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 200.mL flask with 2.4 atm of methane gas and 3.9 atm of water vapor at 46.0°C. She then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 6.5 atm. Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture.

Answer 1

Answer: The pressure equilibrium constant for the reaction is 1473.8

Explanation:

We are given:

Initial partial pressure of methane gas = 2.4 atm

Initial partial pressure of water vapor = 3.9 atm

Equilibrium partial pressure of hydrogen gas = 6.5 atm

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

                        $CH_4+H_2O\rightleftharpoons CO+3H_2$

Initial:               2.4       3.9

At eqllm:        2.4-x    3.9-x        x       3x

Evaluating the value of 'x':

$\Rightarrow 3x=6.5\\\\x=2.167$

So, equilibrium partial pressure of methane gas = (2.4 - x) = [2.4 - 2.167] = 0.233 atm

Equilibrium partial pressure of water vapor = (3.9 - x) = [3.9 - 2.167] = 1.733 atm

Equilibrium partial pressure of carbon monoxide gas = x = 2.167 atm

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{CO}\times (p_{H_2})^3}{p_{CH_4}\times p_{H_2O}}$

Putting values in above equation, we get:

$K_p=\frac{2.167\times (6.5)^3}{0.233\times 1.733}\\\\K_p=1473.8$

Hence, the pressure equilibrium constant for the reaction is 1473.8