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Question 16.CSGP.101: Find the equilibrium constant for the reaction in Problem 16......

Find the equilibrium constant for the reaction in Problem 16.83.

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Reaction:    2 CO +2 H _2 \Leftrightarrow CH _4+ CO _2

For CH_4 at 600 K (formula in Table A.6),  \overline{ C }_{ P 0}=52.22

At 900 K

\begin{aligned}& \overline{ h }_{ CH 4}^0=-74873+52.22(900-298.2)=-43446\,kJ / kmol \\& \overline{ s }_{ CH 4}^0=186.251+52.22 \ln (900 / 298.2)=243.936 \,kJ / kmol K\end{aligned}

(The integrated-equation values are -43 656 and 240.259)

\begin{aligned}& \begin{array}{l}\Delta H _{900 K }^0=1(-43446)+1(-393522+28030) \\\quad-2(-110527+18397)-2(0+17657)=-259993 \,kJ \\\Delta S _{900 K }^0=1(243.936)+1(263.646) \\\quad-2(231.074)-2(163.060)=-280.687\, kJ / K \\\Delta G _{900 K }^0=-259993-900(-280.687)=-7375\, kJ \\\ln K =\frac{+7375}{8.3145 \times 900}=0.9856, \quad K = 2 . 6 7 9\end{array}\end{aligned}

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