Question

Recall the formation of methanol from synthesis gas, the reversible reaction at the heart of a process with great potential for the future production of automotive fuels (page 663 ). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 14.5 \mathrm{at} 483 \mathrm{K} \end{aligned}$$ A particular synthesis gas consisting of 35.0 mole percent \(\mathrm{CO}(g)\) and 65.0 mole percent \(\mathrm{H}_{2}(\mathrm{g})\) at a total pressure of 100.0 atm at \(483 \mathrm{K}\) is allowed to come to equilibrium. Determine the partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture.

Short answer

The partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture is given by the value of \(x\) from Step 4. The exact value will depend on the specific solution approach.

Step-by-step solution

Initial Pressures

Determine the initial pressures of the reactants. The partial pressure of a gas is given by multiplying the total pressure by the mole fraction of the gas. For \(\mathrm{CO}(g)\), this would be \(0.35 \times 100.0\:atm = 35.0\:atm\), and for \(\mathrm{H}_2(g)\), this would be \(0.65 \times 100.0\:atm = 65.0\:atm\). Remember that initially, there is no \(\mathrm{CH}_3\mathrm{OH}(g)\), so its pressure is zero.

Equilibrium Pressures

Write the expressions for the equilibrium pressures of all substances. Since the forward reaction consumes 1 mole of CO and 2 moles of H2 to produce 1 mole of CH3OH, the changes in pressures upon achieving equilibrium are as follows: \(\mathrm{CO}(g)\) changes by \(-x\), \(\mathrm{H}_2(g)\) changes by \(-2x\), and \(\mathrm{CH}_3\mathrm{OH}(g)\) changes by \(+x\). Therefore, the pressure of \(\mathrm{CO}(g)\) at equilibrium will be \(35.0 - x\), the pressure of \(\mathrm{H}_2(g)\) at equilibrium will be \(65.0 - 2x\). The pressure of \(\mathrm{CH}_3\mathrm{OH}(g)\) at equilibrium will simply be \(x\).

Equilibrium Constant Expression

Using the given equilibrium constant, write down the equilibrium expression. The equilibrium constant \(K_c\) is defined as the ratio of the product of the equilibrium concentrations (or pressures) of the products to that of the reactants, each raised to their respective stoichiometric coefficients. In this case, \(K_c = 14.5 = \frac{x}{(35.0-x)(65.0-2x)^2}\).

Solve for x

Solve for \(x\), the equilibrium pressure of \(\mathrm{CH}_3\mathrm{OH}(g)\). This will involve rearranging the equation from Step 3, which is then followed by solving for \(x\) using an appropriate mathematical technique such as root finding or numerical methods.

Key concepts

Synthesis Gas

Synthesis gas, often called "syngas," is a crucial mixture of carbon monoxide (CO) and hydrogen (H₂). It serves as a foundation for producing various chemicals, including methanol. Syngas is typically generated from natural gas, coal, or biomass through processes like steam reforming or partial oxidation.

This mixture is versatile and can be tailored to different industrial needs by adjusting the ratios of CO and H₂. The composition often significantly influences the chemical processes that utilize syngas, such as the formation of methanol. Understanding the components of syngas and their interactions is essential in governing its practical applications in synthesis and fuels.

• Source materials: natural gas, coal, biomass
• Main components: CO and H₂
• Key uses: methanol production, fuel synthesis

Methanol Formation

Methanol is an important chemical in many industries, produced from synthesis gas through a catalytic process. The reaction involves combining carbon monoxide and hydrogen to form methanol, represented by the equation: \[ \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) \]

This reaction is reversible, meaning it can proceed in both directions, forming and decomposing methanol under certain conditions. The production of methanol is influenced by factors such as temperature, pressure, and the presence of a catalyst. The equilibrium constant (\( K_c \)) indicates the extent to which the reaction favors methanol production at a specific temperature, such as 483 K in this case.

• Conversion: carbon monoxide and hydrogen to methanol
• Conditions: temperature, pressure, catalytic activity
• Reversibility: can shift to favor reactants or products

Partial Pressure

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. It's essential in determining the behavior of gases in chemical reactions, like the formation of methanol from syngas. Each gas's partial pressure can be calculated by multiplying its mole fraction by the total pressure.

For the synthesis gas example, with a mole fraction of 35% for CO and 65% for H₂ at a total pressure of 100 atm, the initial partial pressures would be:

• CO: \( 0.35 \times 100 = 35 \) atm
• H₂: \( 0.65 \times 100 = 65 \) atm

Understanding partial pressure helps predict how gases interact during the reaction and achieve equilibrium. It provides insight into the changes each gas undergoes, like those experienced by CO and H₂ shifting to form methanol.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products. In the context of methanol synthesis, equilibrium is achieved when the formation and decomposition rates of methanol balance out.

The equilibrium constant, \( K_c \), provides a quantitative measure of the product-to-reactant ratio at equilibrium. An equilibrium constant of 14.5 indicates a preference toward methanol formation at 483 K. To determine equilibrium concentrations, one must set up and solve equilibrium expressions that incorporate initial pressures and changes. Solving these can be complex but is manageable using algebraic methods or numerical techniques.

• Definition: balance of forward and reverse reactions
• Determined by: temperature, pressure, initial concentrations
• Significance of \( K_c \): ratio reflecting reactant-product favorability