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Equilibrium Constant Expression

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Given the following reaction: C0(g) + H2O(g) <> C02(g) + H2(g) (Please see the attachment for the properly formatted reaction).
a) Write the equilibrium constant expression for this reaction. Note: gaseous-phase H20 is included in the equilibrium expressions, unlike liquid phase H20.
b) The equilibrium constant for the reaction is 8.62 x 10^2 atm^2 at 350°C and under constant pressure. If the initial partial pressures of CH4(g) and H20 (g) are 9.64 atm and 28.37 atm, calculate the pressure of all species at equilibrium. Use the quadratic formulae to solve the second degree equation you will obtain.
c) An approximation can be used to simplify the equilibrium constant equations in certain situations, as demonstrated in the attachment...

Please see the attachment for the rest of the question.

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Solution Summary

The equilibrium constant expressions are examined. The quadratic formulas solved for a second degree are determined.

Solution Preview

Please see the attachment.

K = pCO2 x pH2 / pCO x pH2O
CO + H2O --- CO2 + H2
Initial 9.64 28.37 0 0
Change -p -p +p +p
Equilibrium 9.64-p 28.37-p ...

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