At \(2500 \mathrm{~K}, K_{e}=20\) 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 cquilibrium?

Short Answer

Expert verified
ClF will tend to form as the reaction proceeds towards equilibrium because Q (15.17) is less than Ke (20).

Step by step solution

01

Write the Equation for the Reaction Quotient (Q)

To determine the direction in which the reaction will proceed, calculate the reaction quotient, Q. The reaction quotient, Q, has the same form as the equilibrium constant, Keq, but is calculated for conditions that may not be at equilibrium. For the reaction given, the expression for Q is: Q = [ClF]^2 / ([Cl2] * [F2]) where the concentrations of the gaseous species at the given condition will be plugged into the formula.
02

Substitute Given Concentrations into the Q Expression

Substitute the concentrations given into the expression for Q. So, Q will be: Q = (0.92 mol/L)^2 / ((0.18 mol/L) * (0.31 mol/L))
03

Calculate the Reaction Quotient (Q)

Calculate the numerical value of Q using the substituted values: Q = (0.92^2) / (0.18 * 0.31) = (0.8464) / (0.0558) = 15.17
04

Compare Q with the Equilibrium Constant (Ke)

Compare the calculated Q to the given equilibrium constant Ke (Ke = 20) to predict the direction of the reaction. - If Q < Ke, the reaction will proceed forward, forming more products to reach equilibrium. - If Q > Ke, the reaction will proceed in the reverse direction, forming more reactants to reach equilibrium. In this case, since Q (15.17) is less than Ke (20), the reaction will shift towards the right (forward direction), forming more ClF.

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Key Concepts

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

Understanding the Reaction Quotient (Q)
When dealing with chemical reactions, it's crucial to predict the direction in which a reaction will proceed. This is where the reaction quotient, or Q, comes into play. Imagine Q as a snapshot of where the reaction's concentrations stand at a given moment in time, as opposed to at equilibrium.

For a reaction like \( \text{Cl}_2(g) + \text{F}_2(g) \rightleftharpoons 2\text{ClF}(g) \), the reaction quotient would be expressed as \( Q = \frac{[ClF]^2}{[Cl_2][F_2]} \),where the brackets represent the concentrations of the gases. If Q is less than the equilibrium constant, Ke, it indicates that the reaction will shift to the right to form more products. Conversely, if Q is greater than Ke, the reaction will shift to the left, favoring the formation of reactants.
Exploring the Equilibrium Constant (Ke)
The equilibrium constant, Ke, serves as the pivotal point for a chemical reaction in a state of dynamic equilibrium. It is determined at a specific temperature and remains constant regardless of initial concentrations. The value of Ke is a ratio obtained when the reaction has reached equilibrium, representing the proportion of products to reactants.

To illustrate, for the equilibrium \( \text{Cl}_2(g) + \text{F}_2(g) \rightleftharpoons 2\text{ClF}(g) \),Ke is given as 20 at 2500 K. This number tells us the ratio of products to reactants is favored 20 to 1 at that temperature. Knowing the value of Ke allows us to predict the direction in which the reaction will shift to reach equilibrium.
Performing Equilibrium Calculations
Equilibrium calculations involve determining the concentration of reactants and products at equilibrium. Using the given example, one would start with the equilibrium expression and substitute concentrations or partial pressures for gases into the equation. After this, algebraic methods or approximations are used to solve for the unknowns.

It's like piecing together a puzzle — the given concentrations and the Ke provide clues to the final arrangement at equilibrium. Since our calculated Q (15.17) was less than the Ke (20), we'd expect the reaction to proceed forward, creating more ClF until Q equals Ke, and the system is at equilibrium.
Applying Le Chatelier's Principle
Le Chatelier's Principle is like the universe's way of maintaining balance. It states that if a change is imposed on a system at equilibrium, the system adjusts to counteract that change. This could be changes in concentration, pressure, volume, or temperature.

To put this into context with our reaction, if we were to add more \( \text{Cl}_2 \), the system would respond by forming more \( \text{ClF} \) to reduce the added \( \text{Cl}_2 \) concentration and reach a new equilibrium. Understanding this principle helps chemists optimize conditions for chemical processes, ensuring the most efficient pathway to the desired product.

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Most popular questions from this chapter

Calculate the standard reaction free energy of each of the following reactions: (a) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), K=41\) ar \(400 \mathrm{~K}\) (b) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.0 \times 10^{4}\) at \(700 \mathrm{~K}\)

Let \(\alpha\) be the fraction of \(\mathrm{PCl}_{5}\) molecules that have decomposed to \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) in the reaction \(\mathrm{PCl}_{5}(\mathrm{~g})=\) \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) in a constant- volume container, so that the amount of \(\mathrm{PCl}_{5}\) at equilibrium is \(n(1-\alpha)\), where \(n\) is the pressure present initially. Derive an equation for \(K\) in terms of \(\alpha\) and the total pressure \(P\), and solve it for \(\alpha\) in terms of \(P\). Calculate the fraction decomposed at \(556 \mathrm{~K}\), at which temperature \(K=4.96\), and the total pressure is (a) \(0.50\) har; (b) \(1.00\) bar.

Predict whether each of the following equilibria will shift toward products or reactants with a temperature increase. (a) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\), \(\Delta H^{\circ}=+206 \mathrm{~kJ}\) (b) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\), \(\Delta H^{2}=-41 \mathrm{~kJ}\) (c) \(2 \mathrm{SO}_{2}\) (g) \(+\mathrm{O}_{2}\) (g) \(=2 \mathrm{SO}_{3}\) (g), \(\Delta H^{\circ}=-198 \mathrm{~kJ}\)

Calculate the equilibrium constant ar \(25^{\circ} \mathrm{C}\) for cach of the following reactions from data available in Appendix 2A. (a) the combustion of hydrogen: \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) the oxidation of carbon monoxide: \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})\) (c) the decompostion of limestone: \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\)

Consider the equilibrium \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). (a) If the partial pressure of \(\mathrm{CO}_{2}\) is increased, what happens to the partial pressure of \(\mathrm{H}_{2}\) ? (b) If the partial pressure of \(\mathrm{CO}\) is decreased, what happens to the partial pressure of \(\mathrm{CO}_{2}\) ? (c) If the concentration of \(\mathrm{CO}\) is increased, what happens to the concentration of \(\mathrm{H}_{2}\) ? (d) If the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is decreased, what happens to the equilibrium constant for the reaction?

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