Consider a chemical reaction CO(g) + 2H_2 (g) = CH_3OH(g). When 2 moles of CO we

Question

Consider a chemical reaction CO(g) + 2H_2 (g) = CH_3OH(g). When 2 moles of CO were reacting with 4 moles of H_2 at temperature 500K and pressure p = 100 bar, and chemical equilibrium was reached, n = 0.366 moles of CO was consumed. Based on the given data determine the value of the chemical equilibrium constant K(T) for T = 500K. Consider now the same reaction, when 3 moles of CO are reacting with 3 moles of H_2 at the temperature T = 500K and the pressure p_0 = 1bar. Find the mass m of the product, i.e., CH_3OH , obtained in the reaction.

Explanation / Answer

Given,

Reaction: CO    +   2H2 ----> CH3OH

Initial:       2.............4................0

Change: 2-X..........4-2X..........X

It is aslo known that 0.366 moles of CO was consumed

=> X = 0.366

At equilibrium moles of

CO = 2 - 0.366 = 1.634

H2 = 4 - (2 x 0.366) = 3.268

CH3OH = X = 0.366 moles

Total Moles of Gas = 1.634 + 3.268 + 0.366 = 5.268 moles

=> Mole fraction of CO = 1.634 / 5.268 = 0.31

Mole fraction of H2 = 3.268 / 5.268 = 0.62

Mole fraction of CH3OH = 0.366 / 5.268 = 0.07

Partial Pressure of CO = 0.31 x 100 = 31 bar

Partial Pressure of H2 = 0.62 x 100 = 62 bar

Partial Pressure of CH3OH = 0.07 x 100 = 7 bar

Kp = 7 / 31 x (62)^2 = 5.9 x 10^-5

b)

CO + 2H2 ----> CH3OH

3-X.....3-3X.........X

Total Moles = 3-X + 3-2X + X = 6 - 3X

Partial Pressure of CO = (3-X) / (6-2X) (since Po = 1 bar)

Partial Pressure of H2 = (3-2X) / (6-2X)

Partial Pressure of CH3OH = X / (6-2X)

Kp = (X / (6-2X)) / ((3-X) / (6-2X)) x ((3-2X) / (6-2X))^2

=> 5.9 x 10^-5 = X (6-2X)^2 / (3-X) (3-2X)^2

Neglecting X with reapect to 3 and 6, we get

5.9 x 10^-5 = 36X / 27

=> X = 4.41 x 10^-5 = Moles of CH3OH formed

=> Mass of CH3OH = 4.41 x 10^-5 x 32 = 1.41 x 10^-3 g

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