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 1 in the reactor contents at equilibrium. The feed to a batch shift reactor contains 20.0 mol% CO, 10.0 mol% CO_2, 30.0 mol% H_2O, and the balance an inert gas. The reactor is maintained at T = 1623 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

Q1.

amount of elemental balances:

elements; carbon, hydrogen and Oxygen

C,H,O = 3 elements, 3 balances

Q2.

Variables: T, yco2,yh2o, yh2, yco2

Equations:

y1+y2+y3+y4 = 1; thjerefore, we can deduce all compositions

Keq can be calcualted via equation shown above

Temperature is given

So...

DOF = 0, the systm can be solved right now with 1 solution

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