Hydrogen is made from natural gas (methane) for immediate consumption in indus- trial processes, such as ammonia production. The first step is called the "steam reforming of methane": CH, (g) + H2O(g)=CO(g) + 3 H2 (g) The equilibrium constant for this reaction is 1.8 x 10-7 at 600 K. Gaseous CH4, H,O, and CO are introduced into an evacuated container at 600 K, and their initial partial pressures (before reaction) are 1.40 atm, 2.30 atm, and 1.60 atm, respectively. Deter- mine the partial pressure of H2(g) that will result at equilibrium.

The rate of a forward reaction is equal to the product of the rate constant for the forward reaction and concentration of reactants. Rate constant is dependent on temperature, but not concentration. An increase in temperature increases the rate constant, and hence the rate. An increase in concentration increases the rate, but not the rate constant.

Similarly, there is rate constant for the backward reaction. Equilibrium constant is the ratio of forward reaction rate constant to the backward reaction rate constant. It simplifies to the ratio of concentration of products to the concentration of the reactants.

The reaction taking place is, 

${CH}_4+H_{2}O\rightleftharpoons CO+3 H_2$

All the components are gases. 

Let us assume that all the gases are ideal and the temperature remains constant.

The equilibrium constant is given as, $K_c=1.8\times {10}^{-7}$

we know that ,

$K_p=K_c {\left(RT\right)}^{∆n}$

where,

$K_p$ is the equilibrium constant in terms of partial pressure

$R$ is the universal gas constant $\left(8.314 \frac{J}{K.mol}\right)$

$T$ is the absolute temperature

$∆n$ is the difference between the stoichiometric constants of products and the reactants,

i.e  $$\begin{gathered} ∆n=(1+3)-(1+1) \\ ∆n=2 \end{gathered}$$

Hence,

$$\begin{gathered} K_p=1.8\times {10}^{-7} {\left(8.314 \frac{J}{K.mol}\times 600 K\right)}^2 \\ {K}_p=4.48 \end{gathered}$$

we know,

$K_p=\frac{P_{CO}.{P_{H_2}}^3}{P_{C{H}_4}.P_{H_{2}O}}$

Where,

P  represents the partial pressures of the components.

Let us put up a flow data for the reactants and product.

Let x be the reduce in pressure due to the reaction

We know,

$\begin{array}{rcl}{K}_p& =& \frac{P_{CO}.{P_{H_2}}^3}{P_{C{H}_4}.P_{H_{2}O}}\\ 4.48& =& \frac{P_{CO}.{P_{H_2}}^3}{P_{C{H}_4}.P_{H_{2}O}}\end{array}$

Substituting the equilibrium pressures,

$\begin{array}{rcl}4.48& =& \frac{\left(1.6+x\right){\left(3x\right)}^3}{\left(1.4-x\right)\left(2.3-x\right)}\\ 4.48& =& \frac{43.2x^3+3x^4}{3.22-1.4x-2.3x+x^2}\\ 14.4256-16.576x+4.48x^2& =& 43.2x^3+27x^4\\ 27x^4+43.2x^3-4.48x^2+16.576x-14.4256& =& 0\end{array}$

Solving the above equation,

we get, $\begin{array}{rcl}{x}_1& =& -0.08829+0.7373i\\ x_2& =& -0.0882+-0.7373i\\ {\mathit{x}}_{\mathbf{3}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{5029}\\ x_4& =& -1.9263\end{array}$

Since the value of x cannot be negative or imaginary, we will be considering $\mathit{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5029}\mathbf{.}$

Hence, the equilibrium pressures of the components are:

$\begin{array}{rcl}{P}_{C{H}_4}& =& 1.4-x\\ P_{C{H}_4}& =& 1.4-0.5029\\ {\mathit{P}}_{\mathbf{C}{\mathbf{H}}_{\mathbf{4}}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{8971}\mathbf{ }\mathit{a}\mathit{t}\mathit{m}\\ P_{H_{2}O}& =& 2.3-x\\ P_{H_{2}O}& =& 2.3-0.5029\\ {\mathit{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}& \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{7971}\mathbf{ }\mathit{a}\mathit{t}\mathit{m}\\ P_{CO}& =& 1.6+x\\ P_{CO}& =& 1.6+0.5029\\ {\mathit{P}}_{\mathbf{C}\mathbf{O}}& \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{1029}\mathbf{ }\mathit{a}\mathit{t}\mathit{m}\\ P_{H_2}& =& 3x\\ P_{H_2}& =& 3\times 0.5029\\ {\mathit{P}}_{{\mathbf{H}}_{\mathbf{2}}}& \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{5087}\mathbf{ }\mathit{a}\mathit{t}\mathit{m}\end{array}$