Problem 4

Production of methanol. Methanol (

\ce{CH3OH}

) is used for gasoline additives and many plastics. A factory is based on

\ce{CO + 2H2 -> CH3OH}

; the

\ce{H2}

and CO are obtained by reforming

\ce{CH4 + H2O -> CO + 3H2}

. The plant has three units: the reformer, the methanol reactor, and a separator that removes methanol from CO and

\ce{H2}

. Positions α, β, γ, δ are: α = feed to reformer (

\ce{CH4}

\ce{H2O}

), β = gas entering the methanol reactor (CO,

\ce{H2}

), γ = stream leaving the reactor before separation (CO,

\ce{H2}

\ce{CH3OH}

), δ = recycled/burnt excess CO and

\ce{H2}

after the separator. The methanol flow at γ is

n(\ce{CH3OH}, \gamma) = 1000

mol s

^{-1}

, and the design converts 2/3 of the CO to methanol; the reformer reaction goes to completion. (4.1) Calculate the flows of CO and

\ce{H2}

at β. (4.2) Calculate the flows of CO and

\ce{H2}

at γ. (4.3) Calculate the flows of

\ce{CH4}

and

\ce{H2O}

needed at α. (4.4) All species are gaseous at γ; with total pressure

p = 10

MPa, calculate the partial pressures of CO,

\ce{H2}

and

\ce{CH3OH}

at γ using

p_i = p\, n_i/n_{\mathrm{tot}}

. (4.5) When the reactor is large enough the reaction reaches equilibrium and the partial pressures obey

K_p = \dfrac{p(\ce{CH3OH})\,p^0}{p(\ce{CO})\,p(\ce{H2})^2}

with

p^0 = 0.1

MPa. Calculate

K_p

(the corresponding temperature from the graph of

\log K_p

T

is about 630 K).

Step 4 of 4: Equilibrium constant and temperature

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K_p = \frac{p(\ce{CH3OH})\,p^0}{p(\ce{CO})\,p(\ce{H2})^2} = \frac{2.5\times0.1^2}{1.25\times6.25^2} = 5.12\times10^{-4}\ \Rightarrow\ T \approx 630\ \text{K}

Analysis

Inserting the partial pressures into the given dimensionless form of $K_p$ yields $5.12\times10^{-4}$ ; on the $\log K_p$ versus $T$ diagram this corresponds to about 630 K.

Common pitfall. Watch the standard-pressure factor:

K_p

is dimensionless here, so

p(\ce{CH3OH})

is multiplied by

p^0 = 0.1

MPa, not by 1 MPa.