1=197.674 J K-1 mol-1_$

The value of $ΔrSο$ is $197.674 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is positive. Therefore, the Gibbs free energy will negative at high temperatures. Hence, the reaction will be product-favored at high-temperatures.

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Answer 2

Question

Chapter 16, Problem 50QRT

(a)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

(a)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ and the value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]$

The value of $Hfο(CO2(g))$ is $−393.509 kJ mol−1$ .

The value of $Hfο(H2O(l))$ is $−285.83 kJ mol−1$ .

The value of $Hfο(C6H12O6(s))$ is $−1274.4 kJ mol−1$ .

The value of $Hfο(O2(g))$ is $0 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]=[6×(−393.509 kJ mol−1)+6×(−285.83 kJ mol−1)]−[1×(−1274.4 kJ mol−1)+6×(0 kJ mol−1)]=−4075.98 kJ mol−1+1274.4 kJ mol−1=-2801.58 kJ mol-1_$

The value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]$

The value of $Sfο(CO2(g))$ is $213.74 J k−1 mol−1$ .

The value of $Sfο(H2O(l))$ is $69.91 J k−1 mol−1$ .

The value of $Sfο(C6H12O6(s))$ is $235.9 J k−1 mol−1$ .

The value of $Sfο(O2(g))$ is $205.138 J k−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]=[6×(213.74 J k−1 mol−1)+6×(69.91 J k−1 mol−1)]−[1×(235.9 J k−1 mol−1)+6×(205.138 J k−1 mol−1)]=1701.9 J K−1 mol−1−1466.728 J K−1 mol−1=235.172 J K-1 mol-1_$

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is also negative. Therefore, the Gibbs free energy will be negative. Hence, the reaction will be product-favored.

(b)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

Refer to part (a)

(b)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $-121.316 J K-1 mol-1_$ and the value of $ΔrHο$ is $166.36 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$MgO(s)+C(s,graphite)→Mg(s)+CO(g)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]$

The value of $Hfο(MgO(s))$ is $−601.7 kJ mol−1$ .

The value of $Hfο(C(s))$ is $0 kJ mol−1$ .

The value of $Hfο(Mg(s))$ is $0 kJ mol−1$ .

The value of $Hfο(CO(g))$ is $−110.525 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]=[1×(0 kJ mol−1)+1×(−110.525 kJ mol−1)]−[1×(−601.7 kJ mol−1)+1×(0 kJ mol−1)]=491.175 kJ mol-1_$

The value of $ΔrHο$ is $491.175 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]$

The value of $Sfο(MgO(s))$ is $26.94 J K−1 mol−1$ .

The value of $Sfο(C(s))$ is $5.74 J K−1 mol−1$ .

The value of $Sfο(Mg(s))$ is $32.68 J K−1 mol−1$ .

The value of $Sfο(CO(g))$ is $197.674 J K−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]=[1×(32.68 J K−1 mol−1)+1×(197.674 J K−1 mol−1)]−[1×(26.94 J K−1 mol−1)+1×(5.74 J K−1 mol−1)]=230.354 J K−1 mol−1−32.68 J K−1 mol−1=197.674 J K-1 mol-1_$

The value of $ΔrSο$ is $197.674 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is positive. Therefore, the Gibbs free energy will negative at high temperatures. Hence, the reaction will be product-favored at high-temperatures.

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Answer 3

Question

Chapter 16, Problem 50QRT

(a)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

(a)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ and the value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]$

The value of $Hfο(CO2(g))$ is $−393.509 kJ mol−1$ .

The value of $Hfο(H2O(l))$ is $−285.83 kJ mol−1$ .

The value of $Hfο(C6H12O6(s))$ is $−1274.4 kJ mol−1$ .

The value of $Hfο(O2(g))$ is $0 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]=[6×(−393.509 kJ mol−1)+6×(−285.83 kJ mol−1)]−[1×(−1274.4 kJ mol−1)+6×(0 kJ mol−1)]=−4075.98 kJ mol−1+1274.4 kJ mol−1=-2801.58 kJ mol-1_$

The value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]$

The value of $Sfο(CO2(g))$ is $213.74 J k−1 mol−1$ .

The value of $Sfο(H2O(l))$ is $69.91 J k−1 mol−1$ .

The value of $Sfο(C6H12O6(s))$ is $235.9 J k−1 mol−1$ .

The value of $Sfο(O2(g))$ is $205.138 J k−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]=[6×(213.74 J k−1 mol−1)+6×(69.91 J k−1 mol−1)]−[1×(235.9 J k−1 mol−1)+6×(205.138 J k−1 mol−1)]=1701.9 J K−1 mol−1−1466.728 J K−1 mol−1=235.172 J K-1 mol-1_$

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is also negative. Therefore, the Gibbs free energy will be negative. Hence, the reaction will be product-favored.

(b)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

Refer to part (a)

(b)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $-121.316 J K-1 mol-1_$ and the value of $ΔrHο$ is $166.36 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$MgO(s)+C(s,graphite)→Mg(s)+CO(g)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]$

The value of $Hfο(MgO(s))$ is $−601.7 kJ mol−1$ .

The value of $Hfο(C(s))$ is $0 kJ mol−1$ .

The value of $Hfο(Mg(s))$ is $0 kJ mol−1$ .

The value of $Hfο(CO(g))$ is $−110.525 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]=[1×(0 kJ mol−1)+1×(−110.525 kJ mol−1)]−[1×(−601.7 kJ mol−1)+1×(0 kJ mol−1)]=491.175 kJ mol-1_$

The value of $ΔrHο$ is $491.175 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]$

The value of $Sfο(MgO(s))$ is $26.94 J K−1 mol−1$ .

The value of $Sfο(C(s))$ is $5.74 J K−1 mol−1$ .

The value of $Sfο(Mg(s))$ is $32.68 J K−1 mol−1$ .

The value of $Sfο(CO(g))$ is $197.674 J K−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]=[1×(32.68 J K−1 mol−1)+1×(197.674 J K−1 mol−1)]−[1×(26.94 J K−1 mol−1)+1×(5.74 J K−1 mol−1)]=230.354 J K−1 mol−1−32.68 J K−1 mol−1=197.674 J K-1 mol-1_$

The value of $ΔrSο$ is $197.674 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is positive. Therefore, the Gibbs free energy will negative at high temperatures. Hence, the reaction will be product-favored at high-temperatures.

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Answer 4

Question

Chapter 16, Problem 50QRT

(a)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

(a)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ and the value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]$

The value of $Hfο(CO2(g))$ is $−393.509 kJ mol−1$ .

The value of $Hfο(H2O(l))$ is $−285.83 kJ mol−1$ .

The value of $Hfο(C6H12O6(s))$ is $−1274.4 kJ mol−1$ .

The value of $Hfο(O2(g))$ is $0 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]=[6×(−393.509 kJ mol−1)+6×(−285.83 kJ mol−1)]−[1×(−1274.4 kJ mol−1)+6×(0 kJ mol−1)]=−4075.98 kJ mol−1+1274.4 kJ mol−1=-2801.58 kJ mol-1_$

The value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]$

The value of $Sfο(CO2(g))$ is $213.74 J k−1 mol−1$ .

The value of $Sfο(H2O(l))$ is $69.91 J k−1 mol−1$ .

The value of $Sfο(C6H12O6(s))$ is $235.9 J k−1 mol−1$ .

The value of $Sfο(O2(g))$ is $205.138 J k−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]=[6×(213.74 J k−1 mol−1)+6×(69.91 J k−1 mol−1)]−[1×(235.9 J k−1 mol−1)+6×(205.138 J k−1 mol−1)]=1701.9 J K−1 mol−1−1466.728 J K−1 mol−1=235.172 J K-1 mol-1_$

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is also negative. Therefore, the Gibbs free energy will be negative. Hence, the reaction will be product-favored.

(b)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

Refer to part (a)

(b)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $-121.316 J K-1 mol-1_$ and the value of $ΔrHο$ is $166.36 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$MgO(s)+C(s,graphite)→Mg(s)+CO(g)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]$

The value of $Hfο(MgO(s))$ is $−601.7 kJ mol−1$ .

The value of $Hfο(C(s))$ is $0 kJ mol−1$ .

The value of $Hfο(Mg(s))$ is $0 kJ mol−1$ .

The value of $Hfο(CO(g))$ is $−110.525 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]=[1×(0 kJ mol−1)+1×(−110.525 kJ mol−1)]−[1×(−601.7 kJ mol−1)+1×(0 kJ mol−1)]=491.175 kJ mol-1_$

The value of $ΔrHο$ is $491.175 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]$

The value of $Sfο(MgO(s))$ is $26.94 J K−1 mol−1$ .

The value of $Sfο(C(s))$ is $5.74 J K−1 mol−1$ .

The value of $Sfο(Mg(s))$ is $32.68 J K−1 mol−1$ .

The value of $Sfο(CO(g))$ is $197.674 J K−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]=[1×(32.68 J K−1 mol−1)+1×(197.674 J K−1 mol−1)]−[1×(26.94 J K−1 mol−1)+1×(5.74 J K−1 mol−1)]=230.354 J K−1 mol−1−32.68 J K−1 mol−1=197.674 J K-1 mol-1_$

The value of $ΔrSο$ is $197.674 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is positive. Therefore, the Gibbs free energy will negative at high temperatures. Hence, the reaction will be product-favored at high-temperatures.

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Answer 5

Chapter 16, Problem 50QRT

(a)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

(a)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ and the value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]$

The value of $Hfο(CO2(g))$ is $−393.509 kJ mol−1$ .

The value of $Hfο(H2O(l))$ is $−285.83 kJ mol−1$ .

The value of $Hfο(C6H12O6(s))$ is $−1274.4 kJ mol−1$ .

The value of $Hfο(O2(g))$ is $0 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]=[6×(−393.509 kJ mol−1)+6×(−285.83 kJ mol−1)]−[1×(−1274.4 kJ mol−1)+6×(0 kJ mol−1)]=−4075.98 kJ mol−1+1274.4 kJ mol−1=-2801.58 kJ mol-1_$

The value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]$

The value of $Sfο(CO2(g))$ is $213.74 J k−1 mol−1$ .

The value of $Sfο(H2O(l))$ is $69.91 J k−1 mol−1$ .

The value of $Sfο(C6H12O6(s))$ is $235.9 J k−1 mol−1$ .

The value of $Sfο(O2(g))$ is $205.138 J k−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]=[6×(213.74 J k−1 mol−1)+6×(69.91 J k−1 mol−1)]−[1×(235.9 J k−1 mol−1)+6×(205.138 J k−1 mol−1)]=1701.9 J K−1 mol−1−1466.728 J K−1 mol−1=235.172 J K-1 mol-1_$

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is also negative. Therefore, the Gibbs free energy will be negative. Hence, the reaction will be product-favored.

(b)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

Refer to part (a)

(b)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $-121.316 J K-1 mol-1_$ and the value of $ΔrHο$ is $166.36 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$MgO(s)+C(s,graphite)→Mg(s)+CO(g)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]$

The value of $Hfο(MgO(s))$ is $−601.7 kJ mol−1$ .

The value of $Hfο(C(s))$ is $0 kJ mol−1$ .

The value of $Hfο(Mg(s))$ is $0 kJ mol−1$ .

The value of $Hfο(CO(g))$ is $−110.525 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]=[1×(0 kJ mol−1)+1×(−110.525 kJ mol−1)]−[1×(−601.7 kJ mol−1)+1×(0 kJ mol−1)]=491.175 kJ mol-1_$

The value of $ΔrHο$ is $491.175 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]$

The value of $Sfο(MgO(s))$ is $26.94 J K−1 mol−1$ .

The value of $Sfο(C(s))$ is $5.74 J K−1 mol−1$ .

The value of $Sfο(Mg(s))$ is $32.68 J K−1 mol−1$ .

The value of $Sfο(CO(g))$ is $197.674 J K−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]=[1×(32.68 J K−1 mol−1)+1×(197.674 J K−1 mol−1)]−[1×(26.94 J K−1 mol−1)+1×(5.74 J K−1 mol−1)]=230.354 J K−1 mol−1−32.68 J K−1 mol−1=197.674 J K-1 mol-1_$

The value of $ΔrSο$ is $197.674 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is positive. Therefore, the Gibbs free energy will negative at high temperatures. Hence, the reaction will be product-favored at high-temperatures.

Answer 6

Chapter 16, Problem 50QRT

(a)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

(a)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ and the value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]$

The value of $Hfο(CO2(g))$ is $−393.509 kJ mol−1$ .

The value of $Hfο(H2O(l))$ is $−285.83 kJ mol−1$ .

The value of $Hfο(C6H12O6(s))$ is $−1274.4 kJ mol−1$ .

The value of $Hfο(O2(g))$ is $0 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]=[6×(−393.509 kJ mol−1)+6×(−285.83 kJ mol−1)]−[1×(−1274.4 kJ mol−1)+6×(0 kJ mol−1)]=−4075.98 kJ mol−1+1274.4 kJ mol−1=-2801.58 kJ mol-1_$

The value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]$

The value of $Sfο(CO2(g))$ is $213.74 J k−1 mol−1$ .

The value of $Sfο(H2O(l))$ is $69.91 J k−1 mol−1$ .

The value of $Sfο(C6H12O6(s))$ is $235.9 J k−1 mol−1$ .

The value of $Sfο(O2(g))$ is $205.138 J k−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]=[6×(213.74 J k−1 mol−1)+6×(69.91 J k−1 mol−1)]−[1×(235.9 J k−1 mol−1)+6×(205.138 J k−1 mol−1)]=1701.9 J K−1 mol−1−1466.728 J K−1 mol−1=235.172 J K-1 mol-1_$

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is also negative. Therefore, the Gibbs free energy will be negative. Hence, the reaction will be product-favored.

Answer 7

Chapter 16, Problem 50QRT

(a)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer 8

(a)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ and the value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The given reaction is shown below.

$C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]$

The value of $Hfο(CO2(g))$ is $−393.509 kJ mol−1$ .

The value of $Hfο(H2O(l))$ is $−285.83 kJ mol−1$ .

The value of $Hfο(C6H12O6(s))$ is $−1274.4 kJ mol−1$ .

The value of $Hfο(O2(g))$ is $0 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[6⋅Hfο(CO2(g))+6⋅Hfο(H2O(l))]−[1⋅Hfο(C6H12O6(s))+6⋅Hfο(O2(g))]=[6×(−393.509 kJ mol−1)+6×(−285.83 kJ mol−1)]−[1×(−1274.4 kJ mol−1)+6×(0 kJ mol−1)]=−4075.98 kJ mol−1+1274.4 kJ mol−1=-2801.58 kJ mol-1_$

The value of $ΔrHο$ is $-2801.58 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]$

The value of $Sfο(CO2(g))$ is $213.74 J k−1 mol−1$ .

The value of $Sfο(H2O(l))$ is $69.91 J k−1 mol−1$ .

The value of $Sfο(C6H12O6(s))$ is $235.9 J k−1 mol−1$ .

The value of $Sfο(O2(g))$ is $205.138 J k−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[6⋅Sfο(CO2(g))+6⋅Sfο(H2O(l))]−[1⋅Sfο(C6H12O6(s))+6⋅Sfο(O2(g))]=[6×(213.74 J k−1 mol−1)+6×(69.91 J k−1 mol−1)]−[1×(235.9 J k−1 mol−1)+6×(205.138 J k−1 mol−1)]=1701.9 J K−1 mol−1−1466.728 J K−1 mol−1=235.172 J K-1 mol-1_$

The value of $ΔrSο$ is $235.172 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is also negative. Therefore, the Gibbs free energy will be negative. Hence, the reaction will be product-favored.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

Refer to part (a)

(b)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $-121.316 J K-1 mol-1_$ and the value of $ΔrHο$ is $166.36 kJ mol-1_$ .

Explanation of Solution

The given reaction is shown below.

$MgO(s)+C(s,graphite)→Mg(s)+CO(g)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]$

The value of $Hfο(MgO(s))$ is $−601.7 kJ mol−1$ .

The value of $Hfο(C(s))$ is $0 kJ mol−1$ .

The value of $Hfο(Mg(s))$ is $0 kJ mol−1$ .

The value of $Hfο(CO(g))$ is $−110.525 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]=[1×(0 kJ mol−1)+1×(−110.525 kJ mol−1)]−[1×(−601.7 kJ mol−1)+1×(0 kJ mol−1)]=491.175 kJ mol-1_$

The value of $ΔrHο$ is $491.175 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]$

The value of $Sfο(MgO(s))$ is $26.94 J K−1 mol−1$ .

The value of $Sfο(C(s))$ is $5.74 J K−1 mol−1$ .

The value of $Sfο(Mg(s))$ is $32.68 J K−1 mol−1$ .

The value of $Sfο(CO(g))$ is $197.674 J K−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]=[1×(32.68 J K−1 mol−1)+1×(197.674 J K−1 mol−1)]−[1×(26.94 J K−1 mol−1)+1×(5.74 J K−1 mol−1)]=230.354 J K−1 mol−1−32.68 J K−1 mol−1=197.674 J K-1 mol-1_$

The value of $ΔrSο$ is $197.674 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is positive. Therefore, the Gibbs free energy will negative at high temperatures. Hence, the reaction will be product-favored at high-temperatures.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The values of $ΔrHο$ and $ΔrSο$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

Refer to part (a)

Answer 15

(b)

Expert Solution

Answer to Problem 50QRT

The value of $ΔrSο$ is $-121.316 J K-1 mol-1_$ and the value of $ΔrHο$ is $166.36 kJ mol-1_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The given reaction is shown below.

$MgO(s)+C(s,graphite)→Mg(s)+CO(g)$

The standard enthalpy change for the reaction ( $ΔrHο$ ) is calculated by using the expression shown below.

$ΔrHο=∑m⋅Hο(products)−∑n⋅Hο(reactants)$

Where,

$Hο(products)$ is the standard heat of formation of products.
$Hο(reactants)$ is the standard heat of formation of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]$

The value of $Hfο(MgO(s))$ is $−601.7 kJ mol−1$ .

The value of $Hfο(C(s))$ is $0 kJ mol−1$ .

The value of $Hfο(Mg(s))$ is $0 kJ mol−1$ .

The value of $Hfο(CO(g))$ is $−110.525 kJ mol−1$ .

Substitute the values in the above expression.

$ΔrHο=[1⋅Hfο(Mg(s))+1⋅Hfο(CO(g))]−[1⋅Hfο(MgO(s))+1⋅Hfο(C(s))]=[1×(0 kJ mol−1)+1×(−110.525 kJ mol−1)]−[1×(−601.7 kJ mol−1)+1×(0 kJ mol−1)]=491.175 kJ mol-1_$

The value of $ΔrHο$ is $491.175 kJ mol-1_$ .

The standard entropy change for the reaction ( $ΔrSο$ ) is calculated by using the expression shown below.

$ΔrSο=∑m⋅Sο(products)−∑n⋅Sο(reactants)$

Where,

$Sο(products)$ is the absolute molar entropy of products.
$Sο(reactants)$ is the absolute molar entropy of reactants.
$m$ is the total moles of products.
$n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]$

The value of $Sfο(MgO(s))$ is $26.94 J K−1 mol−1$ .

The value of $Sfο(C(s))$ is $5.74 J K−1 mol−1$ .

The value of $Sfο(Mg(s))$ is $32.68 J K−1 mol−1$ .

The value of $Sfο(CO(g))$ is $197.674 J K−1 mol−1$ .

Substitute the values in the above expression.

$ΔrSο=[1⋅Sfο(Mg(s))+1⋅Sfο(CO(g))]−[1⋅Sfο(MgO(s))+1⋅Sfο(C(s))]=[1×(32.68 J K−1 mol−1)+1×(197.674 J K−1 mol−1)]−[1×(26.94 J K−1 mol−1)+1×(5.74 J K−1 mol−1)]=230.354 J K−1 mol−1−32.68 J K−1 mol−1=197.674 J K-1 mol-1_$

The value of $ΔrSο$ is $197.674 J K-1 mol-1_$ .

The value of entropy is positive; therefore, the term $TΔrSο$ will be negative. Enthalpy is positive. Therefore, the Gibbs free energy will negative at high temperatures. Hence, the reaction will be product-favored at high-temperatures.