91Ó°ÊÓ

The reaction $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$ occurs at high temperatures. At \(700^{\circ} \mathrm{C},\) a 200.0 - \(\mathrm{L}\) tank contains 1.0 mol of \(\mathrm{CO}, 0.20\) mol of \(\mathrm{CO}_{2}\), and 0.40 mol of \(\mathrm{C}\) at equilibrium. (a) Calculate \(K_{c}\) for the reaction at \(700^{\circ} \mathrm{C}\) (b) Calculate \(K_{c}\) for the reaction, also at \(700^{\circ} \mathrm{C},\) if the amounts at equilibrium in the 200.0 -L tank are 1.0 mol of \(\mathrm{CO}, 0.20\) mol of \(\mathrm{CO}_{2}\), and \(0.80 \mathrm{mol}\) of \(\mathrm{C}\) (c) Compare the results of (a) and (b). Does the quantity of carbon affect the value of \(K_{c} ?\) Explain.

Step by step solution

01

Write the Expression for Kc

For the reaction \( \mathrm{C} (\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \), \( K_c \) is given by the formula: \[ K_c = \frac{[ \mathrm{CO}]^2}{[ \mathrm{CO}_{2}]} \]Note that solids, like \(\mathrm{C}(\mathrm{s})\), do not appear in the expression for \(K_c\). Only gaseous species are considered.

02

Calculate Kc for Scenario (a)

For scenario (a), the given equilibrium concentrations are \([\mathrm{CO}] = \frac{1.0}{200.0} = 0.005 \, \text{M}\) and \([\mathrm{CO}_2] = \frac{0.20}{200.0} = 0.001 \, \text{M}\). Substitute these values into the \(K_c\) expression:\[ K_c = \frac{(0.005)^2}{0.001} = \frac{0.000025}{0.001} = 0.025 \]

03

Calculate Kc for Scenario (b)

In scenario (b), equilibrium concentrations are the same with \([\mathrm{CO}] = 0.005 \, \text{M}\) and \([\mathrm{CO}_2] = 0.001 \, \text{M}\), as in scenario (a). Therefore, \[ K_c = \frac{(0.005)^2}{0.001} = 0.025 \]

04

Compare and Analyse the Results

Both scenario (a) and (b) yield \(K_c = 0.025\). The value of \(K_c\) is independent of the quantity of solid carbon \(\mathrm{C}\) because solids do not appear in the equilibrium expression. Therefore, changing the amount of \(\mathrm{C}\) has no effect on \(K_c\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant

When we talk about chemical equilibrium, the equilibrium constant, denoted by \( K_c \), plays a pivotal role. It provides a snapshot of the relative concentrations of reactants and products in a chemical reaction. For the reaction \( \mathrm{C} (\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \), the expression for \( K_c \) is given by: \[ K_c = \frac{[ \mathrm{CO}]^2}{[ \mathrm{CO}_{2}]} \]This equation highlights an important point: solids, such as \( \mathrm{C}(\mathrm{s}) \), do not appear in the \( K_c \) expression. Only gaseous components matter because they can change in concentration during the reaction.

• \( K_c \) reflects the ratio of product concentrations to reactant concentrations at equilibrium.
• Values of \( K_c \) are temperature-dependent and only valid for the specific temperature at which they were calculated.

Reaction Quotient

The reaction quotient, \( Q_c \), is similar to \( K_c \) but applies at any point during the reaction, not just at equilibrium. To calculate \( Q_c \), you use the same formula intended for \( K_c \), substituting the actual concentrations at any given time:\[ Q_c = \frac{[ \mathrm{CO}]^2}{[ \mathrm{CO}_{2}]} \]Comparison of the values of \( Q_c \) and \( K_c \) helps in determining the direction in which the reaction needs to proceed:

• If \( Q_c < K_c \), the reaction will proceed forward to create more products.
• If \( Q_c > K_c \), the reaction will go in reverse, favoring the formation of reactants.
• If \( Q_c = K_c \), the system is at equilibrium.

Understanding \( Q_c \) helps chemists control reactions to reach equilibrium efficiently.

Gaseous Equilibrium

Gaseous equilibrium pertains to reactions where gases are both reactants and products. The focus is on the concentrations of the gaseous substances. For the given reaction \( \mathrm{C} (\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \), gaseous equilibrium is concerned with the gases \( \mathrm{CO}_{2} \) and \( \mathrm{CO} \).

• In gaseous equilibrium, changes in pressure, temperature, and concentration directly affect the gases involved.
• The reaction may shift to counteract these changes, following Le Chatelier's Principle.
• The equilibrium position is the point at which the rate of the forward and backward reactions are equal, as reflected in the constant \( K_c \).

Recognizing when a system has reached its gaseous equilibrium is crucial for chemists to understand reaction behaviors and outcomes.

Solid Components

Solid components in reactions, like \( \mathrm{C}(\mathrm{s}) \) in our scenario, present a unique consideration. Solids are omitted from the \( K_c \) expression because their concentration is constant throughout the reaction.

• Solids have a fixed density, which means they do not change in molar concentration.
• Regardless of the quantity of solid present, it does not affect the equilibrium position.
• This is why, in the exercise, even when the amount of \( \mathrm{C}(\mathrm{s}) \) changes, the \( K_c \) remains the same.

This principle simplifies equilibrium calculations since we only focus on the aqueous or gaseous components that truly impact the equilibrium state.