Question 13.6: Analyzing Combustion of Methane with Oxygen at Constant Volu...

Analyzing Combustion of Methane with Oxygen at Constant Volume

A mixture of 1 kmol of gaseous methane and 2 kmol of oxygen initially at 25°C and 1 atm burns completely in a closed, rigid container. Heat transfer occurs until the products are cooled to 900 K. If the reactants and products each form ideal gas mixtures, determine (a) the amount of heat transfer, in kJ, and (b) the final pressure, in atm.

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Known A mixture of gaseous methane and oxygen, initially at 25°C and 1 atm, burns completely within a closed rigid container. The products are cooled to 900 K.

Find Determine the amount of heat transfer, in kJ, and the final pressure of the combustion products, in atm.

Schematic and Given Data:

Engineering Model

1. The contents of the closed, rigid container are taken as the system.

2. Kinetic and potential energy effects are absent, and W = 0.

3. Combustion is complete.

4. The initial mixture and the products of combustion each form ideal gas mixtures.

5. The initial and final states are equilibrium states.

Analysis The chemical reaction equation for the complete combustion of methane with oxygen is

 

CH4+2O2CO2+2H2O(g)CH _{4}+2 O _{2} \rightarrow CO _{2}+2 H _{2} O ( g )

 

a. With assumptions 2 and 3, the closed system energy balance takes the form

 

UPUR=QW0U_{ P }-U_{ R }=Q- W ^{\nearrow0}

 

or

 

Q=UPUR=(1uˉCO2+2uˉH2O(g))(1uˉCH4(g)+2uˉO2)Q=U_{ P }-U_{ R }=\left(1 \bar{u}_{ CO _{2}}+2 \bar{u}_{ H _{2} O ( g )}\right)-\left(1 \bar{u}_{ CH _{4}( g )}+2 \bar{u}_{ O _{2}}\right)

 

Each coefficient in this equation is the same as the corresponding term of the balanced chemical equation.

Since each reactant and product behaves as an ideal gas, the respective specific internal energies can be evaluated as uˉ=hˉRˉT\bar{u}=\bar{h}-\bar{R} T. The energy balance then becomes

 

Q=[1(hˉCO2RˉT2)+2(hˉH2O(g)RˉT2)][1(hˉCH4(g)RˉT1)+2(hˉO2RˉT1)]\begin{aligned}Q=&\left[1\left(\bar{h}_{ CO _{2}}-\bar{R} T_{2}\right)+2\left(\bar{h}_{ H _{2} O ( g )}-\bar{R} T_{2}\right)\right] \\&-\left[1\left(\bar{h}_{ CH _{4}( g )}-\bar{R} T_{1}\right)+2\left(\bar{h}_{ O _{2}}-\bar{R} T_{1}\right)\right]\end{aligned}

 

where T1 and T2T_{1} \text { and } T_{2} denote, respectively, the initial and final temperatures. Collecting like terms

 

Q=(hˉCO2+2hˉH2O(g)hˉCH4(g)2hˉO2)+3Rˉ(T1T2)Q=\left(\bar{h}_{ CO _{2}}+2 \bar{h}_{ H _{2} O ( g )}-\bar{h}_{ CH _{4}( g )}-2 \bar{h}_{ O _{2}}\right)+3 \bar{R}\left(T_{1}-T_{2}\right)

 

The specific enthalpies are evaluated in terms of the respective enthalpies of formation to give

 

1 Q=[(hˉf+Δhˉ)CO2+2(hˉf+Δhˉ)H2O(g)(hˉf+Δhˉ0)CH4(g)2(hˉf0+Δhˉ0)O2]+3Rˉ(T1T2)\begin{aligned}Q=&\left[\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{ CO _{2}}+2\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{ H _{2} O ( g )}\right.\\&\left.-\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}^{\nearrow0}\right)_{ CH _{4}( g )}-2\left(\bar{h}_{ f }^{\circ\nearrow0}+\Delta \bar{h}^{\nearrow0}\right)_{ O _{2}}\right]+3 \bar{R}\left(T_{1}-T_{2}\right)\end{aligned}

 

Since the methane and oxygen are initially at 25°C, Δhˉ=0\Delta \bar{h}=0 for each of these reactants. Also, hˉf=0\bar{h}_{ f }^{\circ}=0 for oxygen.

With enthalpy of formation values for CO2,H2O(g)CO _{2}, H _{2} O ( g ) and CH4(g)CH _{4}( g ) from Table A-25 and enthalpy values for H2O and CO2H _{2} O \text { and } CO _{2} from Table A-23

 

Q=[393,520+(37,4059364)]+2[241,820+(31,8289904)](74,850)+3(8.314)(298900)\begin{aligned}Q=&[-393,520+(37,405-9364)]+2[-241,820\\&+(31,828-9904)]-(-74,850) \\&+3(8.314)(298-900)\end{aligned}

= -745,436 kJ

 

b. By assumption 4, the initial mixture and the products of combustion each form ideal gas mixtures. Thus, for the reactants

 

p1V=nRRˉT1p_{1} V=n_{ R } \bar{R} T_{1}

 

where nRn_{ R } is the total number of moles of reactants and p1p_{1} is the initial pressure. Similarly, for the products

 

p2V=nPRˉT2p_{2} V=n_{ P } \bar{R} T_{2}

 

where nPn_{ P } is the total number of moles of products and p2p_{2} is the final pressure.

 

Since nR=nP=3n_{ R }=n_{ P }=3 and volume is constant, these equations combine to give

 

p2=T2T1p1=(900K298K)(1atm)=3.02atmp_{2}=\frac{T_{2}}{T_{1}} p_{1}=\left(\frac{900 K }{298 K }\right)(1 atm )=3.02 atm

 

1 This expression corresponds to Eq. 13.17b.

 

QW=Pn(hˉf+ΔhˉRˉTP)Rn(hˉf+ΔhˉRˉTR)=Pn(hˉf+Δhˉ)Rn(hˉf+Δhˉ)RˉTPPn+RˉTRRn\begin{aligned}Q-W &=\sum_{ P } n\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}-\bar{R} T_{ P }\right)-\sum_{ R } n\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}-\bar{R} T_{ R }\right) \\&=\sum_{ P } n\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)-\sum_{ R } n\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)-\bar{R} T_{ P } \sum_{ P } n+\bar{R} T_{ R } \sum_{ R } n\end{aligned} (13.17b)

 

Skills Developed 

Ability to…

• apply the closed system energy balance to a reacting system.

• evaluate property data appropriately.

• apply the ideal gas equation of state.

Quick Quiz

Calculate the volume of the system, in m3m ^{3}. Ans. 73.36 m3m ^{3}.

6

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