Hydrogen is made from natural gas (methane) for immediate consumption in industr

Question

Hydrogen is made from natural gas (methane) for immediate consumption in industrial processes (for example the production of ammonia). The first step in the process is called the steam reforming of methane.

CH4(g) + H2O (g) CO(g) + 3H2
The equilibrium constant Kp  for this reaction is 1.8 x 10-7 at 600K. Gases CH4, H2O and CO are introduced into an evacuated container at 600K, their initial pressures (before reaction) are 1.30 atm, 2.30 atm and 1.60 atm. Determine the partial pressure of H2 at equilibrium.

Explanation / Answer

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Kp = PCO *PH2^3 / (PCH4)(PH2O)

Kp = 1.8*10^-7

initially

PCH4 = 1.3

PH2O = 2.3

PCO = 1.6

PH2 = 0

in equilbirium

PCH4 = 1.3 - x

PH2O = 2.3 - x

PCO = 1.6 + x

PH2 = 0 +3x

substitute

Kp = PCO *PH2^3 / (PCH4)(PH2O)

(1.8*10^-7) = (1.6+x) * (3x)^3 / ((1.3 - x)(2.3-x))

1.3*2.3 - (2.3+1.3)x + x^2 = 27/(1.8*10^-7) * (x^4 + 1.6x^3)

150000000*x^4 + 240000000x^3 - x^2 + 3.6x - 2.99 = 0

x = 0.002315

PCH4 = 1.3 - 0.002315 = 1.297685

PH2O = 2.3 - 0.002315 = 2.297685

PCO = 1.6 + 0.002315 = 1.602315

PH2 = 0 +3*0.002315 = 0.006945 atm

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