Chapter 4, Problem 4.24P

Interpretation Introduction

(a)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 600 ⁰C

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 600 ⁰C = 873 K

$\tau =\frac{T}{T_o}$ = 2.92953

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-109799$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(2.92953-1)+\frac{0.004181}{2}\times {298}_{}^2({2.92953}^2-1)+0+\frac{-66100}{298}(\frac{2.92953-1}{2.92953})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-2114.1$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -2114.1$

  $\Delta H_R^0=-17575.6\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

Heat of formation of ammonia is, $\Delta H^0{}_{298}=$ -46110 J/mol

For 2 moles of NH_3,$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-17575.6-92220=-109799$ J

Interpretation Introduction

(b)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 50 ⁰C

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 50 ⁰C = 323 K

$\tau =\frac{T}{T_o}$ = 1.08389

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-93313.2$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(1.083892-1)+\frac{0.004181}{2}\times {298}_{}^2({1.083892}^2-1)+0+\frac{-66100}{298}(\frac{1.083892-1}{1.083892})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-131.488$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -131.488$

  $\Delta H_R^0=-1093.19\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-1093.19-92220=-93313.2$ J

Interpretation Introduction

(f)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 650 ⁰C

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 650 ⁰C = 923 K

$\tau =\frac{T}{T_o}$ = 3.0973154

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-110712$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(3.097315-1)+\frac{0.004181}{2}\times {298}_{}^2({3.097315}^2-1)+0+\frac{-66100}{298}(\frac{3.097315-1}{3.097315})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-2224.2$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -2224.2$

  $\Delta H_R^0=-18492.5\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-18492.5-92220=-110712$ J

Interpretation Introduction

(i)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 700 ⁰C

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 700 ⁰C = 973 K

$\tau =\frac{T}{T_o}$ = 3.2651

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-111536$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(3.2651-1)+\frac{0.004181}{2}\times {298}_{}^2({3.2651}^2-1)+0+\frac{-66100}{298}(\frac{3.2651-1}{3.2651})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-2323.31$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -2323.31$

  $\Delta H_R^0=-19316\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-19316-92220=-111536$ J

Interpretation Introduction

(j)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 590 F

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 590 F = $\left(\left.(590-32)\times \frac{5}{9}\right)\right.+273$ = 583 K

$\tau =\frac{T}{T_o}$ = 1.9563

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-102669$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(1.9563-1)+\frac{0.004181}{2}\times {298}_{}^2({1.9563}^2-1)+0+\frac{-66100}{298}(\frac{1.9563-1}{1.9563})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-1256.77$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -1256.77$

  $\Delta H_R^0=-10448.8\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-10448.8-92220=-102669$ J

Interpretation Introduction

(l)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 770 F

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 770 F = $\left(\left.(770-32)\times \frac{5}{9}\right)\right.+273$ = 683 K

$\tau =\frac{T}{T_o}$ = 2.2919

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-105788$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

A₃= 3.280

>n₃ = 1

₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(2.2919-1)+\frac{0.004181}{2}\times {298}_{}^2({2.2919}^2-1)+0+\frac{-66100}{298}(\frac{2.2919-1}{2.2919})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-1595.82$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -1595.82$

  $\Delta H_R^0=-13267.6\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-13267.6-92220=-105488$ J

Interpretation Introduction

(m)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 850 K

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 850 K

$\tau =\frac{T}{T_o}$ = 2.8523

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-109348$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(2.8523-1)+\frac{0.004181}{2}\times {298}_{}^2({2.8523}^2-1)+0+\frac{-66100}{298}(\frac{2.8523-1}{2.8523})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-2060.1$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -2060.1$

  $\Delta H_R^0=-17127.7\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

  $\Delta H^0=-17127.7-92220=-109348$ J

Interpretation Introduction

(n)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 1300 K

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A, B, C, D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 1300 K

$\tau =\frac{T}{T_o}$ = 4.3624

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-114721$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(4.3624-1)+\frac{0.004181}{2}\times {298}_{}^2({4.3624}^2-1)+0+\frac{-66100}{298}(\frac{4.3624-1}{4.3624})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-2706.41$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -2706.41$

  $\Delta H_R^0=-22501.1\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-22501.1-92220=-114721$ J

Interpretation Introduction

(o)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 800 ⁰C

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A, B, C, D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 800 ⁰C = 1073 K

$\tau =\frac{T}{T_o}$ = 3.6

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-112914$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(3.6-1)+\frac{0.004181}{2}\times {298}_{}^2({3.6}^2-1)+0+\frac{-66100}{298}(\frac{3.6-1}{3.6})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-2489$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -2489$

  $\Delta H_R^0=-20693.8\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-20693.8-92220=-112914$ J

Interpretation Introduction

(r)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 450 ⁰C

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 450 ⁰C = 723 K

$\tau =\frac{T}{T_o}$ = 2.426

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-106507$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(2.426-1)+\frac{0.004181}{2}\times {298}_{}^2({2.426}^2-1)+0+\frac{-66100}{298}(\frac{2.426-1}{2.426})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-1718$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -1718$

  $\Delta H_R^0=-14287.1\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-14287.1-92220=-106507$ J

Interpretation Introduction

(t)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 860 F

In this case,

N₂ + 3H₂ ? 2 NH₃

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 860 F = $\left(\left.(860-32)\times \frac{5}{9}\right)\right.+273$ = 733 K

$\tau =\frac{T}{T_o}$ = 2.46

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-106753$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(2.46-1)+\frac{0.004181}{2}\times {298}_{}^2({2.46}^2-1)+0+\frac{-66100}{298}(\frac{2.46-1}{2.46})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-1747$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -1718$

  $\Delta H_R^0=-14532\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-14287.1-92220=-106753$ J

Interpretation Introduction

(u)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 750 K

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 750 K

$\tau =\frac{T}{T_o}$ = 2.5167

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-107161$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(2.51677-1)+\frac{0.004181}{2}\times {298}_{}^2({2.51677}^2-1)+0+\frac{-66100}{298}(\frac{2.51677-1}{2.51677})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-1797.11$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -1797.11$

  $\Delta H_R^0=-14941.2\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-14941.2-92220=-107161$ J

Interpretation Introduction

(v)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 900 K

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 900 K

$\tau =\frac{T}{T_o}$ = 3.02

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-110303$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(3.02-1)+\frac{0.004181}{2}\times {298}_{}^2({3.02}^2-1)+0+\frac{-66100}{298}(\frac{3.02-1}{3.02})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-2175$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -2175$

  $\Delta H_R^0=-18083.4\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-18083-92220=-110303$ J

Interpretation Introduction

(w)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 400 ⁰C

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 400 ⁰C = 673 K

$\tau =\frac{T}{T_o}$ = 2.2583

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-105223$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(2.2583-1)+\frac{0.004181}{2}\times {298}_{}^2({2.2583}^2-1)+0+\frac{-66100}{298}(\frac{2.2583-1}{2.2583})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-1564$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -1564$

  $\Delta H_R^0=-13003.2\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-13003.2-92220=-105223$ J

Interpretation Introduction

(x)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 375 ⁰C

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 375 ⁰C = 648 K

$\tau =\frac{T}{T_o}$ = 2.1745

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-104545$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(2.1745-1)+\frac{0.004181}{2}\times {298}_{}^2({2.1745}^2-1)+0+\frac{-66100}{298}(\frac{2.1745-1}{2.1745})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-1482.5$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -1482.5$

  $\Delta H_R^0=-12325.4\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-12325.4-92220=-104545$ J

Interpretation Introduction

(y)

Interpretation:

To determine the standard heat of reaction of one of the reactions at 1490 F

In this case,

N₂ + 3H₂ ? 2 NH₃

Concept Introduction:

The integral of rate of change of molar heat capacity is the enthalpy change as given below: -

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where,

A,B,C,D are constants

T_o = Intial temperature = 298 K

T = Final temperature = 1490 F = $\left(\left.(1490-32)\times \frac{5}{9}\right)\right.+273$ = 1083 K

$\tau =\frac{T}{T_o}$ = 3.634

C_p= Molar heat capacity

R= Universal Gas constant

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H_{298}^0$ = Heat of formation at 298 K

$\Delta H^o$ = Standard Heat of Reaction

Answer to Problem 4.24P

$\Delta H=-113032$ J

Explanation of Solution

N₂ + 3H₂ ? 2 NH₃

Basis of Number of Moles of N₂ = 1

Number of Moles of $N{H}_3$ = 2

Number of Moles of H₂ = 3

For NH₃

n₁ = 2

A₁= 3.578

B₁= 0.00302

D₁ = -18600

For H₂

n₂ = 3

A₂= 3.249

B₂= 0.000422

D₂ = 8300

For N₂

n₃ = 1

A₃= 3.280

B₃= 0.000593

D₃ = 4000

i = 1,2,3 and substituting the values for each of the reactants and products below

  $A={\displaystyle \sum _i^{}(n_i\times A_i)}$

  $A=(2\times 3.578)-(1\times 3.28)-(3\times 3.249)$ = -5.871

  $B={\displaystyle \sum _i^{}(n_i\times B_i)}$

  $B=(2\times 0.00302)-(1\times 0.000593)-(3\times 0.000422)$ = 0.004181

  $D={\displaystyle \sum _i^{}(n_i\times D_i)}$

  $D=(2\times -18600)-(1\times 4000)-(3\times 8300)$ = -66100

Total mean heat capacity of combined stream is as given below,

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-5.871\times 298(3.634-1)+\frac{0.004181}{2}\times {298}_{}^2({3.634}^2-1)+0+\frac{-66100}{298}(\frac{3.634-1}{3.634})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=-2503.23$

  $\Delta H_R^0=R{\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=8.314\times -2503.23$

  $\Delta H_R^0=-20811\frac{kJ}{mol}$

$\Delta H^o={\displaystyle \sum _i{v}_i{H}_i^o}$ = Heat of formation of products − Heat of formation of reactants

From the table and the reaction we get,

$\Delta H^0{}_{298}=$ -46110 J/mol

$\Delta H^0{}_{298}=2\times (-46110)=-92220$ J

We get,

$\Delta H^o=\Delta H_{298}^0+R{\displaystyle {\int }_{T_0}^T\frac{\Delta C_P}{R}}dT$

$\Delta H^0=-20811.9-92220=-113032$ J