The equilibrium constant at 25 o C has to be determined for the given reaction. CH 4 (g)+Cl 2 (g) ⇌ CH 3 Cl(g)+HCl(g) Concept Introduction: The equilibrium constant of a reaction can be calculated using the given expression, lnK= -ΔG r o RT where, R=Gas constant (8 .314 JK -1 mol -1 ) ΔG r ° = Standard Gibb's free energy T = Temperature K= Equilibrium constant

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Answer 1

Question

Chapter 5, Problem 5G.22E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

$lnK= -ΔGroRT where,R=Gas constant (8.314 JK-1mol-1)ΔGr° = Standard Gibb's free energyT = TemperatureK= Equilibrium constant$

(a)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.0×10−1$ .

Explanation of Solution

Given reaction is

The synthesis of trichloromethane (chloroform) from natural gas (methane):

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $CH3Cl$ is $48.5 k J⋅mol-1$

The $ΔGf°$ value of $HCl$ is $−95.30 k J⋅mol-1$

The $ΔGf°$ value of $Cl2$ is $0$

The $ΔGf°$ value of $CH4$ is $−50.72 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° CH3Cl(g)+ΔGf° HCl(g)]-[ΔGf° CH4(g)+ΔGf° Cl2(g)]=(48.5 ×103−95.30 ×103)−(−50.72 ×103−0)=4.0 ×103$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =-4.0×103 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=−1.61$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e−1.61=2.0×10−1$

Therefore, the $K$ value of given reaction is $2.0×10−1$ .

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is.

Explanation of Solution

Given reaction is

The hydrogenation of acetylene to ethane:

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $C2H6$ is $−32.82 k J⋅mol-1$

The $ΔGf°$ value of $C2H2$ is $+209.20 k J⋅mol-1$

The $ΔGf°$ value of $H2$ is $0$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =ΔGf° (C2H6(g))-[ΔGf° (C2H2(g))+2ΔGf° (H2(g))]=(-32.82 ×103)−[209.20 ×103−0]=−2.42 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =2.42 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=97.7$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e97.7=2.7×1042$

Therefore, the $K$ value of given reaction is $1×1090$ .

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Concept introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $1.8×109$ .

Explanation of Solution

Given reaction is

The final step in the industrial production of nitric acid:

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $HNO3$ is $−111.25 k J⋅mol-1$

The $ΔGf°$ value of $NO$ is $+86.55 k J⋅mol-1$

The $ΔGf°$ value of $NO2$ is $+51.31$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° 2HNO3(aq)+ΔGf° NO(g)]-[3ΔGf° NO2(g)+ΔGf° H2O(aq)]=[2×(-111.25 ×103)+86.55×103]−[3×(51.31×103)−237.13×103]=−5.28 ×104$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =5.28 ×104 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=21.31$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e21.31=1.8×109$

Therefore, the $K$ value of given reaction is $1.8×109$ .

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

Concept introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.4×10109$ .

Explanation of Solution

Given reaction is

The reaction of hydrazine and oxygen in a rocket:

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

The $ΔGf°$ value of $N2$ is $0$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

The $ΔGf°$ value of $N2H4$ is $+149.34 k J⋅mol-1$

The $ΔGf°$ value of $O2$ is $0$

Temperature of the reaction is $298 K$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° N2(g)+2ΔGf° H2O(l)]-[ΔGf° N2H4(l)+ΔGf° O2(g)]=[0+2×(−237.13×103)]−[149.34×103)−0]=−6.24 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =6.24 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=251.84$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e251.84=2.4×10109$

Therefore, the $K$ value of given reaction is $2.4×10109$ .

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0.500 moles of NOCl are placed into a 1.00 L vessesl at 700K and after the system comes to equilibrium, the consentration of NOCl is 0.440 M. Calculate the equilibrium constant Kc for the reaction: 2NOCL (g) --> 2NO (g) + Cl2 (g)

Answer 2

Question

Chapter 5, Problem 5G.22E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

$lnK= -ΔGroRT where,R=Gas constant (8.314 JK-1mol-1)ΔGr° = Standard Gibb's free energyT = TemperatureK= Equilibrium constant$

(a)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.0×10−1$ .

Explanation of Solution

Given reaction is

The synthesis of trichloromethane (chloroform) from natural gas (methane):

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $CH3Cl$ is $48.5 k J⋅mol-1$

The $ΔGf°$ value of $HCl$ is $−95.30 k J⋅mol-1$

The $ΔGf°$ value of $Cl2$ is $0$

The $ΔGf°$ value of $CH4$ is $−50.72 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° CH3Cl(g)+ΔGf° HCl(g)]-[ΔGf° CH4(g)+ΔGf° Cl2(g)]=(48.5 ×103−95.30 ×103)−(−50.72 ×103−0)=4.0 ×103$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =-4.0×103 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=−1.61$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e−1.61=2.0×10−1$

Therefore, the $K$ value of given reaction is $2.0×10−1$ .

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is.

Explanation of Solution

Given reaction is

The hydrogenation of acetylene to ethane:

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $C2H6$ is $−32.82 k J⋅mol-1$

The $ΔGf°$ value of $C2H2$ is $+209.20 k J⋅mol-1$

The $ΔGf°$ value of $H2$ is $0$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =ΔGf° (C2H6(g))-[ΔGf° (C2H2(g))+2ΔGf° (H2(g))]=(-32.82 ×103)−[209.20 ×103−0]=−2.42 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =2.42 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=97.7$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e97.7=2.7×1042$

Therefore, the $K$ value of given reaction is $1×1090$ .

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Concept introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $1.8×109$ .

Explanation of Solution

Given reaction is

The final step in the industrial production of nitric acid:

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $HNO3$ is $−111.25 k J⋅mol-1$

The $ΔGf°$ value of $NO$ is $+86.55 k J⋅mol-1$

The $ΔGf°$ value of $NO2$ is $+51.31$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° 2HNO3(aq)+ΔGf° NO(g)]-[3ΔGf° NO2(g)+ΔGf° H2O(aq)]=[2×(-111.25 ×103)+86.55×103]−[3×(51.31×103)−237.13×103]=−5.28 ×104$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =5.28 ×104 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=21.31$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e21.31=1.8×109$

Therefore, the $K$ value of given reaction is $1.8×109$ .

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

Concept introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.4×10109$ .

Explanation of Solution

Given reaction is

The reaction of hydrazine and oxygen in a rocket:

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

The $ΔGf°$ value of $N2$ is $0$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

The $ΔGf°$ value of $N2H4$ is $+149.34 k J⋅mol-1$

The $ΔGf°$ value of $O2$ is $0$

Temperature of the reaction is $298 K$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° N2(g)+2ΔGf° H2O(l)]-[ΔGf° N2H4(l)+ΔGf° O2(g)]=[0+2×(−237.13×103)]−[149.34×103)−0]=−6.24 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =6.24 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=251.84$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e251.84=2.4×10109$

Therefore, the $K$ value of given reaction is $2.4×10109$ .

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Answer 3

Question

Chapter 5, Problem 5G.22E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

$lnK= -ΔGroRT where,R=Gas constant (8.314 JK-1mol-1)ΔGr° = Standard Gibb's free energyT = TemperatureK= Equilibrium constant$

(a)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.0×10−1$ .

Explanation of Solution

Given reaction is

The synthesis of trichloromethane (chloroform) from natural gas (methane):

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $CH3Cl$ is $48.5 k J⋅mol-1$

The $ΔGf°$ value of $HCl$ is $−95.30 k J⋅mol-1$

The $ΔGf°$ value of $Cl2$ is $0$

The $ΔGf°$ value of $CH4$ is $−50.72 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° CH3Cl(g)+ΔGf° HCl(g)]-[ΔGf° CH4(g)+ΔGf° Cl2(g)]=(48.5 ×103−95.30 ×103)−(−50.72 ×103−0)=4.0 ×103$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =-4.0×103 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=−1.61$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e−1.61=2.0×10−1$

Therefore, the $K$ value of given reaction is $2.0×10−1$ .

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is.

Explanation of Solution

Given reaction is

The hydrogenation of acetylene to ethane:

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $C2H6$ is $−32.82 k J⋅mol-1$

The $ΔGf°$ value of $C2H2$ is $+209.20 k J⋅mol-1$

The $ΔGf°$ value of $H2$ is $0$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =ΔGf° (C2H6(g))-[ΔGf° (C2H2(g))+2ΔGf° (H2(g))]=(-32.82 ×103)−[209.20 ×103−0]=−2.42 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =2.42 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=97.7$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e97.7=2.7×1042$

Therefore, the $K$ value of given reaction is $1×1090$ .

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Concept introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $1.8×109$ .

Explanation of Solution

Given reaction is

The final step in the industrial production of nitric acid:

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $HNO3$ is $−111.25 k J⋅mol-1$

The $ΔGf°$ value of $NO$ is $+86.55 k J⋅mol-1$

The $ΔGf°$ value of $NO2$ is $+51.31$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° 2HNO3(aq)+ΔGf° NO(g)]-[3ΔGf° NO2(g)+ΔGf° H2O(aq)]=[2×(-111.25 ×103)+86.55×103]−[3×(51.31×103)−237.13×103]=−5.28 ×104$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =5.28 ×104 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=21.31$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e21.31=1.8×109$

Therefore, the $K$ value of given reaction is $1.8×109$ .

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

Concept introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.4×10109$ .

Explanation of Solution

Given reaction is

The reaction of hydrazine and oxygen in a rocket:

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

The $ΔGf°$ value of $N2$ is $0$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

The $ΔGf°$ value of $N2H4$ is $+149.34 k J⋅mol-1$

The $ΔGf°$ value of $O2$ is $0$

Temperature of the reaction is $298 K$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° N2(g)+2ΔGf° H2O(l)]-[ΔGf° N2H4(l)+ΔGf° O2(g)]=[0+2×(−237.13×103)]−[149.34×103)−0]=−6.24 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =6.24 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=251.84$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e251.84=2.4×10109$

Therefore, the $K$ value of given reaction is $2.4×10109$ .

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Answer 4

Question

Chapter 5, Problem 5G.22E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

$lnK= -ΔGroRT where,R=Gas constant (8.314 JK-1mol-1)ΔGr° = Standard Gibb's free energyT = TemperatureK= Equilibrium constant$

(a)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.0×10−1$ .

Explanation of Solution

Given reaction is

The synthesis of trichloromethane (chloroform) from natural gas (methane):

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $CH3Cl$ is $48.5 k J⋅mol-1$

The $ΔGf°$ value of $HCl$ is $−95.30 k J⋅mol-1$

The $ΔGf°$ value of $Cl2$ is $0$

The $ΔGf°$ value of $CH4$ is $−50.72 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° CH3Cl(g)+ΔGf° HCl(g)]-[ΔGf° CH4(g)+ΔGf° Cl2(g)]=(48.5 ×103−95.30 ×103)−(−50.72 ×103−0)=4.0 ×103$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =-4.0×103 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=−1.61$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e−1.61=2.0×10−1$

Therefore, the $K$ value of given reaction is $2.0×10−1$ .

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is.

Explanation of Solution

Given reaction is

The hydrogenation of acetylene to ethane:

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $C2H6$ is $−32.82 k J⋅mol-1$

The $ΔGf°$ value of $C2H2$ is $+209.20 k J⋅mol-1$

The $ΔGf°$ value of $H2$ is $0$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =ΔGf° (C2H6(g))-[ΔGf° (C2H2(g))+2ΔGf° (H2(g))]=(-32.82 ×103)−[209.20 ×103−0]=−2.42 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =2.42 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=97.7$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e97.7=2.7×1042$

Therefore, the $K$ value of given reaction is $1×1090$ .

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Concept introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $1.8×109$ .

Explanation of Solution

Given reaction is

The final step in the industrial production of nitric acid:

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $HNO3$ is $−111.25 k J⋅mol-1$

The $ΔGf°$ value of $NO$ is $+86.55 k J⋅mol-1$

The $ΔGf°$ value of $NO2$ is $+51.31$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° 2HNO3(aq)+ΔGf° NO(g)]-[3ΔGf° NO2(g)+ΔGf° H2O(aq)]=[2×(-111.25 ×103)+86.55×103]−[3×(51.31×103)−237.13×103]=−5.28 ×104$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =5.28 ×104 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=21.31$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e21.31=1.8×109$

Therefore, the $K$ value of given reaction is $1.8×109$ .

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

Concept introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.4×10109$ .

Explanation of Solution

Given reaction is

The reaction of hydrazine and oxygen in a rocket:

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

The $ΔGf°$ value of $N2$ is $0$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

The $ΔGf°$ value of $N2H4$ is $+149.34 k J⋅mol-1$

The $ΔGf°$ value of $O2$ is $0$

Temperature of the reaction is $298 K$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° N2(g)+2ΔGf° H2O(l)]-[ΔGf° N2H4(l)+ΔGf° O2(g)]=[0+2×(−237.13×103)]−[149.34×103)−0]=−6.24 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =6.24 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=251.84$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e251.84=2.4×10109$

Therefore, the $K$ value of given reaction is $2.4×10109$ .

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Answer 5

Chapter 5, Problem 5G.22E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

$lnK= -ΔGroRT where,R=Gas constant (8.314 JK-1mol-1)ΔGr° = Standard Gibb's free energyT = TemperatureK= Equilibrium constant$

(a)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.0×10−1$ .

Explanation of Solution

Given reaction is

The synthesis of trichloromethane (chloroform) from natural gas (methane):

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $CH3Cl$ is $48.5 k J⋅mol-1$

The $ΔGf°$ value of $HCl$ is $−95.30 k J⋅mol-1$

The $ΔGf°$ value of $Cl2$ is $0$

The $ΔGf°$ value of $CH4$ is $−50.72 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° CH3Cl(g)+ΔGf° HCl(g)]-[ΔGf° CH4(g)+ΔGf° Cl2(g)]=(48.5 ×103−95.30 ×103)−(−50.72 ×103−0)=4.0 ×103$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =-4.0×103 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=−1.61$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e−1.61=2.0×10−1$

Therefore, the $K$ value of given reaction is $2.0×10−1$ .

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is.

Explanation of Solution

Given reaction is

The hydrogenation of acetylene to ethane:

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $C2H6$ is $−32.82 k J⋅mol-1$

The $ΔGf°$ value of $C2H2$ is $+209.20 k J⋅mol-1$

The $ΔGf°$ value of $H2$ is $0$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =ΔGf° (C2H6(g))-[ΔGf° (C2H2(g))+2ΔGf° (H2(g))]=(-32.82 ×103)−[209.20 ×103−0]=−2.42 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =2.42 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=97.7$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e97.7=2.7×1042$

Therefore, the $K$ value of given reaction is $1×1090$ .

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Concept introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $1.8×109$ .

Explanation of Solution

Given reaction is

The final step in the industrial production of nitric acid:

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $HNO3$ is $−111.25 k J⋅mol-1$

The $ΔGf°$ value of $NO$ is $+86.55 k J⋅mol-1$

The $ΔGf°$ value of $NO2$ is $+51.31$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° 2HNO3(aq)+ΔGf° NO(g)]-[3ΔGf° NO2(g)+ΔGf° H2O(aq)]=[2×(-111.25 ×103)+86.55×103]−[3×(51.31×103)−237.13×103]=−5.28 ×104$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =5.28 ×104 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=21.31$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e21.31=1.8×109$

Therefore, the $K$ value of given reaction is $1.8×109$ .

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

Concept introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.4×10109$ .

Explanation of Solution

Given reaction is

The reaction of hydrazine and oxygen in a rocket:

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

The $ΔGf°$ value of $N2$ is $0$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

The $ΔGf°$ value of $N2H4$ is $+149.34 k J⋅mol-1$

The $ΔGf°$ value of $O2$ is $0$

Temperature of the reaction is $298 K$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° N2(g)+2ΔGf° H2O(l)]-[ΔGf° N2H4(l)+ΔGf° O2(g)]=[0+2×(−237.13×103)]−[149.34×103)−0]=−6.24 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =6.24 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=251.84$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e251.84=2.4×10109$

Therefore, the $K$ value of given reaction is $2.4×10109$ .

Answer 6

Chapter 5, Problem 5G.22E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

$lnK= -ΔGroRT where,R=Gas constant (8.314 JK-1mol-1)ΔGr° = Standard Gibb's free energyT = TemperatureK= Equilibrium constant$

(a)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.0×10−1$ .

Explanation of Solution

Given reaction is

The synthesis of trichloromethane (chloroform) from natural gas (methane):

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $CH3Cl$ is $48.5 k J⋅mol-1$

The $ΔGf°$ value of $HCl$ is $−95.30 k J⋅mol-1$

The $ΔGf°$ value of $Cl2$ is $0$

The $ΔGf°$ value of $CH4$ is $−50.72 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° CH3Cl(g)+ΔGf° HCl(g)]-[ΔGf° CH4(g)+ΔGf° Cl2(g)]=(48.5 ×103−95.30 ×103)−(−50.72 ×103−0)=4.0 ×103$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =-4.0×103 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=−1.61$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e−1.61=2.0×10−1$

Therefore, the $K$ value of given reaction is $2.0×10−1$ .

Answer 7

Chapter 5, Problem 5G.22E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

$lnK= -ΔGroRT where,R=Gas constant (8.314 JK-1mol-1)ΔGr° = Standard Gibb's free energyT = TemperatureK= Equilibrium constant$

Answer 8

(a)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.0×10−1$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given reaction is

The synthesis of trichloromethane (chloroform) from natural gas (methane):

$CH4(g)+Cl2(g)⇌ CH3Cl(g)+HCl(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $CH3Cl$ is $48.5 k J⋅mol-1$

The $ΔGf°$ value of $HCl$ is $−95.30 k J⋅mol-1$

The $ΔGf°$ value of $Cl2$ is $0$

The $ΔGf°$ value of $CH4$ is $−50.72 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° CH3Cl(g)+ΔGf° HCl(g)]-[ΔGf° CH4(g)+ΔGf° Cl2(g)]=(48.5 ×103−95.30 ×103)−(−50.72 ×103−0)=4.0 ×103$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =-4.0×103 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=−1.61$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e−1.61=2.0×10−1$

Therefore, the $K$ value of given reaction is $2.0×10−1$ .

Answer 13

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is.

Explanation of Solution

Given reaction is

The hydrogenation of acetylene to ethane:

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $C2H6$ is $−32.82 k J⋅mol-1$

The $ΔGf°$ value of $C2H2$ is $+209.20 k J⋅mol-1$

The $ΔGf°$ value of $H2$ is $0$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =ΔGf° (C2H6(g))-[ΔGf° (C2H2(g))+2ΔGf° (H2(g))]=(-32.82 ×103)−[209.20 ×103−0]=−2.42 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =2.42 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=97.7$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e97.7=2.7×1042$

Therefore, the $K$ value of given reaction is $1×1090$ .

Answer 14

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Concept introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given reaction is

The hydrogenation of acetylene to ethane:

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $C2H6$ is $−32.82 k J⋅mol-1$

The $ΔGf°$ value of $C2H2$ is $+209.20 k J⋅mol-1$

The $ΔGf°$ value of $H2$ is $0$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =ΔGf° (C2H6(g))-[ΔGf° (C2H2(g))+2ΔGf° (H2(g))]=(-32.82 ×103)−[209.20 ×103−0]=−2.42 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =2.42 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=97.7$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e97.7=2.7×1042$

Therefore, the $K$ value of given reaction is $1×1090$ .

Answer 20

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Concept introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $1.8×109$ .

Explanation of Solution

Given reaction is

The final step in the industrial production of nitric acid:

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $HNO3$ is $−111.25 k J⋅mol-1$

The $ΔGf°$ value of $NO$ is $+86.55 k J⋅mol-1$

The $ΔGf°$ value of $NO2$ is $+51.31$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° 2HNO3(aq)+ΔGf° NO(g)]-[3ΔGf° NO2(g)+ΔGf° H2O(aq)]=[2×(-111.25 ×103)+86.55×103]−[3×(51.31×103)−237.13×103]=−5.28 ×104$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =5.28 ×104 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=21.31$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e21.31=1.8×109$

Therefore, the $K$ value of given reaction is $1.8×109$ .

Answer 21

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Concept introduction:

Refer to part (a).

Answer 22

(c)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $1.8×109$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given reaction is

The final step in the industrial production of nitric acid:

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Temperature of the reaction is $298 K$ ,

The $ΔGf°$ value of $HNO3$ is $−111.25 k J⋅mol-1$

The $ΔGf°$ value of $NO$ is $+86.55 k J⋅mol-1$

The $ΔGf°$ value of $NO2$ is $+51.31$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° 2HNO3(aq)+ΔGf° NO(g)]-[3ΔGf° NO2(g)+ΔGf° H2O(aq)]=[2×(-111.25 ×103)+86.55×103]−[3×(51.31×103)−237.13×103]=−5.28 ×104$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =5.28 ×104 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=21.31$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e21.31=1.8×109$

Therefore, the $K$ value of given reaction is $1.8×109$ .

Answer 27

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

Concept introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.4×10109$ .

Explanation of Solution

Given reaction is

The reaction of hydrazine and oxygen in a rocket:

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

The $ΔGf°$ value of $N2$ is $0$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

The $ΔGf°$ value of $N2H4$ is $+149.34 k J⋅mol-1$

The $ΔGf°$ value of $O2$ is $0$

Temperature of the reaction is $298 K$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° N2(g)+2ΔGf° H2O(l)]-[ΔGf° N2H4(l)+ΔGf° O2(g)]=[0+2×(−237.13×103)]−[149.34×103)−0]=−6.24 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =6.24 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=251.84$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e251.84=2.4×10109$

Therefore, the $K$ value of given reaction is $2.4×10109$ .

Answer 28

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

Concept introduction:

Refer to part (a).

Answer 29

(d)

Expert Solution

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $2.4×10109$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given reaction is

The reaction of hydrazine and oxygen in a rocket:

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

The $ΔGf°$ value of $N2$ is $0$

The $ΔGf°$ value of $H2O$ is $−237.13 k J⋅mol-1$

The $ΔGf°$ value of $N2H4$ is $+149.34 k J⋅mol-1$

The $ΔGf°$ value of $O2$ is $0$

Temperature of the reaction is $298 K$

$R= 8.3145 J ⋅K-1⋅mol-1$

$T=298K$

Here, $ΔGr°$ is calculated using given formula,

$ΔGr° =[ΔGf° N2(g)+2ΔGf° H2O(l)]-[ΔGf° N2H4(l)+ΔGf° O2(g)]=[0+2×(−237.13×103)]−[149.34×103)−0]=−6.24 ×105$

Now, the equilibrium constant of the reaction is calculated as

$lnK= -ΔGroRT =6.24 ×105 J⋅mol-1(8.3145 J⋅K-1⋅mol-1)×(298 K)=251.84$

To calculate $K$ value, inverse logarithm is taken for the above reaction.

$K=e251.84=2.4×10109$

Therefore, the $K$ value of given reaction is $2.4×10109$ .

Answer 34

Ch. 5 - Prob. 5A.1AST Ch. 5 - Prob. 5A.1BST Ch. 5 - Prob. 5A.2AST Ch. 5 - Prob. 5A.2BST Ch. 5 - Prob. 5A.3AST Ch. 5 - Prob. 5A.3BST Ch. 5 - Prob. 5A.1E Ch. 5 - Prob. 5A.2E Ch. 5 - Prob. 5A.3E Ch. 5 - Prob. 5A.4E

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Answer to Problem 5G.22E

Explanation of Solution

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$C2H2(g)+2H2(g)⇌ C2H6(g)$

Concept introduction:

Refer to part (a).

Answer to Problem 5G.22E

Explanation of Solution

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$3NO2(g)+H2O(l)⇌ 2HNO3(aq)+NO(g)$

Concept introduction:

Refer to part (a).

Answer to Problem 5G.22E

Explanation of Solution

Interpretation:

The equilibrium constant at $25oC$ has to be determined for the given reaction.

$N2H4(l)+O2(g)⇌ N2(g)+2H2O(l)$

Concept introduction:

Refer to part (a).

Answer to Problem 5G.22E

Explanation of Solution

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Chapter 5 Solutions