One of the key reactions in the gasification of coal is the methanation reaction, in which methane is produced from synthesis gas-a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons & \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta H &=-230 \mathrm{kJ} ; K_{\mathrm{c}}=190 \mathrm{at} 1000 \mathrm{K} \end{aligned}$$ (a) Is the equilibrium conversion of synthesis gas to methane favored at higher or lower temperatures? Higher or lower pressures? (b) Assume you have 4.00 mol of synthesis gas with a 3:1 mol ratio of \(\mathrm{H}_{2}(\mathrm{g})\) to \(\mathrm{CO}(\mathrm{g})\) in a 15.0 L flask. What will be the mole fraction of \(\mathrm{CH}_{4}(\mathrm{g})\) at equilibrium at \(1000 \mathrm{K} ?\)

Step by step solution

01

Understanding the Temperature effect

From part (a), we need to understand the effect of temperature on the equilibrium. Since the reaction is exothermic (\( \Delta H < 0 \)), according to Le Chatelier's principle, increasing the temperature will shift the equilibrium to the left (favoring the reverse reaction or making less methane). So, the equilibrium conversion of synthesis gas to methane is favored at lower temperatures.

02

Understanding the Pressure effect

The reaction involves 4 moles of gases on the left side (reactants) and 2 moles of gases on the right side (products). According to Le Chatelier's principle, increasing the pressure will shift the equilibrium towards the side with fewer moles of gases. So, the equilibrium conversion of synthesis gas to methane is favored at higher pressures.

03

Calculate Initial Moles and Initial Composition

The problem states that the initial moles of synthesis gas is a 3:1 ratio mix of \( \mathrm{H}_{2}(\mathrm{g}) \) and \( \mathrm{CO}(\mathrm{g}) \) totalling 4 moles. So, we have 3 moles of \( \mathrm{H}_{2}(\mathrm{g}) \) and 1 mole of \( \mathrm{CO}(\mathrm{g}) \). No \( \mathrm{CH}_{4}(\mathrm{g}) \) or \( \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \) is initially present.

04

Calculate Equilibrium Concentration

Using the stoichiometry of the reaction \( \mathrm{CO}(\mathrm{g}) + 3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \), we know that for every mole of \( \mathrm{CO}(\mathrm{g}) \) consumed, 1 mole of \( \mathrm{CH}_{4}(\mathrm{g}) \) is formed. Let 'x' be the moles of \( \mathrm{CO}(\mathrm{g}) \) and \( \mathrm{H}_{2}(\mathrm{g}) \) consumed. At equilibrium, we will have (1-x) moles of \( \mathrm{CO}(\mathrm{g}) \), (3-3x) moles of \( \mathrm{H}_{2}(\mathrm{g}) \), x moles of \( \mathrm{CH}_{4}(\mathrm{g}) \), and x moles of \( \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \). The total moles will be \( 4 - 2x \). The mole fraction of \( \mathrm{CH}_{4}(\mathrm{g}) \) at equilibrium will be x divided by the total moles.

05

Use Equilibrium Constant to Solve for x

Write down the expression for the equilibrium constant. Substitute the equilibrium concentrations into the equilibrium constant expression and solve for 'x'. The value of 'x' is then used to find the mole fraction of \( \mathrm{CH}_{4}(\mathrm{g}) \) at equilibrium.

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