Question

At \(327^{\circ} \mathrm{C}\), the equilibrium concentrations are \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=\) \(0.15 M,[\mathrm{CO}]=0.24 M\), and \(\left[\mathrm{H}_{2}\right]=1.1 M\) for the reaction $$ \mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) $$ Calculate \(K_{\mathrm{p}}\) at this temperature.

Short answer

At a temperature of \(327^{\circ} \mathrm{C}\), the value of \(K_p\) for the gaseous reaction CH3OH(g) ⇌ CO(g) + 2H2(g) is approximately 1.93.

Step-by-step solution

Write the Kp expression for the reaction

For the gaseous reaction CH3OH(g) ⇌ CO(g) + 2H2(g), the expression for Kp can be written as: \(K_p = \frac{[\mathrm{CO}] \times [\mathrm{H}_2]^2}{[\mathrm{CH}_3 \mathrm{OH}]} \) Where [CH3OH], [CO], and [H2] represent the equilibrium molar concentrations of the respective species.

Insert the given equilibrium concentrations

We are given the equilibrium concentrations as: [CH3OH] = 0.15 M [CO] = 0.24 M [H2] = 1.1 M Insert these values into the Kp expression: \(K_p = \frac{(0.24 \,\text{M}) \times (1.1 \,\text{M})^2}{(0.15 \,\text{M})}\)

Calculate Kp

Now, compute the Kp value using the inserted concentrations: \(K_p = \frac{(0.24 \,\text{M}) \times (1.1 \,\text{M})^2}{(0.15 \,\text{M})} = \frac{(0.24 \,\text{M}) \times (1.21 \,\text{M^2})}{(0.15 \,\text{M})} = 1.93\) Therefore, at a temperature of \(327^{\circ} \mathrm{C}\), the value of \(K_p\) for the given reaction is approximately 1.93.

Key concepts

Equilibrium Constant

The equilibrium constant, often represented as \(K\), is a crucial concept in chemistry that helps us understand how a chemical reaction behaves under equilibrium conditions. When a chemical reaction reaches a state where the forward and reverse reactions occur at the same rate, it is said to be at equilibrium.

Here's what you need to know about the equilibrium constant:

• For reactions involving gases, we often use \(K_p\) instead of \(K_c\), as it expresses the ratios in terms of partial pressures instead of concentrations. This is pertinent when dealing with gaseous chemical equations.
• It's calculated using the equilibrium concentrations or pressures of the reactants and products involved in the reaction.
• The formula for \(K_p\) is structured in such a way that it focuses on the products over the reactants. For a reaction \( aA + bB \rightleftharpoons cC + dD \), it would be \(K_p = \frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b}\).

Understanding \(K_p\) and how it is calculated allows chemists to predict how changes in conditions can affect the position of equilibrium, helping in designing processes that maximize yield.

Reaction Quotient

The reaction quotient, symbolized as \(Q\), is another important concept that works hand in hand with the equilibrium constant. Unlike \(K\), which uses equilibrium concentrations, \(Q\) can be determined at any point in the reaction.

• \(Q\) is calculated with the same expression as \(K\), but using the concentrations or pressures at any given moment, not necessarily at equilibrium.
• If \(Q < K\), the system will shift towards products to reach equilibrium. If \(Q > K\), the system will move towards reactants.
• When \(Q = K\), the system is at equilibrium, meaning no net change in concentrations.

Understanding \(Q\) is vital as it provides insight into how far a reaction has progressed relative to its equilibrium state. This insight equips chemists to take corrective actions if a system is not at equilibrium.

Heterogeneous Equilibrium

Heterogeneous equilibrium refers to reactions where reactants and products are present in different phases. In these reactions, we only include the concentrations or pressures of species in the gaseous or aqueous phase when writing equilibrium expressions.

Here's what's important about heterogeneous equilibrium:

• Solid and liquid pure substances are omitted from the equilibrium expression because their concentrations are constants and do not change the rate of reaction.
• Even when multiple phases are present, the principles of calculating \(K\) remain consistent, focusing solely on the phases that change concentrations (often gases and solutions).
• Understanding how heterogeneous equilibria work is crucial for predicting and managing reactions in processes like catalysis and separation technologies.

By focusing on the active phases, chemists can simplify their calculations and still accurately predict how a reaction behaves under equilibrium conditions.