Steam reforming of methane ( ) 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 tank with of methane gas and of water vapor at . He then raises the temperature, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be . Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Steam reforming of methane 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 125 L tank with 20 mol of methane gas and 10 mol of water vapor at 38 degrees celsius. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be 18 mol.

Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 3 significant digits.

Answer: The value of equilibrium constant for given equation is 0.0399

Explanation:

We are given:

Initial moles of methane gas = 20 moles

Initial moles of water vapor = 10 moles

Volume of solution = 125 L

Equilibrium moles of hydrogen gas = 18 moles

Molarity is calculated by using the equation:

[tex]\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]

[tex]\text{Initial concentration of methane}=\frac{20}{125}=0.16M[/tex]

[tex]\text{Initial concentration of water vapor}=\frac{10}{125}=0.08M[/tex]

[tex]\text{Equilibrium concentration of hydrogen gas}=\frac{18}{125}=0.144M[/tex]

The chemical equation for the reaction of methane and water follows:

[tex]CH_4+H_2O\rightarrow CO+3H_2[/tex]

Initial: 0.16 0.08

At eqllm: 0.16-x 0.08-x x 3x

Evaluating the value of 'x'

[tex]\Rightarrow 3x=0.144\\\\x=0.048M[/tex]

So, equilibrium concentration of methane gas = (0.16-x) = (0.16 - 0.048) = 0.112 M

Equilibrium concentration of water vapor = (0.08-x) = (0.08 - 0.048) = 0.032 M

Equilibrium concentration of carbon monoxide gas = x = 0.048 M

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[CO][H_2]^3}{[CH_4][H_2O]}[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{0.048\times (0.144)^3}{0.112\times 0.032}\\\\K_c=0.0399[/tex]

Hence, the value of equilibrium constant for given equation is 0.0399