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Chapter 14: Problem 49

The following pictures represent the initial and equilibrium states for the exothermic reaction of solid A (red) with gaseous \(\mathrm{B}_{2}\) (blue) to give gaseous AB: (a) Write a balanced chemical equation for the reaction. (b) Will the number of AB molecules in the equilibrium mixture increase, decrease, or remain the same after each of the following changes? Explain. (1) Increasing the partial pressure of \(\mathrm{B}_{2}\) (2) Adding more solid \(A\) (3) Increasing the volume (4) Increasing the temperature

Short Answer

Expert verified
(a) \(\mathrm{A} + \mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\). (b) AB increases after (1), unaffected by (2) and (3), and decreases after (4).

Step by step solution

01

Write the Balanced Equation

To write a balanced chemical equation, identify the reactants and products. Here, solid A (red) reacts with gaseous \(\mathrm{B}_{2}\) (blue) to form gaseous AB. Thus, the balanced chemical equation is: \[ \mathrm{A} + \mathrm{B}_{2} \rightarrow 2 \mathrm{AB} \]
02

Effect of Increasing Partial Pressure of \(\mathrm{B}_{2}\)

Increasing the partial pressure of \(\mathrm{B}_{2}\) will shift the equilibrium to the right. According to Le Chatelier's principle, increasing the concentration of a reactant will drive the reaction toward forming more products. Thus, the number of AB molecules will increase.
03

Adding More Solid A

Solid \(A\) does not affect the equilibrium position because it is not in the gaseous phase. Therefore, adding more solid \(A\) will not change the number of AB molecules in the equilibrium mixture.
04

Increasing the Volume

Increasing the volume of the container decreases the pressure. The system will shift the equilibrium toward the side with more moles of gas to counteract this change. Since \(\mathrm{B}_{2}\) and \(2\mathrm{AB}\) both have 2 moles of gas, the reaction is not affected by a change in volume, assuming the amount of gases remains the same.
05

Increasing the Temperature

For an exothermic reaction, increasing the temperature shifts the equilibrium to the left (toward the reactants) to absorb the added heat. Therefore, the number of AB molecules in the equilibrium mixture will decrease.

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Key Concepts

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

Le Chatelier's Principle
Understanding Le Chatelier's Principle helps us predict how a chemical equilibrium will respond to changes in conditions. Imagine a seesaw that represents the equilibrium. When you change something like concentration, pressure, or temperature, the system will "tilt" to restore balance. Here鈥檚 how it works in simpler terms:

  • If you increase the concentration of a reactant, the reaction will "shift" to make more products, tipping the seesaw to the right.
  • If you increase the pressure by adding more gas molecules, the equilibrium will adjust by favoring fewer gas moles on one side to decrease the pressure.
  • When the temperature changes, the equation reacts as if it鈥檚 trying to "undo" this energy change.
For example, in our exercise, increasing the partial pressure of \(\mathrm{B}_2\) would cause more AB molecules to form as the reaction shifts right.
Conversely, increasing the temperature in an exothermic reaction shifts it left because the reaction "wants" to reduce the heat by forming more reactants.
Exothermic Reaction
Exothermic reactions release heat energy as they proceed. You can think of them like a heater; when they occur, they give off heat to the surroundings. During these reactions, the sum of bond energies of the products is lower than that of the reactants, resulting in the release of energy.

  • An increase in temperature will favor the reverse reaction, consuming heat to move back toward the reactants.
  • A decrease in temperature makes the forward reaction favorable, leading to more heat being produced.
In the given example, the reaction between solid A and \(\mathrm{B}_2\) forms gaseous AB in an exothermic process. Increasing temperature, therefore, counters the forward direction, reducing the amount of AB at equilibrium, as the system works to "undo" this added heat. It uses the energy to drive the reaction in the opposite direction.
Balanced Chemical Equation
A balanced chemical equation ensures that the same number of each type of atom is present on both sides of the equation, adhering to the law of conservation of mass. In simpler terms, what you start with (reactants) must equal what you end up with (products) in terms of atom count.

In the exercise, reactants are solid A and gaseous \(B_2\), leading to the product gaseous AB. The equation is balanced when it reads: \(A + B_2 \rightarrow 2AB\).

  • Each side of the equation has the same number of A and B atoms, maintaining balance.
  • The coefficients (numbers in front) in the equation ensure that molecules are using the correct proportions.
This balance is crucial for correctly understanding how much of each reactant is needed and how much product will form, allowing predictions of the reaction鈥檚 behavior under different conditions.

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Most popular questions from this chapter

At \(500^{\circ} \mathrm{C}, \mathrm{F}_{2}\) gas is stable and does not dissociate, but at \(840^{\circ} \mathrm{C}\), some dissociation occurs: \(\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{~F}(g) .\) A flask filled with \(0.600\) atm of \(\mathrm{F}_{2}\) at \(500^{\circ} \mathrm{C}\) was heated to \(840^{\circ} \mathrm{C}\), and the pressure at equilibrium was measured to be \(0.984\) atm. What is the equilibrium constant \(K_{\mathrm{p}}\) for the dissociation of \(\mathrm{F}_{2}\) gas at \(840^{\circ} \mathrm{C} ?\)

For the reaction \(\mathrm{A}_{2}+2 \mathrm{~B} \rightleftharpoons 2 \mathrm{AB}\), the rate of the forward reaction is \(18 \mathrm{M} / \mathrm{s}\) and the rate of the reverse reaction is \(12 \mathrm{M} / \mathrm{s}\). The reaction is not at equilibrium. Will the reaction proceed in the forward or reverse direction to attain equilibrium?

For the decomposition reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\) \(\mathrm{Cl}_{2}(g), K_{\mathrm{p}}=381\) at \(600 \mathrm{~K}\) and \(K_{c}=46.9\) at \(700 \mathrm{~K}\) (a) Is the reaction endothermic or exothermic? Explain. Does your answer agree with what you would predict based on bond energies? (b) If \(1.25 \mathrm{~g}\) of \(\mathrm{PCl}_{5}\) is introduced into an evacuated \(0.500 \mathrm{~L}\) flask at \(700 \mathrm{~K}\) and the decomposition reaction is allowed to reach equilibrium, what percent of the \(\mathrm{PCl}_{5}\) will decompose and what will be the total pressure in the flask? (c) Write electron-dot structures for \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\), and indicate whether these molecules have a dipole moment. Explain.

At \(1000 \mathrm{~K}, K_{\mathrm{p}}=2.1 \times 10^{6}\) and \(\Delta H^{\circ}=-107.7 \mathrm{~kJ}\) for the reac- tion \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) (a) A \(0.974\) mol quantity of \(\mathrm{Br}_{2}\) is added to a \(1.00 \mathrm{~L}\) reaction vessel that contains \(1.22 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) gas at \(1000 \mathrm{~K}\). What are the partial pressures of \(\mathrm{H}_{2}, \mathrm{Br}_{2}\), and \(\mathrm{HBr}\) at equilibrium? (b) For the equilibrium in part (a), each of the following changes will increase the equilibrium partial pressure of \(\mathrm{HBr}\). Choose the change that will cause the greatest increase in the pressure of \(\mathrm{HBr}\), and explain your choice. (i) Adding \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) (ii) Adding \(0.10 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) (iii) Decreasing the temperature to \(700 \mathrm{~K}\).

Will the concentration of \(\mathrm{NO}_{2}\) increase, decrease, or remain the same when the equilibrium $$ \mathrm{NO}_{2} \mathrm{Cl}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NOCl}(g)+\mathrm{NO}_{2}(g) $$ is disturbed by the following changes? (a) Adding \(\mathrm{NOCl}\) (b) Adding \(\mathrm{NO}\) (c) Removing \(\mathrm{NO}\) (d) Adding \(\mathrm{NO}_{2} \mathrm{Cl} ;\) also account for the change using the reaction quotient \(Q_{c}\)
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