Question

Question:Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures. \(C{H_4}(g) + {H_2}O(g) \rightleftharpoons 3{H_2}(g) + CO(g)\)

Short answer

The reaction of the elevated temperatures \(K = 6.28\)

Step-by-step solution

Define Hydrogen atom

A hydrogen atom is an atom of the chemical element hydrogen. The electrically neutral atom contains a single positively charged proton and a single negatively charged electron bound to the nucleus by the Coulomb force. Atomic hydrogen constitutes about 75% of the baryonic mass of the universe

The reaction of equilibrium constant reaction

We need to calculate equilibrium constant for following reaction:

\(C{H_4}(g) + {H_2}O(g) \to 3{H_2}(g) + CO(g)\)

this is a reverse reaction so it should be written with a double arrow. Equilibrium constant is calculated like this:

\(K = \frac{{c\left( {{H_2}} \right)_{eq}^3 \cdot c{{(CO)}_{eq}}}}{{c{{\left( {C{H_4}} \right)}_{eq}} \cdot c{{\left( {{H_2}O} \right)}_{eq}}}}\)

Solving the concentration of the compound \({C_{eq}}\) 

\(\begin{array}{*{20}{c}}{c{{\left( {{H_2}} \right)}_{eq}} = 1.15M}\\{c{{(CO)}_{eq}} = 0.126M}\\{c{{\left( {C{H_4}} \right)}_{eq}} = 0.126M}\\{c{{\left( {{H_2}O} \right)}_{eq}} = 0.242M}\end{array}\)

The constant equation\(\begin{array}{*{20}{c}}{K = \frac{{{{1.15}^3} \cdot 0.126}}{{0.126 \cdot 0.242}}}\\{K = 6.28}\end{array}\)