Problem 17
Question
What is the equilibrium constant \(\left(K_{c}\right)\) expression for the following reversible reaction? $$ 2 \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons 2 \mathrm{C}(g) $$
Step-by-Step Solution
Verified Answer
Answer: The equilibrium constant expression, \(K_c\), for the given reaction is: \(K_c = \frac{[\mathrm{C}]^{2}}{[\mathrm{A}]^{2}[\mathrm{B}]}\).
1Step 1: Write down the balanced equation
The balanced equation for the reversible reaction is given by:
$$
2 \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons 2 \mathrm{C}(g)
$$
2Step 2: Understand the equilibrium constant expression
The equilibrium constant, \(K_c\), is the ratio of the concentrations of the products raised to the power of their stoichiometric coefficients, divided by the concentrations of the reactants raised to the power of their stoichiometric coefficients. This can be generally represented as:
$$
K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}
$$
3Step 3: Write the equilibrium constant expression for the given reaction
The equilibrium constant expression for the provided reversible reaction would be:
$$
K_c = \frac{[\mathrm{C}]^{2}}{[\mathrm{A}]^{2}[\mathrm{B}]}
$$
The equilibrium constant expression, \(K_c\), for the given reaction is:
$$
K_c = \frac{[\mathrm{C}]^{2}}{[\mathrm{A}]^{2}[\mathrm{B}]}
$$
Key Concepts
Chemical EquilibriumReaction QuotientStoichiometry
Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, the concentrations of the reactants and products remain constant over time, but are not necessarily equal. This dynamic state is crucial as it determines the proportion of reactants and products present in a reaction mixture at any given time.
When you write an equation such as \(2 \text{A}(g) + \text{B}(g) \rightleftharpoons 2 \text{C}(g)\), it suggests that \text{A} and \text{B} react to form \text{C}, but \text{C} can also decompose back into \text{A} and \text{B}. At equilibrium, the amount of \text{A}, \text{B}, and \text{C} stops changing. It's important to realize that a reaction at equilibrium is not 'stopped'—the reactants and products continue to interconvert, but at a rate where their amounts do not change.
To grasp chemical equilibrium thoroughly, one should also understand the law of mass action, which states that for a reversible reaction at equilibrium and at a constant temperature, a certain ratio of the concentration of products to reactants, each raised to the power of their respective stoichiometric coefficients, remains constant. This constant is the equilibrium constant, \(K_c\).
When you write an equation such as \(2 \text{A}(g) + \text{B}(g) \rightleftharpoons 2 \text{C}(g)\), it suggests that \text{A} and \text{B} react to form \text{C}, but \text{C} can also decompose back into \text{A} and \text{B}. At equilibrium, the amount of \text{A}, \text{B}, and \text{C} stops changing. It's important to realize that a reaction at equilibrium is not 'stopped'—the reactants and products continue to interconvert, but at a rate where their amounts do not change.
To grasp chemical equilibrium thoroughly, one should also understand the law of mass action, which states that for a reversible reaction at equilibrium and at a constant temperature, a certain ratio of the concentration of products to reactants, each raised to the power of their respective stoichiometric coefficients, remains constant. This constant is the equilibrium constant, \(K_c\).
Reaction Quotient
The reaction quotient, denoted as \(Q_c\), is a concept that plays a pivotal role in comparing the current state of a reaction to its equilibrium position. You can think of it as a 'snapshot' of the system at a particular moment in time. The reaction quotient is calculated using the same expression as the equilibrium constant, \(K_c\):
$$Q_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}$$
However, while \(K_c\) uses the concentrations of reactants and products at equilibrium, \(Q_c\) uses their concentrations at any point in the reaction. Comparing \(Q_c\) to \(K_c\) can tell you which direction the reaction will shift to reach equilibrium:
$$Q_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}$$
However, while \(K_c\) uses the concentrations of reactants and products at equilibrium, \(Q_c\) uses their concentrations at any point in the reaction. Comparing \(Q_c\) to \(K_c\) can tell you which direction the reaction will shift to reach equilibrium:
- If \(Q_c < K_c\), the reaction will proceed in the forward direction to produce more products.
- If \(Q_c = K_c\), the system is already at equilibrium.
- If \(Q_c > K_c\), the reaction will shift in the reverse direction to produce more reactants.
Stoichiometry
Stoichiometry is the study of the quantitative relationships between the substances involved in chemical reactions. It is based on the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. Hence, stoichiometry involves using balanced chemical equations to calculate the mass, mole, or volume relationships of reactants and products.
For example, in the balanced reaction of \(2 \text{A}(g) + \text{B}(g) \rightleftharpoons 2 \text{C}(g)\), the stoichiometric coefficients are the numbers in front of each species. These coefficients show that 2 moles of \(\text{A}\) react with 1 mole of \(\text{B}\) to produce 2 moles of \(\text{C}\). Not only do these coefficients govern the ratios in which substances react and are produced, but they are also exponentiated in the equilibrium constant expression to reflect the impact of each substance's concentration on the position of equilibrium.
Therefore, when we write the equilibrium constant expression based on this stoichiometry, such as \(K_c = \frac{[\text{C}]^{2}}{[\text{A}]^{2}[\text{B}]}\), the squared terms indicate the direct relationship of the change in concentration of \(\text{A}\) and \(\text{C}\) on the equilibrium constant value. Proper understanding of stoichiometry is vital, as it forms the foundation for analyzing and predicting the outcomes of chemical reactions.
For example, in the balanced reaction of \(2 \text{A}(g) + \text{B}(g) \rightleftharpoons 2 \text{C}(g)\), the stoichiometric coefficients are the numbers in front of each species. These coefficients show that 2 moles of \(\text{A}\) react with 1 mole of \(\text{B}\) to produce 2 moles of \(\text{C}\). Not only do these coefficients govern the ratios in which substances react and are produced, but they are also exponentiated in the equilibrium constant expression to reflect the impact of each substance's concentration on the position of equilibrium.
Therefore, when we write the equilibrium constant expression based on this stoichiometry, such as \(K_c = \frac{[\text{C}]^{2}}{[\text{A}]^{2}[\text{B}]}\), the squared terms indicate the direct relationship of the change in concentration of \(\text{A}\) and \(\text{C}\) on the equilibrium constant value. Proper understanding of stoichiometry is vital, as it forms the foundation for analyzing and predicting the outcomes of chemical reactions.
Other exercises in this chapter
Problem 15
Under what conditions are the numerical values of \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}}\) equal?
View solution Problem 16
At \(298 \mathrm{K},\) is \(K_{\mathrm{p}}\) greater than or less than \(K_{\mathrm{c}}\) if there is a net increase in the number of moles of gas in the reacti
View solution Problem 18
What is the equilibrium constant \(\left(K_{\mathrm{p}}\right)\) expression for the following reversible reaction? $$ 2 \mathrm{A}(g)+3 \mathrm{B}(g) \rightleft
View solution Problem 19
Write \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}}\) expressions for the following gas- phase reactions. a. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \rig
View solution