Chapter 4, Problem 4.21P

(a)

Interpretation Introduction

Interpretation:

Calculate the final flame temperature of ethylene if ethylene is heated from initial temperature ${\text{25}}^{\text{o}}\text{C}$ with burned theoretical amount of air at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

Since we need to calculate flame temperature, hence we will consider composition of product gas.

The transfer of heat is defined as:

  $Q=\Delta H_P=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$

                        .....(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

                        .....(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.21P

  $T=2533.38\text{K}$

Explanation of Solution

Given information:

  $T_0={\text{25}}^{\text{o}}\text{C=25+273}\text{.15}\text{K=298}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_4}=52510\frac{\text{J}}{\text{mol}}$, $\Delta H^{\circ }{}_{C{O}_2}=-393509\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(l)}=-285830\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{O_2}=0\frac{\text{J}}{\text{mol}}$

The ethylene is burned with theoretical amount of air according to reaction:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(g)+3O}}_{\text{2}}\to {\text{2CO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(g)}$

                        .....(A)

Applying energy balance and considering steady state process,

  $\begin{array}{l}\Delta H^{\circ }{}_{298K}+\Delta H^{\circ }{}_P=0\\ \\ \Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\Delta H^{\circ }{}_P-\Delta H^{\circ }{}_R\\ \Delta H^{\circ }{}_{298K} =n_{C{O}_2}\times \Delta H^{\circ }{}_{C{O}_2}+n_{H_{2}O}\times \Delta H^{\circ }{}_{H_{2}O}-n_{C_2{H}_4}\times \Delta H^{\circ }{}_{C_2{H}_4}-n_{O_2}\times \Delta H^{\circ }{}_{O_2}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\left(2\times -393509\frac{\text{J}}{\text{mol}}\right)+\left(2\times -241818\frac{\text{J}}{\text{mol}}\right)-\left(1\times -52510\frac{\text{J}}{\text{mol}}\right)-\left(3\times 0\frac{\text{J}}{\text{mol}}\right)\\ \Delta H^{\circ }{}_{298K} =-1323164\frac{\text{J}}{\text{mol}}\text{×}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}=-1323.164\frac{\text{kJ}}{\text{mol}}\end{array}$

Hence $\Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}=-1323164\frac{\text{J}}{\text{mol}}$

Now,

The product gas has nitrogen, carbon dioxide and water gas if complete combustion takes place with theoretical amount of air.

The one mole of ethylene burns with three moles of air for the complete combustion and the ratio of Nitrogen: Oxygen in the air is $79:21$, So the total moles of unreacted nitrogen come with product gas is $3\text{mol}\times \frac{79}{21}=11.286\text{mol}$

The one mole of ethylene gives two moles of carbon dioxide and water gas after the complete combustion.

So, composition of product gas is:

  $\begin{array}{l}{n}_{N_2}=11.286\text{mol} \\ n_{C{O}_2}=2\text{mol}\\ n_{H_{2}O}=2\text{mol}\end{array}$

  $n_{O_2}=0$

To find value of constants in equation (2),

  $\begin{array}{l}\Delta A={\nu }^i{A}^i\\ \Delta B={\nu }^i{B}^i\\ \Delta C={\nu }^i{C}^i\\ \Delta D={\nu }^i{D}^i\end{array}$

Values of above constants are given in appendix C table C.1 and noted down below:

So,

  $\begin{array}{l}\Delta A={\nu }^i{A}^i=2\times 5.457+11.286\times 3.280+2\times 3.47=54.8721\\ \Delta A=54.8721\\ \Delta B={\nu }^i{B}^i=2\times 1.045\times {10}^{-3}+11.286\times 0.593\times {10}^{-3}+2\times 1.45\times {10}^{-3}=11.68\times {10}^{-3}\\ \Delta B=11.68\times {10}^{-3}\\ \Delta C={\nu }^i{C}^i=2\times 0+11.286\times 0+2\times 0=0\\ \Delta C=0\\ \Delta D={\nu }^i{D}^i=2\times -1.157\times {10}^5+11.286\times 0.04\times {10}^5+2\times 0.121\times {10}^5=-162056\\ \Delta D=-162056\end{array}$

So, from equation (2)

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

Put the values

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=54.8721\times 298.15\times (\tau -1)+\frac{11.68\times {10}^{-3}}{2}{298.15}^2({\tau }^2-1)\\ +\frac{0}{3}{298.15}^3({\tau }^3-1)+\frac{-162056}{298.15}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=16360.12(\tau -1)+519.14({\tau }^2-1)-543.54(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}16360.12\tau -16360.12+519.14{\tau }^2-519.14-543.54+\frac{543.54}{\tau }\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}519.14{\tau }^2+16360.12\tau -17422.8+\frac{543.54}{\tau }\end{array}$

Now from equation (1),

  $\begin{array}{l}1323.164\frac{\text{kJ}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1323164\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 159148.9\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in above

  $\begin{array}{l}159148.9\text{K=}{\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}519.14{\tau }^2+16360.12\tau -17422.8+\frac{543.54}{\tau }\\ 176571.7=519.14{\tau }^2+16360.12\tau +\frac{543.54}{\tau }\end{array}$

On solving,

  $\tau =8.497$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{25+273.15}=8.497\\ \\ T=2533.38\text{K}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Calculate the final flame temperature of ethylene if ethylene is burned from initial temperature ${\text{25}}^{\text{o}}\text{C}$ with $\text{25\%}$ excess air at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

Since we need to calculate flame temperature, hence we will consider composition of product gas.

The transfer of heat is defined as:

  $Q=\Delta H_P=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$

                        .....(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

                        .....(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.21P

  $T=2577.036\text{K}$

Explanation of Solution

Given information:

  $T_0={\text{25}}^{\text{o}}\text{C=25+273}\text{.15}\text{K=298}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_4}=52510\frac{\text{J}}{\text{mol}}$, $\Delta H^{\circ }{}_{C{O}_2}=-393509\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(l)}=-285830\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{O_2}=0\frac{\text{J}}{\text{mol}}$

The ethylene is burned with theoretical amount of air according to reaction:

  $C_2{H}_4(g)+3O_2\to 2C{O}_2(g)+2H_{2}O(g)$

                        .....(A)

Applying energy balance and considering steady state process,

  $\begin{array}{l}\Delta H^{\circ }{}_{298K}+\Delta H^{\circ }{}_P=0\\ \\ \Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\Delta H^{\circ }{}_P-\Delta H^{\circ }{}_R\\ \Delta H^{\circ }{}_{298K} =n_{C{O}_2}\times \Delta H^{\circ }{}_{C{O}_2}+n_{H_{2}O}\times \Delta H^{\circ }{}_{H_{2}O}-n_{C_2{H}_4}\times \Delta H^{\circ }{}_{C_2{H}_4}-n_{O_2}\times \Delta H^{\circ }{}_{O_2}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\left(2\times -393509\frac{\text{J}}{\text{mol}}\right)+\left(2\times -241818\frac{\text{J}}{\text{mol}}\right)-\left(1\times -52510\frac{\text{J}}{\text{mol}}\right)-\left(3\times 0\frac{\text{J}}{\text{mol}}\right)\\ \Delta H^{\circ }{}_{298K} =-1323164\frac{\text{J}}{\text{mol}}\text{×}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}=-1323.164\frac{\text{kJ}}{\text{mol}}\end{array}$

Hence $\Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}=-1323164\frac{\text{J}}{\text{mol}}$

Now,

For $\text{25\%}$ excess air, number of moles of ethylene will be same as it will be limiting reagents for fully combustion reaction and hence fully consumed in the reaction. So only $n_{O_2}=3$ moles of oxygen are required for it, rest excess moles will be come out from furnace. Same nitrogen moles will not consume in the furnace, so it will be also in product stream with extra moles of nitrogen fed with feed.

  $n_{O_2}\text{with}\text{excess}\text{air}\text{25\%=3+3×0}\text{.25=3}\text{.75}$

  $n_{O_2}=3$ moles of oxygen are required for the reaction so, In product stream   $n_{O_2}=3.75-3=0.75$

The one mole of ethylene gives two moles of carbon dioxide and water gas after the complete combustion.

  $n_{C{O}_2}=2\text{mol}$

  $n_{H_{2}O}=2\text{mol}$

  $n_{N_2}=\text{Number}\text{of}\text{moles}\text{of}\text{nitrogen}\text{due}\text{to 25\%}\text{excess}\text{air}$

The proportion of composition of nitrogen: oxygen in air is $N_2:O_2=79:21$, So  $n_{N_2}=\frac{79}{21}\times 3.75=14.107$

To find value of constants in equation (2),

  $\begin{array}{l}\Delta A={\nu }^i{A}^i\\ \Delta B={\nu }^i{B}^i\\ \Delta C={\nu }^i{C}^i\\ \Delta D={\nu }^i{D}^i\end{array}$

Values of above constants are given in appendix C table C.1 and noted down below:

So,

  $\begin{array}{l}\Delta A={\nu }^i{A}^i=2\times 5.457+14.107\times 3.280+0.75\times 3.639+2\times 3.47=66.8542\\ \Delta A=66.8542\\ \Delta B={\nu }^i{B}^i=2\times 1.045\times {10}^{-3}+14.107\times 0.593\times {10}^{-3}+0.75\times 0.506\times {10}^{-3}+2\times 1.45\times {10}^{-3}=13.73\times {10}^{-3}\\ \Delta B=13.73\times {10}^{-3}\\ \Delta C={\nu }^i{C}^i=2\times 0+14.107\times 0+0.75\times 0+2\times 0=0\\ \Delta C=0\\ \Delta D={\nu }^i{D}^i=2\times -1.157\times {10}^5+14.107\times 0.04\times {10}^5+0.75\times -0.227\times {10}^5+2\times 0.121\times {10}^5=-167797\\ \Delta D=-167797\end{array}$

So, from equation (2)

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

Put the values

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=66.8542\times 298.15\times (\tau -1)+\frac{13.73\times {10}^{-3}}{2}{298.15}^2({\tau }^2-1)\\ +\frac{0}{3}{298.15}^3({\tau }^3-1)+\frac{-167797}{298.15}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=19932.58(\tau -1)+610.25({\tau }^2-1)-562.794(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}19932.58\tau -19932.58+610.25{\tau }^2-610.25-562.794+\frac{562.794}{\tau }\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}610.25{\tau }^2+19932.58\tau -21105.624+\frac{562.794}{\tau }\end{array}$

Now from equation (1),

  $\begin{array}{l}1323.164\frac{\text{kJ}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1323164\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 159148.9\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in above

  $\begin{array}{l}159148.9\text{K=}{\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}610.25{\tau }^2+19932.58\tau -21105.624+\frac{562.794}{\tau }\\ 180254.524=519.14{\tau }^2+16360.12\tau +\frac{543.54}{\tau }\end{array}$

On solving,

  $\tau =8.64342$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{25+273.15}=8.64342\\ \\ T=2577.036\text{K}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Calculate the final flame temperature of ethylene if ethylene is burned from initial temperature ${\text{25}}^{\text{o}}\text{C}$ with $\text{50\%}$ excess air at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

Since we need to calculate flame temperature, hence we will consider composition of product gas.

The transfer of heat is defined as:

  $Q=\Delta H_P=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$

                        .....(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

                        .....(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.21P

  $T=2626.7\text{K}$

Explanation of Solution

Given information:

  $T_0={\text{25}}^{\text{o}}\text{C=25+273}\text{.15}\text{K=298}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_4}=52510\frac{\text{J}}{\text{mol}}$, $\Delta H^{\circ }{}_{C{O}_2}=-393509\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(l)}=-285830\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{O_2}=0\frac{\text{J}}{\text{mol}}$

The ethylene is burned with theoretical amount of air according to reaction:

  $C_2{H}_4(g)+3O_2\to 2C{O}_2(g)+2H_{2}O(g)$

                        .....(A)

Applying energy balance and considering steady state process,

  $\begin{array}{l}\Delta H^{\circ }{}_{298K}+\Delta H^{\circ }{}_P=0\\ \\ \Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\Delta H^{\circ }{}_P-\Delta H^{\circ }{}_R\\ \Delta H^{\circ }{}_{298K} =n_{C{O}_2}\times \Delta H^{\circ }{}_{C{O}_2}+n_{H_{2}O}\times \Delta H^{\circ }{}_{H_{2}O}-n_{C_2{H}_4}\times \Delta H^{\circ }{}_{C_2{H}_4}-n_{O_2}\times \Delta H^{\circ }{}_{O_2}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\left(2\times -393509\frac{\text{J}}{\text{mol}}\right)+\left(2\times -241818\frac{\text{J}}{\text{mol}}\right)-\left(1\times -52510\frac{\text{J}}{\text{mol}}\right)-\left(3\times 0\frac{\text{J}}{\text{mol}}\right)\\ \Delta H^{\circ }{}_{298K} =-1323164\frac{\text{J}}{\text{mol}}\text{×}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}=-1323.164\frac{\text{kJ}}{\text{mol}}\end{array}$

Hence $\Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}=-1323164\frac{\text{J}}{\text{mol}}$

Now,

For $\text{50\%}$ excess air, number of moles of ethylene will be same as it will be limiting reagents for fully combustion reaction and hence fully consumed in the reaction. So only $n_{O_2}=3$ moles of oxygen are required for it, rest excess moles will be come out from furnace. Same nitrogen moles will not consume in the furnace, so it will be also in product stream with extra moles of nitrogen fed with feed.

  $n_{O_2}\text{with}\text{excess}\text{air}\text{50\%=3+3×0}\text{.5=4}\text{.5}$

  $n_{O_2}=3$ moles of oxygen are required for the reaction so, In product stream   $n_{O_2}=4.5-3=1.5$

The one mole of ethylene gives two moles of carbon dioxide and water gas after the complete combustion.

  $n_{C{O}_2}=2\text{mol}$

  $n_{H_{2}O}=2\text{mol}$

  $n_{N_2}=\text{Number}\text{of}\text{moles}\text{of}\text{nitrogen}\text{due}\text{to50\%}\text{excess}\text{air}$

The proportion of composition of nitrogen:oxygen in air is $N_2:O_2=79:21$, So  $n_{N_2}=\frac{79}{21}\times 4.5=16.929$

To find value of constants in equation (2),

  $\begin{array}{l}\Delta A={\nu }^i{A}^i\\ \Delta B={\nu }^i{B}^i\\ \Delta C={\nu }^i{C}^i\\ \Delta D={\nu }^i{D}^i\end{array}$

Values of above constants are given in appendix C table C.1 and noted down below:

So,

  $\begin{array}{l}\Delta A={\nu }^i{A}^i=2\times 5.457+16.929\times 3.280+1.5\times 3.639+2\times 3.47=78.8396\\ \Delta A=78.8396\\ \Delta B={\nu }^i{B}^i=2\times 1.045\times {10}^{-3}+16.929\times 0.593\times {10}^{-3}+1.5\times 0.506\times {10}^{-3}+2\times 1.45\times {10}^{-3}=15.79\times {10}^{-3}\\ \Delta B=15.79\times {10}^{-3}\\ \Delta C={\nu }^i{C}^i=2\times 0+16.929\times 0+1.5\times 0+2\times 0=0\\ \Delta C=0\\ \Delta D={\nu }^i{D}^i=2\times -1.157\times {10}^5+16.929\times 0.04\times {10}^5+1.5\times -0.227\times {10}^5+2\times 0.121\times {10}^5=-173534\\ \Delta D=-173534\end{array}$

So, from equation (2)

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

Put the values

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=78.8396\times 298.15\times (\tau -1)+\frac{15.79\times {10}^{-3}}{2}{298.15}^2({\tau }^2-1)\\ +\frac{0}{3}{298.15}^3({\tau }^3-1)+\frac{-173534}{298.15}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=23506.027(\tau -1)+701.81({\tau }^2-1)-582.036(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}23506.027\tau -23506.027+701.81{\tau }^2-701.81-582.036+\frac{582.036}{\tau }\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}701.81{\tau }^2+23506.027\tau -24789.873+\frac{582.036}{\tau }\end{array}$

Now from equation (1),

  $\begin{array}{l}1323.164\frac{\text{kJ}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1323164\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 159148.9\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in above

  $\begin{array}{l}159148.9\text{K=}{\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}701.81{\tau }^2+23506.027\tau -24789.873+\frac{582.036}{\tau }\\ 184538.773=519.14{\tau }^2+16360.12\tau +\frac{543.54}{\tau }\end{array}$

On solving,

  $\tau =8.81$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{25+273.15}=8.81\\ \\ T=2626.7\text{K}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Calculate the final flame temperature of ethylene if ethylene is burned from initial temperature ${\text{25}}^{\text{o}}\text{C}$ with $\text{100\%}$ excess air at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

Since we need to calculate flame temperature, hence we will consider composition of product gas.

The transfer of heat is defined as:

  $Q=\Delta H_P=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$

                        .....(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

                        .....(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.21P

  $T=2704.22\text{K}$

Explanation of Solution

Given information:

  $T_0={\text{25}}^{\text{o}}\text{C=25+273}\text{.15}\text{K=298}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_4}=52510\frac{\text{J}}{\text{mol}}$, $\Delta H^{\circ }{}_{C{O}_2}=-393509\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(l)}=-285830\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{O_2}=0\frac{\text{J}}{\text{mol}}$

The ethylene is burned with theoretical amount of air according to reaction:

  $C_2{H}_4(g)+3O_2\to 2C{O}_2(g)+2H_{2}O(g)$

                        .....(A)

Applying energy balance and considering steady state process,

  $\begin{array}{l}\Delta H^{\circ }{}_{298K}+\Delta H^{\circ }{}_P=0\\ \\ \Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\Delta H^{\circ }{}_P-\Delta H^{\circ }{}_R\\ \Delta H^{\circ }{}_{298K} =n_{C{O}_2}\times \Delta H^{\circ }{}_{C{O}_2}+n_{H_{2}O}\times \Delta H^{\circ }{}_{H_{2}O}-n_{C_2{H}_4}\times \Delta H^{\circ }{}_{C_2{H}_4}-n_{O_2}\times \Delta H^{\circ }{}_{O_2}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\left(2\times -393509\frac{\text{J}}{\text{mol}}\right)+\left(2\times -241818\frac{\text{J}}{\text{mol}}\right)-\left(1\times -52510\frac{\text{J}}{\text{mol}}\right)-\left(3\times 0\frac{\text{J}}{\text{mol}}\right)\\ \Delta H^{\circ }{}_{298K} =-1323164\frac{\text{J}}{\text{mol}}\text{×}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}=-1323.164\frac{\text{kJ}}{\text{mol}}\end{array}$

Hence $\Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}=-1323164\frac{\text{J}}{\text{mol}}$

Now,

For $\text{100\%}$ excess air, number of moles of ethylene will be same as it will be limiting reagents for fully combustion reaction and hence fully consumed in the reaction. So only $n_{O_2}=3$ moles of oxygen are required for it, rest excess moles will be come out from furnace. Same nitrogen moles will not consume in the furnace, so it will be also in product stream with extra moles of nitrogen fed with feed.

  $n_{O_2}\text{with}\text{excess}\text{air}\text{50\%=3+3×1=6}$

  $n_{O_2}=3$ moles of oxygen are required for the reaction so, In product stream   $n_{O_2}=6-3=3$

The one mole of ethylene gives two moles of carbon dioxide and water gas after the complete combustion.

  $n_{C{O}_2}=2\text{mol}$

  $n_{H_{2}O}=2\text{mol}$

  $n_{N_2}=\text{Number}\text{of}\text{moles}\text{of}\text{nitrogen}\text{due}\text{to50\%}\text{excess}\text{air}$

The proportion of composition of nitrogen: oxygen in air is $N_2:O_2=79:21$, So  $n_{N_2}=\frac{79}{21}\times 6=22.57$

To find value of constants in equation (2),

  $\begin{array}{l}\Delta A={\nu }^i{A}^i\\ \Delta B={\nu }^i{B}^i\\ \Delta C={\nu }^i{C}^i\\ \Delta D={\nu }^i{D}^i\end{array}$

Values of above constants are given in appendix C table C.1 and noted down below:

So,

  $\begin{array}{l}\Delta A={\nu }^i{A}^i=2\times 5.457+22.57\times 3.280+3\times 3.639+2\times 3.47=102.8006\\ \Delta A=102.8\\ \Delta B={\nu }^i{B}^i=2\times 1.045\times {10}^{-3}+22.57\times 0.593\times {10}^{-3}+3\times 0.506\times {10}^{-3}+2\times 1.45\times {10}^{-3}=19.89\times {10}^{-3}\\ \Delta B=19.89\times {10}^{-3}\\ \Delta C={\nu }^i{C}^i=2\times 0+22.57\times 0+3\times 0+2\times 0=0\\ \Delta C=0\\ \Delta D={\nu }^i{D}^i=2\times -1.157\times {10}^5+22.57\times 0.04\times {10}^5+3\times -0.227\times {10}^5+2\times 0.121\times {10}^5=-185020\\ \Delta D=-185020\end{array}$

So, from equation (2)

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

Put the values

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=102.8\times 298.15\times (\tau -1)+\frac{19.89\times {10}^{-3}}{2}{298.15}^2({\tau }^2-1)\\ +\frac{0}{3}{298.15}^3({\tau }^3-1)+\frac{-185020}{298.15}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=30649.82(\tau -1)+884.045({\tau }^2-1)-620.56(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}30649.82\tau -30649.82+884.045{\tau }^2-884.045-620.56+\frac{620.56}{\tau }\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}884.045{\tau }^2+30649.82\tau -32154.425+\frac{620.56}{\tau }\end{array}$

Now from equation (1),

  $\begin{array}{l}1323.164\frac{\text{kJ}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1323164\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 159148.9\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in above

  $\begin{array}{l}159148.9\text{K=}{\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}884.045{\tau }^2+30649.82\tau -32154.425+\frac{620.56}{\tau }\\ 191303.32=519.14{\tau }^2+16360.12\tau +\frac{543.54}{\tau }\end{array}$

On solving,

  $\tau =9.07$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{25+273.15}=9.07\\ \\ T=2704.22\text{K}\end{array}$

(e)

Interpretation Introduction

Interpretation:

Calculate the final flame temperature of ethylene if ethylene is burned from initial temperature ${\text{25}}^{\text{o}}\text{C}$ with $\text{50\%}$ excess air preheated to ${\text{500}}^{\text{o}}\text{C}$ .

Concept Introduction:

Since we need to calculate flame temperature, hence we will consider composition of product gas.

The transfer of heat is defined as:

  $Q=\Delta H_P=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$

                        .....(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

                        .....(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.21P

  $T=3044.1115\text{K}$

Explanation of Solution

Given information:

  $T_0={\text{25}}^{\text{o}}\text{C=25+273}\text{.15}\text{K=298}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_4}=52510\frac{\text{J}}{\text{mol}}$, $\Delta H^{\circ }{}_{C{O}_2}=-393509\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(l)}=-285830\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{O_2}=0\frac{\text{J}}{\text{mol}}$

The ethylene is burned with theoretical amount of air according to reaction:

  $C_2{H}_4(g)+3O_2\to 2C{O}_2(g)+2H_{2}O(g)$

                        .....(A)

Applying energy balance and considering steady state process,

  $\Delta H^{\circ }{}_{air}+\Delta H^{\circ }{}_{298K}+\Delta H^{\circ }{}_P=0$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\Delta H^{\circ }{}_P-\Delta H^{\circ }{}_R\\ \Delta H^{\circ }{}_{298K} =n_{C{O}_2}\times \Delta H^{\circ }{}_{C{O}_2}+n_{H_{2}O}\times \Delta H^{\circ }{}_{H_{2}O}-n_{C_2{H}_4}\times \Delta H^{\circ }{}_{C_2{H}_4}-n_{O_2}\times \Delta H^{\circ }{}_{O_2}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\left(2\times -393509\frac{\text{J}}{\text{mol}}\right)+\left(2\times -241818\frac{\text{J}}{\text{mol}}\right)-\left(1\times -52510\frac{\text{J}}{\text{mol}}\right)-\left(3\times 0\frac{\text{J}}{\text{mol}}\right)\\ \Delta H^{\circ }{}_{298K} =-1323164\frac{\text{J}}{\text{mol}}\text{×}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}=-1323.164\frac{\text{kJ}}{\text{mol}}\end{array}$

Temperature of air is increased to ${\text{500}}^{\text{o}}\text{C+273}\text{.15}\text{K=773}\text{.15}\text{K}$

So, $\Delta H^{\circ }{}_{air}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$

For air, Values of constants used in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ air3.3550.5750-0.016\end{array}$

Hence,

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

  $\tau =\frac{T}{T_0}=\frac{773.15}{298.15}=2.593$

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}x=3.355\times 298.15\times (2.593-1)+\frac{0.575\times {10}^{-3}}{2}{298.15}^2({2.593}^2-1)\\ +\frac{0}{3}{298.15}^3({2.593}^3-1)+\frac{-0.016\times {10}^5}{298.15}(\frac{2.593-1}{2.593})\\ {\displaystyle \underset{298.15}{\overset{773.15}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=-1736.45\text{K}\end{array}$

  $n_{air}\text{with}\text{excess}\text{air}\text{50\%=3+3×0}\text{.5=4}\text{.5}$

The number of moles of oxygen are therefore, $n_{O_2}\text{=}\frac{\text{4}\text{.5}}{0.21}=21.429$

For

  $\begin{array}{l}\Delta H^{\circ }{}_{air}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=21.429\times 8.314\frac{\text{J}}{\text{mol K}}\times -1736.45\text{K}\\ \\ \Delta H^{\circ }{}_{air}=-309367.16\frac{\text{J}}{\text{mol}}\end{array}$

Now,

For $\text{50\%}$ excess air, number of moles of ethylene will be same as it will be limiting reagents for fully combustion reaction and hence fully consumed in the reaction. So only $n_{O_2}=3$ moles of oxygen are required for it, rest excess moles will be come out from furnace. Same nitrogen moles will not consume in the furnace, so it will be also in product stream with extra moles of nitrogen fed with feed.

  $n_{O_2}\text{with}\text{excess}\text{air}\text{50\%=3+3×0}\text{.5=4}\text{.5}$

  $n_{O_2}=3$ moles of oxygen are required for the reaction so, In product stream   $n_{O_2}=4.5-3=1.5$

The one mole of ethylene gives two moles of carbon dioxide and water gas after the complete combustion.

  $n_{C{O}_2}=2\text{mol}$

  $n_{H_{2}O}=2\text{mol}$

  $n_{N_2}=\text{Number}\text{of}\text{moles}\text{of}\text{nitrogen}\text{due}\text{to50\%}\text{excess}\text{air}$

The proportion of composition of nitrogen:oxygen in air is $N_2:O_2=79:21$, So  $n_{N_2}=\frac{79}{21}\times 4.5=16.929$

To find value of constants in equation (2),

  $\begin{array}{l}\Delta A={\nu }^i{A}^i\\ \Delta B={\nu }^i{B}^i\\ \Delta C={\nu }^i{C}^i\\ \Delta D={\nu }^i{D}^i\end{array}$

So,

  $\begin{array}{l}\Delta A={\nu }^i{A}^i=2\times 5.457+16.929\times 3.280+1.5\times 3.639+2\times 3.47=78.8396\\ \Delta A=78.8396\\ \Delta B={\nu }^i{B}^i=2\times 1.045\times {10}^{-3}+16.929\times 0.593\times {10}^{-3}+1.5\times 0.506\times {10}^{-3}+2\times 1.45\times {10}^{-3}=15.79\times {10}^{-3}\\ \Delta B=15.79\times {10}^{-3}\\ \Delta C={\nu }^i{C}^i=2\times 0+16.929\times 0+1.5\times 0+2\times 0=0\\ \Delta C=0\\ \Delta D={\nu }^i{D}^i=2\times -1.157\times {10}^5+16.929\times 0.04\times {10}^5+1.5\times -0.227\times {10}^5+2\times 0.121\times {10}^5=-173534\\ \Delta D=-173534\end{array}$

So, from equation (2)

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

Put the values

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=78.8396\times 298.15\times (\tau -1)+\frac{15.79\times {10}^{-3}}{2}{298.15}^2({\tau }^2-1)\\ +\frac{0}{3}{298.15}^3({\tau }^3-1)+\frac{-173534}{298.15}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=23506.027(\tau -1)+701.81({\tau }^2-1)-582.036(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}23506.027\tau -23506.027+701.81{\tau }^2-701.81-582.036+\frac{582.036}{\tau }\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}701.81{\tau }^2+23506.027\tau -24789.873+\frac{582.036}{\tau }\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{air}+\Delta H^{\circ }{}_{298K}+\Delta H^{\circ }{}_P=0\\ \\ \Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{air}-\Delta H^{\circ }{}_{298K}\end{array}$

  $\begin{array}{l}309367.16\frac{\text{J}}{\text{mol}}+1323.164\frac{\text{kJ}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1632531.158\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 196359.29\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in above

  $\begin{array}{l}196359.29\text{K=}{\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}701.81{\tau }^2+23506.027\tau -24789.873+\frac{582.036}{\tau }\\ 221149.17=519.14{\tau }^2+16360.12\tau +\frac{543.54}{\tau }\end{array}$

On solving,

  $\tau =10.21$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{25+273.15}=10.21\\ \\ T=3044.1115\text{K}\end{array}$

(f)

Interpretation Introduction

Interpretation:

Calculate the final flame temperature of ethylene if ethylene is burned from initial temperature ${\text{25}}^{\text{o}}\text{C}$ with theoretical amount of pure oxygen at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

Since we need to calculate flame temperature, hence we will consider composition of product gas.

The transfer of heat is defined as:

  $Q=\Delta H_P=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$

                        .....(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

                        .....(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.21P

  $T=2399.81\text{K}$

Explanation of Solution

Given information:

  $T_0={\text{25}}^{\text{o}}\text{C=25+273}\text{.15}\text{K=298}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_4}=52510\frac{\text{J}}{\text{mol}}$, $\Delta H^{\circ }{}_{C{O}_2}=-393509\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(l)}=-285830\frac{\text{J}}{\text{mol}}$ and $\Delta H^{\circ }{}_{O_2}=0\frac{\text{J}}{\text{mol}}$

The ethylene is burned with theoretical amount of air according to reaction:

  $C_2{H}_4(g)+3O_2\to 2C{O}_2(g)+2H_{2}O(g)$

                        .....(A)

Applying energy balance and considering steady state process,

  $\begin{array}{l}\Delta H^{\circ }{}_{298K}+\Delta H^{\circ }{}_P=0\\ \\ \Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\Delta H^{\circ }{}_P-\Delta H^{\circ }{}_R\\ \Delta H^{\circ }{}_{298K} =n_{C{O}_2}\times \Delta H^{\circ }{}_{C{O}_2}+n_{H_{2}O}\times \Delta H^{\circ }{}_{H_{2}O}-n_{C_2{H}_4}\times \Delta H^{\circ }{}_{C_2{H}_4}-n_{O_2}\times \Delta H^{\circ }{}_{O_2}\end{array}$

  $\begin{array}{l}\Delta H^{\circ }{}_{298K} =\left(2\times -393509\frac{\text{J}}{\text{mol}}\right)+\left(2\times -241818\frac{\text{J}}{\text{mol}}\right)-\left(1\times -52510\frac{\text{J}}{\text{mol}}\right)-\left(3\times 0\frac{\text{J}}{\text{mol}}\right)\\ \Delta H^{\circ }{}_{298K} =-1323164\frac{\text{J}}{\text{mol}}\text{×}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}=-1323.164\frac{\text{kJ}}{\text{mol}}\end{array}$

Hence $\Delta H^{\circ }{}_P=-\Delta H^{\circ }{}_{298K}=-1323164\frac{\text{J}}{\text{mol}}$

Now,

The product gas has nitrogen, carbon dioxide and water gas if complete combustion takes place with theoretical amount of air.

The one mole of ethylene gives two moles of carbon dioxide and water gas after the complete combustion.

So, composition of product gas is:

  $\begin{array}{l}{n}_{C{O}_2}=2\text{mol}\\ n_{H_{2}O}=2\text{mol}\end{array}$

  $n_{O_2}=0$

To find value of constants in equation (2),

  $\begin{array}{l}\Delta A={\nu }^i{A}^i\\ \Delta B={\nu }^i{B}^i\\ \Delta C={\nu }^i{C}^i\\ \Delta D={\nu }^i{D}^i\end{array}$

Values of above constants are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}i{\nu }^{i}A{10}^{3}B{10}^{6}C{10}^{-5}D\\ C{O}_{2}25.4571.0450-1.157\\ H_{2}O23.4701.45000.121\end{array}$

So,

  $\begin{array}{l}\Delta A={\nu }^i{A}^i=2\times 5.457+2\times 3.47=17.854\\ \Delta A=17.854\\ \Delta B={\nu }^i{B}^i=2\times 1.045\times {10}^{-3}+2\times 1.45\times {10}^{-3}=5\times {10}^{-3}\\ \Delta B=5\times {10}^{-3}\\ \Delta C={\nu }^i{C}^i=2\times 0+2\times 0=0\\ \Delta C=0\\ \Delta D={\nu }^i{D}^i=2\times -1.157\times {10}^5+2\times 0.121\times {10}^5=-207200\\ \Delta D=-207200\end{array}$

So, from equation (2)

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=(\Delta A)T_0(\tau -1)+\frac{\Delta B}{2}{T}_0{}^2({\tau }^2-1)+\frac{\Delta C}{3}{T}_0{}^3({\tau }^3-1)+\frac{\Delta D}{T_0}(\frac{\tau -1}{\tau })$

Put the values

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=17.854\times 298.15\times (\tau -1)+\frac{5\times {10}^{-3}}{2}{298.15}^2({\tau }^2-1)\\ +\frac{0}{3}{298.15}^3({\tau }^3-1)+\frac{-207200}{298.15}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=5323.17(\tau -1)+222.23({\tau }^2-1)-694.95(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}5323.17\tau -5323.17+222.23{\tau }^2-222.23-694.95+\frac{694.95}{\tau }\\ {\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}222.23{\tau }^2+5323.17\tau -6240.35+\frac{694.95}{\tau }\end{array}$

Now from equation (1),

  $\begin{array}{l}1323.164\frac{\text{kJ}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1323164\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 159148.9\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in above

  $\begin{array}{l}159148.9\text{K=}{\displaystyle \underset{298.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}222.23{\tau }^2+5323.17\tau -6240.35+\frac{694.95}{\tau }\\ 165389.25=519.14{\tau }^2+16360.12\tau +\frac{543.54}{\tau }\end{array}$

On solving,

  $\tau =8.049$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{25+273.15}=8.049\\ \\ T=2399.81\text{K}\end{array}$