Question

Steam reforming of methane ( CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 1.5 L flask with 0.60 atm of methane gas and 2.6 atm of water vapor at 47. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 1.4 atm. Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits x10

Answer 1

Answer:

1.5

Explanation:

The reaction between methane and water vapor to produce carbon monoxide and hydrogen gas is

CH4(g) + 2H2O(g) ⇄ CO2(g) + 4H2(g)

The reaction is reversible and achieves the equilibrium when the velocity of the formation of the products and the velocity of the formation of the reagents are the same. In this case, both concentration and pressure remain constant.

The pressure equilibrium constant is the value of the ratio of the multiplication of the partial pressure of the products by the multiplication of the partial pressure of the reagents, each one elevated by the coefficient of the substance.

Thus, doing an equilibrium chart:

CH4(g) + 2H2O(g) ⇄ CO2(g) + 4H2(g)

0.60 atm 2.6 atm 0 0 Initial

-x -2x +x +4x Reacts (stoichiometry is 1:2:1:4)

0.60-x 2.6-2x x 4x Equilibrium

pH2 = 1.4 atm

4x = 1.4

x = 0.35 atm

So, the partial pressures in equilibrium are:

pCH4 = 0.60 - 0.35 = 0.25 atm

pH2O = 2.6 - 2*0.35 = 1.9 atm

pCO2 = 0.35 atm

pH2 = 1.4 atm

$Kp = \frac{(pCO_2)*(pH_2)^4}{(pCH_4)*p(H_2O)^2}$

$Kp = \frac{0.35*(1.4)^4}{0.25*(1.9)^2}$

Kp = 1.5

               

Answer 2

A 1.5 L flask is filled with 0.60 atm CH₄(g) and 2.6 atm H₂O(g). The temperature is raised and the partial pressure of H₂(g) at equilibrium is 1.4 atm. The pressure equilibrium constant for the steam reforming of methane at the final temperature is 4.7.

Let's consider the reaction between methane gas and water vapor to form carbon monoxide gas and hydrogen gas.

CH₄(g) + H₂O(g) ⇄ CO(g) + 3 H₂(g)

A 1.5 L flask is filled with 0.60 atm CH₄(g) and 2.6 atm H₂O(g). When the equilibrium is reached, the partial pressure of H₂(g) is 1.4 atm.

We can calculate the concentrations at equilibrium by making an ICE chart.

        CH₄(g)    +    H₂O(g)    ⇄    CO(g)    +    3 H₂(g)

I         0.60              2.6                  0                   0

C         -x                  -x                   +x                +3x

E       0.60-x          2.6-x                  x                  3x

Since the partial pressure at the equilibrium of H₂(g) is 1.4 atm, we get,

$3x = 1.4 atm\\\\x = 0.47atm$

The concentrations at equilibrium are:

$[CH_4] = 0.60-0.47 = 0.13atm\\[H_2O] = 2.6-0.47 = 2.1atm\\[CO] = 0.47 atm\\[H_2] = 1.4 atm$

The pressure equilibrium constant (Kp) for the steam reforming of methane is:

$Kp=\frac{[CO][H_2]^{3} }{[CH_4][H_2O]} = \frac{(0.47)(1.4)^{3} }{(0.13)(2.1)} = 4.7$

A 1.5 L flask is filled with 0.60 atm CH₄(g) and 2.6 atm H₂O(g). The temperature is raised and the partial pressure of H₂(g) at equilibrium is 1.4 atm. The pressure equilibrium constant for the steam reforming of methane at the final temperature is 4.7.

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