A coal gasifier produces a mixture of 1 CO and 2H2 that is then fed to a catalytic converter to produce methane. A chemical-equilibrium gas mixture containing CH4, CO, H2, and H2O exits the reactor at 600 K, 600 kPa. Determine the mole fraction of methane in the mixture.

Question

A coal gasifier produces a mixture of 1 CO and [latex]2H_2[/latex] that is then fed to a catalytic converter to produce methane. A chemical-equilibrium gas mixture containing [latex]CH _4, CO , H _2 ext {, and } H _2 O[/latex] exits the reactor at 600 K, 600 kPa. Determine the mole fraction of methane in the mixture.

Accepted Answer

[latex]\begin{array}{lllcc}& CO & +3 H _2 & \leftrightarrow CH _4 & + H _2 O \ ext { Initial } & \quad 1 & 2 & 0 & 0 \ ext { Change } & \quad - x & -3 x & x & x \ ext { Equil. } & \quad 1- x & 2-3 x & x & x\end{array}[/latex]

Total: n = (1 – x) + (2 – 3x) + x + x = 3 – 2x

[latex]K =\frac{ y _{ CH 4} y _{ H 2 O }}{ y _{ CO } y _{ H 2}^3}\left\lgroup \frac{ P }{ P _{ o }} ight group^{(1+1-1-3)=-2}=\frac{ x ^2}{(1- x )(2-3 x )^3}\left\lgroup \frac{ P }{ P _{ o }} ight group^{-2}[/latex]

[latex]\begin{aligned}& \ln K =-\Delta G ^{ O } / \overline{ R } ; \quad \Delta G ^{ O }=\Delta H ^{ O }- T \Delta S ^{ O } \& \begin{aligned}H _{ P } & = n _{ CH 4}\left[\overline{ h }_{ f }^{ o }+\overline{ C }_{ P }\left( T - T _{ o } ight) ight]+ n _{ H 2 O }\left(\overline{ h }_{ f }^{ o }+\Delta \overline{ h } ight) \& =[-74873+2.254 imes 16.04(600-298.15)]+(-241826+10499)=-295290 \H _{ R } & = n _{ CO }\left(\overline{ h }_{ f }^{ o }+\Delta \overline{ h } ight)+ n _{ H 2}\left(\overline{ h }_{ f }^{ o }+\Delta \overline{ h } ight)=1(-110527+8942)+3(0+8799) \& =-75188 \,kJ\end{aligned}\end{aligned}[/latex]

[latex]\begin{aligned}& \Delta H _{600}^{ o }= H _{ P }- H _{ R }=-295290-(-75188)=-220102 \,kJ \& \left(\overline{ s }_{ T }^{ o } ight)_{ CH 4}=\overline{ s }_{ To }^{ o }+\overline{ C }_{ P } \ln \left( T / T _{ o } ight)=186.251+2.254 imes 16.04 \ln (600 / 298.2)=211.549 \& \left(\overline{ s }_{ T }^{ o } ight)_{ H 2 O }=213.051 \,kJ / kmol - K ; \quad S _{ P }=424.6 \,kJ / K \& \left(\overline{ s }_{ T }^{ o } ight)_{ CO }=218.321 \,kJ / kmol - K , \quad\left(\overline{ s }_{ T }^{ o } ight)_{ H 2}=151.078 \,kJ / kmol - K\end{aligned}[/latex]

[latex]\begin{aligned}& \Delta S _{600}^{ o }= S _{ P }- S _{ R }=\left( n \overline{ s }_{ T }^{ o } ight)_{ CH 4}+\left( n \overline{ s }_{ T }^{ o } ight)_{ H 2 O }-\left( n \overline{ s }_{ T }^{ o } ight)_{ CO }-\left( n \overline{ s }_{ T }^{ o } ight)_{ H 2} \& =(211.549+213.051)-(218.321+3 imes 151.078)=-246.955 \,kJ / K \& \Delta G ^{ O }=\Delta H ^{ O }- T \Delta S ^{ O }=-220102-600(-246.955)=-71929 \,kJ ext {, } \& \ln K =-(-71915) /(8.31451 imes 600)=14.418 \quad \Rightarrow K =1.827 imes 10^6 \& ext { Solve for } x. x =0.6667, n _{ ext {tot }}=1.6667, \quad y _{ C H 4}= 0 . 4 \end{aligned}[/latex]

Question 16.CSGP.105: A coal gasifier produces a mixture of 1 CO and 2H2 that is t......

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