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Chapter 15: Problem 69

(a) Is the dissociation of fluorine molecules into atomic fluorine, \(F_{2}(g) \rightleftharpoons 2 \mathrm{F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{K},\) does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by 100 \(\mathrm{K}\) , does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

Short Answer

Expert verified
(a) The dissociation of fluorine molecules into atomic fluorine is an endothermic process. (b) If the temperature is raised by 100 K, the equilibrium constant for this reaction increases. (c) If the temperature is raised by 100 K, the forward rate constant \(k_f\) increases by a larger amount than the reverse rate constant \(k_r\).

Step by step solution

01

(a) Exothermic or Endothermic process?

Recall that an exothermic process releases heat, while an endothermic process absorbs heat. In the dissociation of fluorine molecules into atomic fluorine, the bond between two fluorine atoms is broken. Breaking a bond requires energy, which means that energy is absorbed in this process. Therefore, the given reaction is an endothermic process.
02

(b) Effect of raising temperature on equilibrium constant

For an endothermic reaction, increasing temperature will shift the equilibrium towards the products side. The reason is that the heat absorbed by the reaction can be thought of as a reactant. When the temperature is increased, the "reactant" is favored, causing the reaction to shift towards the side with more products. Consequently, the equilibrium constant K for the reaction will increase.
03

(c) Effect of raising temperature on the forward and reverse rate constants

Raising the temperature generally increases both the forward and reverse rate constants. However, since this reaction is endothermic, the forward reaction, which is the dissociation of fluorine molecules into atomic fluorine, has a larger activation energy than the reverse reaction. When the temperature is raised by 100 K, the forward rate constant \(k_f\) will increase by a larger amount than the reverse rate constant \(k_r\).

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is central to grasping why reactions do not always proceed to full completion where all reactants are converted into products. At equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time.

This state of balance doesn't mean the reactions have stopped; rather, they continue to take place at the same rate in both directions. It's a dynamic process, where individual molecules keep reacting, but the overall effect is one of stasis due to the equal rates of the forward and backward reactions.
Le Chatelier's Principle
Le Chatelier's principle is a critical concept that predicts how a system at equilibrium will respond to changes in concentration, temperature, and pressure. It essentially states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change.

For example, as seen in the solution to the exercise, increasing the temperature of an endothermic reaction will shift the equilibrium position towards the products, as the system adapts to absorb the excess heat. This principle helps chemists control the outcomes of reactions to favor the desired product.
Activation Energy
Activation energy is the minimum amount of energy required to initiate a chemical reaction. It can be likened to the energy needed to climb a hill before rolling down the other side. This concept is pivotal when considering reaction rates because molecules need enough energy to reach the transition state where bonds can break and new bonds can form.

An endothermic reaction, such as the dissociation of fluorine molecules, typically requires a substantial input of energy to overcome the activation energy barrier compared to exothermic reactions where energy is released.
Equilibrium Constant
The equilibrium constant, denoted as K, is a number that provides the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their coefficients in the balanced equation. It's a snapshot of the position of equilibrium.

When K is high, it indicates a greater concentration of products and that the equilibrium lies to the right. Conversely, a low K means more reactants are present at equilibrium, and the position lies to the left. Changes in temperature can affect K significantly, especially for reactions sensitive to heat, as in endothermic processes.
Temperature Effects on Reaction Rates
Temperature is a vital factor affecting reaction rates. An increase in temperature generally causes molecules to move faster, increasing the frequency of collision and the energy at which they collide. This leads to a higher likelihood of surpassing the activation energy barrier.

In the case of an endothermic reaction, raising the temperature provides more energy to the reacting molecules, thus increasing the rate of the forward reaction disproportionately compared to the reverse. As observed in the exercise, this is reflected by a larger increase in the forward rate constant, denoted by k_f, compared to the reverse rate constant, k_r, when the temperature is increased by 100 K.

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2} :\) $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use thermochemical data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) 2 \(\mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\) (f) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (g) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C},\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} M^{-1} \mathrm{s}^{-1}\) and \(9.3 \times 10^{10} M^{-1} \mathrm{s}^{-1}\) respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium?

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?

Methane, \(\mathrm{CH}_{4},\) reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g) .\) At \(630 \mathrm{K}, K_{p}\) for this reaction is \(2.26 \times 10^{-4} .\) A reaction was set up at 630 \(\mathrm{K}\) with initial partial pressures of methane of 105.1 torr and of 7.96 torr for \(\mathrm{I}_{2}\) . Calculate the pressures, in torr, of all reactants and products at equilibrium.
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