At low to moderate pressures, the equilibrium state of the water-gas shift react

Question

At low to moderate pressures, the equilibrium state of the water-gas shift reaction CO + H_2O doubleheadarrow CO_2 + H_2 is approximately described by the relation Y_CO_2 Y_H_2/Y_CO Y_H_2O = K_eq (T) = 0.0247 exp[4020/T (K)] where T is the reactor temperature, K_eq is the reaction equilibrium constant, and y_i is the mole fraction of species i in the reactor contents at equilibrium. The feed to a batch shift reactor contains 30.0 mol% CO, 15.0 mol% CO_2, 30.0 mol% H_2O, and the balance an inert gas. The reactor is maintained at T = 1423 K. For the batch reaction as specified above, carry out a degree-of-freedom analysis. How many elemental balances may be written? How many degrees of freedom remain?

Explanation / Answer

a)

Elemental balances:

C,O,H

there are 3 elements, so 3 balances can be performed

B)

DOF = Variables - Equations

Variables:

T, Keq, Yi,

Yi= known, since it is given

Ke = we got expresion dependnt on T

Temperature must be SET

Therefore, only one degree is left

That is, if we set T, then the system can be solved with unique solution

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