Question

At 2500 . \(K\), the equilibrium constant is \(K_{c}=20\). for the reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}(\mathrm{g})\). An analysis of a reaction vessel at 2500 . \(\mathrm{K}\) revealed the presence of \(0.18 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{Cl}_{2}\), \(0.31 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~F}_{2}\), and \(0.92 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{ClF}\). Will \(\mathrm{ClF}\) tend to form or to decompose as the reaction proceeds toward equilibrium?

Short answer

The reaction will proceed in the forward direction, forming more \(\mathrm{ClF}\) as it moves toward equilibrium.

Step-by-step solution

Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the reaction \(\mathrm{Cl}_2(\mathrm{g}) + \mathrm{F}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{ClF}(\mathrm{g})\) is given by \[K_c = \frac{[\mathrm{ClF}]^2}{[\mathrm{Cl}_2][\mathrm{F}_2]}\] where brackets denote the molar concentration of the gases.

Calculate the Reaction Quotient, Q

Calculate the reaction quotient, Q, using the initial concentrations provided to determine the direction the reaction will move to reach equilibrium. \[Q = \frac{[\mathrm{ClF}]^2_{\text{initial}}}{[\mathrm{Cl}_2]_{\text{initial}}[\mathrm{F}_2]_{\text{initial}}} = \frac{(0.92)^2}{(0.18)(0.31)}\]

Compare Q with Kc

Compare the calculated Q value with the given Kc value to determine if the reaction will proceed forward or reverse to achieve equilibrium. If Q < Kc, the forward reaction is favored and more \(\mathrm{ClF}\) will form. If Q > Kc, the reverse reaction is favored and \(\mathrm{ClF}\) will decompose.

Perform the Calculation

Perform the calculation to find Q: \[Q = \frac{(0.92)^2}{(0.18)(0.31)} = \frac{0.8464}{0.0558} \approx 15.17\]

Determine the Direction of the Reaction

Since Q (15.17) < Kc (20), the reaction will proceed in the forward direction, causing more \(\mathrm{ClF}\) to form.

Key concepts

Equilibrium Constant

In chemistry, understanding the nature of chemical reactions is crucial, especially when it comes to reactions that can reach a state of balance or equilibrium. The equilibrium constant, denoted as \(K\), provides a way to quantify this balance point in a reversible chemical reaction. It is a value that reflects the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium, with each concentration raised to the power of its coefficient in the balanced equation.

For the reaction \(\mathrm{Cl}_{2}(\mathrm{g}) + \mathrm{F}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{ClF}(\mathrm{g})\), the equilibrium expression is written as:
\[K_c = \frac{[\mathrm{ClF}]^2}{[\mathrm{Cl}_2][\mathrm{F}_2]}\]
This tells us how the system will behave at equilibrium at a given temperature (2500 K in our exercise). A high \(K_c\) value indicates a greater concentration of products at equilibrium, whereas a low value favors the reactants.

Reaction Quotient

The reaction quotient, or \(Q\), is a measure that is used to predict the direction in which a reaction must shift to reach equilibrium. It has the same form as the equilibrium constant expression but is calculated using the initial concentrations before equilibrium is reached. You can think of \(Q\) as a 'snapshot' of a reaction’s status at a point in time.

To calculate \(Q\) for our given reaction, we use:
\[Q = \frac{[\mathrm{ClF}]^2_{\text{initial}}}{[\mathrm{Cl}_2]_{\text{initial}}[\mathrm{F}_2]_{\text{initial}}}\]
Comparing \(Q\) to \(K_c\) helps us understand whether the reaction is at equilibrium (\(Q = K_c\)) or which direction it needs to go to get there. If \(Q < K_c\), more product will form as the reaction proceeds. If \(Q > K_c\), we can expect the reaction to go in the reverse direction.

Chemical Reaction

A chemical reaction involves the conversion of one or more substances, the reactants, into one or more different substances, known as the products. Reactions can either go to completion, where the reactants are entirely converted into products, or they can be reversible, meaning the products can also revert back to the original reactants.

In the reversible reaction from our exercise, \(\mathrm{Cl}_{2}(g) + \mathrm{F}_{2}(g) \rightleftharpoons 2\mathrm{ClF}(g)\), the reactants, chlorine and fluorine gases, combine to form chlorine monofluoride. Because the given reaction is reversible and occurs at a high temperature (2500 K), it's essential to understand that equilibrium can be reached and maintained, where the rate of the forward reaction equals the rate of the reverse reaction.

Le Chatelier's Principle

Le Chatelier's Principle states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way as to partially counteract the stress and reestablish equilibrium. Stresses include changes in concentration, pressure, volume, or temperature.

In the context of our chemical equilibrium exercise, if we were to add more \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{F}_{2}(g)\), according to Le Chatelier's Principle, the system would respond by forming more \(\mathrm{ClF}(g)\) to reduce the stress of the added reactants. Likewise, if the system loses \(\mathrm{ClF}(g)\) through some disturbance, the equilibrium will shift to produce more \(\mathrm{ClF}(g)\) from the reactants \(\mathrm{Cl}_{2}(g)\) and \(\mathrm{F}_{2}(g)\) to compensate. This principle is a fundamental concept to predict how equilibrium will respond to changes and guides chemists in optimizing reactions.