The reaction \(\mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) is run in a 10.0-L vessel. The vessel is loaded with 1 mole of \(\mathrm{CO}\) and 3 moles of \(\mathrm{H}_{2}\). At equilibrium, the amounts are \(0.613\) mole of \(\mathrm{CO}, 1.839\) moles of \(\mathrm{H}_{2}, 0.387\) mole of \(\mathrm{CH}_{4}\), and \(0.387\) mole of \(\mathrm{H}_{2} \mathrm{O}\). What is the value of the equilibrium constant for this reaction? Describe the position of the equilibrium.

First, let's find the initial concentrations of all the species in the reaction. Since the reaction is run in a 10.0 L vessel: \[Initial\: concentration\: of\: CO = \frac{1\: mole}{10.0\: L} = 0.1 M\] \[Initial\: concentration\: of\: H_{2} = \frac{3\: moles}{10.0\: L} = 0.3 M\] Initially, there are no products present. So, the initial concentrations of CH4 and H2O are 0 M.

Next, let's find out how the concentrations of the species change as the reaction proceeds to equilibrium. The reaction has a stoichiometry of 1:3:1:1, which means that 1 mole of CO reacts with 3 moles of H2 to produce 1 mole of CH4 and 1 mole of H2O. Since the equilibrium amount of CO is 0.613 moles and the initial amount is 1 mole, the change in the amount of CO is: \[Change\: in\: CO = (0.613 - 1) = -0.387\: moles\] For every mole of CO that reacted, 3 moles of H2 reacted. Therefore, the change in the amount of H2 is: \[Change\: in\: H_{2} = \frac{-0.387}{1} \times 3 = -1.161\: moles\] Similarly, for every mole of CO that reacted, 1 mole of CH4 and 1 mole of H2O were produced. Therefore, the change in the amounts of CH4 and H2O is: \[Change\: in\: CH_{4} = Change\: in\: H_{2}O = 0.387\: moles\]

Now, let's find the equilibrium concentrations of all the species. \[Equilibrium\: concentration\: of\: CO = \frac{0.613\: moles}{10.0\: L} = 0.0613 M\] \[Equilibrium\: concentration\: of\: H_{2} = \frac{1.839\: moles}{10.0\: L} = 0.1839 M\] \[Equilibrium\: concentration\: of\: CH_{4} = \frac{0.387\: moles}{10.0\: L} = 0.0387 M\] \[Equilibrium\: concentration\: of\: H_{2}O = \frac{0.387\: moles}{10.0\: L} = 0.0387 M\]

Using the equilibrium concentrations, we can now calculate the equilibrium constant (K) for the reaction: \[K = \frac{[CH_4][H_2O]}{[CO][H_{2}]^3}\] \[K = \frac{(0.0387)(0.0387)}{(0.0613)(0.1839)^3} \approx 0.042\]

With the calculated equilibrium constant (K = 0.042), we can determine the position of the equilibrium. Since K is less than 1, the reaction favors the reactants, meaning that most of the CO and H2 are not converted to CH4 and H2O. The equilibrium is positioned more to the left, favoring the reactants. In conclusion, the equilibrium constant for this reaction is approximately 0.042 and the reaction favors the reactants, with equilibrium positioned more to the left.