Question

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 \(1000 \mathrm{K} ?\)

Short answer

So, the equilibrium conversion of synthesis gas to methane is favored at lower temperatures and higher pressures. The mole fraction of \( \mathrm{CH}_{4}(\mathrm{g}) \) at equilibrium will be calculated using the Equilibrium Constant \( K_c \).

Step-by-step solution

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.

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.

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.

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.

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.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a dynamic equilibrium responds to changes in conditions, such as temperature and pressure. This principle helps predict the shift in equilibrium position for the methanation reaction \( \mathrm{CO} + 3 \mathrm{H}_2 \rightleftharpoons \mathrm{CH}_4 + \mathrm{H}_2\mathrm{O} \) when certain conditions are altered.

In the case of temperature, since the methanation reaction is exothermic (\( \Delta H = -230 \mathrm{kJ} \)), increasing temperature adds heat to the system, effectively shifting the equilibrium to the left. This means fewer products (methane and water) are formed. Conversely, a decrease in temperature favors the forward reaction, resulting in more methane production.

Similarly, the impact of pressure can be understood through this principle. As the reaction involves a reduction in the number of gas moles (going from 4 moles in the reactants to 2 moles in the products), an increase in pressure will shift the equilibrium toward producing more methane and water. This is because the system seeks to minimize pressure by favoring the side with fewer gas moles. Therefore, higher pressure favors the conversion of synthesis gas to methane.

Equilibrium Constant

The equilibrium constant, \( K_c \), is a crucial parameter in understanding the position of equilibrium for a chemical reaction at a given temperature. It is defined by the ratio of concentration of products to reactants, each raised to the power of their stoichiometric coefficients.

For the methanation reaction \( \mathrm{CO} + 3 \mathrm{H}_2 \rightleftharpoons \mathrm{CH}_4 + \mathrm{H}_2\mathrm{O} \), the equilibrium constant expression is given by: \[ K_c = \frac{[\mathrm{CH}_4][\mathrm{H}_2\mathrm{O}]}{[\mathrm{CO}][\mathrm{H}_2]^3} \]

At \( 1000 \mathrm{K} \), \( K_c \) is specified as 190, indicating a high preference for the formation of products under these conditions. This constant allows us to calculate equilibrium concentrations and is key to determining how much of each substance is present at equilibrium. By plugging in known concentrations into the equation, we can solve for unknowns, like the concentration of methane and water, thus illustrating how effectively synthesis gas is converted to methane at equilibrium.

Synthesis Gas

Synthesis gas, commonly referred to as syngas, is a mixture primarily composed of carbon monoxide (\( \mathrm{CO} \)) and hydrogen (\( \mathrm{H}_2 \)). This versatile feedstock is a key intermediate in producing various chemicals, with one important application being the methanation process to create methane.

In the context of the methanation reaction, syngas undergoes a chemical transformation in which \( \mathrm{CO} \) and \( \mathrm{H}_2 \) react to form methane (\( \mathrm{CH}_4 \)) and water (\( \mathrm{H}_2\mathrm{O} \)).

Syngas is typically produced through processes like coal gasification or steam reforming of natural gas. Its composition can vary, but it is often processed to have specific ratios to optimize reactions like methanation. For instance, in the given scenario, the synthesis gas has a 3:1 \( \mathrm{H}_2 \) to \( \mathrm{CO} \) molar ratio, meaning there are three times as many hydrogen molecules as carbon monoxide molecules.

This ratio directly affects the reaction's stoichiometry and equilibrium, dictating how much methane can be formed based on the initial syngas composition.