Equimolar methane and steam are sealed in a flask. Equilibria are C H 4 ( g ) + H 2 O ( v ) ⇌ CO ( g ) + 3 H 2 ( g ) and 2 H 2 ( g ) + CO ( g ) ⇌ C H 3 OH ( g ) , the second having K p = 2

Question

Equimolar methane and steam are sealed in a flask. Equilibria are C H 4 ​ ( g ) + H 2 ​ O ( v ) ⇌ CO ( g ) + 3 H 2 ​ ( g ) and 2 H 2 ​ ( g ) + CO ( g ) ⇌ C H 3 ​ OH ( g ) , the second having K p ​ = 2/7 at m − 2 . At equilibrium p CO ​ = 0.5 atm and total pressure is 5 atm. Find K p ​ of the first reaction.

TestHub · Accepted Answer

Let p_CH₃OH=q and p_CO=0.5 atm. Stoichiometry gives p_H₂=3p_CO+p_CH₃OH=1.5+q and p_CH₄=p_H₂O. Total pressure 5 atm gives p_CH₄=p_H₂O=1.5-q. The methanol equilibrium requires q/[0.5(1.5+q)²]=2/7, so q²-4q+2.25=0. The physical root is q=(4-√7)/2=0.6771 atm; the other root makes reactant pressures negative. K_p for reforming=0.5(2.1771)³/(0.8229)²=7.62 atm², inside the printed range 6-8.

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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