/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 The following reaction is found ... [FREE SOLUTION] | 91Ó°ÊÓ

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The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) Which of the following would cause a reduction in the value for the equilibrium constant? (A) Increasing the amount of \(\mathrm{SO}_{3}\) (B) Reducing the amount of \(\mathrm{O}_{2}\) (C) Raising the temperature (D) Lowering the temperature

Short Answer

Expert verified
Option (C) - Raising the temperature would cause a reduction in the value for the equilibrium constant.

Step by step solution

01

- Analyze Option (A)

Increasing the amount of \(SO_{3}\) would shift the reaction toward the products to regain equilibrium. However, this doesn't impact the equilibrium constant as it is only dependent on temperature, not concentrations.
02

- Analyze Option (B)

Reducing the amount of \(O_{2}\) would also shift the reaction toward the products. But again, the equilibrium constant isn't affected by concentrations, it remains the same.
03

- Analyze Option (C)

Raising the temperature will shift an exothermic reaction toward the reactants, to absorb the extra heat. This would indeed decrease the value of the equilibrium constant K.
04

- Analyze Option (D)

Lowering the temperature will shift an exothermic reaction towards the products, as it will try to generate heat. This would increase the value of the equilibrium constant K, not decrease it.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction, leading to the concentrations of reactants and products remaining constant over time. It's important to note that equilibrium does not mean the reactants and products are present in equal amounts but instead that their rates are balanced.

This state can be represented by the equilibrium constant, typically denoted as \(K\). The equilibrium constant is a vital component in understanding how far a reaction proceeds before reaching equilibrium. For the reaction \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g) + 2 \mathrm{SO}_{2}(g)\), the equilibrium constant expresses the ratio of the concentrations of the products to reactants at equilibrium.
  • The value of \(K\) is temperature-dependent.
  • If \(K\) is large, a significant amount of reactants has converted to products at equilibrium.
  • A small \(K\) value indicates that not much reactant has been converted to products.
Le Chatelier's Principle
Le Chatelier's Principle provides insight into how a system at equilibrium responds to external changes. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts itself to partially counteract the effect of the disturbance and a new equilibrium is established.
  • Adding more reactants or removing products will shift the equilibrium towards the products.
  • Conversely, removing reactants or adding more products will shift the equilibrium towards the reactants.
  • Temperature changes have a unique effect because they alter the equilibrium constant itself, not just the position of the equilibrium.
Understanding how equilibrium shifts helps to predict the outcome of various scenarios. However, as seen in this reaction, only temperature changes can genuinely affect the equilibrium constant's value.
Exothermic Reactions
Exothermic reactions are chemical reactions that release heat. The reaction in question, \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g) + 2 \mathrm{SO}_{2}(g)\), is exothermic, as evident from its negative enthalpy change (\(\Delta H = -198 \mathrm{kJ/mol}\)).

For exothermic reactions:
  • Heat is essentially a product of the reaction.
  • Increasing temperature adds heat to the system, shifting the equilibrium towards the reactants to absorb the heat, according to Le Chatelier's Principle.
  • Lowering the temperature favors the formation of more products to replace the lost heat.
The response of the equilibrium to temperature changes is essential to understanding how the equilibrium constant behaves in exothermic reactions. Lowering the temperature benefits product formation and increases \(K\), while raising temperature does the opposite.
Thermodynamics
Thermodynamics plays a crucial role in understanding chemical reactions, including aspects like enthalpy, entropy, and free energy changes.

In our example, the exothermic reaction shows a negative enthalphy change, which reflects the release of energy. Thermodynamics also involves the understanding of how energy and matter interact on a molecular level.
  • Temperature is a critical factor affecting reaction spontaneity and equilibrium constant \(K\).
  • A negative \(\Delta H\) generally indicates that the reaction releases energy (exothermic).
  • According to Gibb's free energy, for a reaction to be spontaneous at constant temperature and pressure, the free energy change \(\Delta G\) must be negative, linking directly to the concepts of enthalpy and entropy.
Through a thermodynamic lens, chemical equilibrium is not just about concentrations but how temperature along with energy variations impact the stability and progression of chemical reactions.

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Most popular questions from this chapter

\(2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{AlCl}_{3}(s)\) The reaction above is not thermodynamically favored under standard conditions, but it becomes thermodynamically favored as the temperature decreases toward absolute zero. Which of the following is true at standard conditions? (A) \(\Delta S\) and \(\Delta H\) are both negative. (B) \(\Delta S\) and \(\Delta H\) are both positive. (C) \(\Delta S\) is negative, and \(\Delta H\) is positive. (D) \(\Delta S\) is positive, and \(\Delta H\) is negative.

Which of the following substances has an asymmetrical molecular structure? (A) \(\mathrm{SF}_{4}\) (B) \(\mathrm{PCl}_{5}\) (C) \(\mathrm{BF}_{3}\) (D) \(\mathrm{CO}_{2}\)

A solution of \(\mathrm{Co}^{2+}\) ions appears red when viewed under white light. Which of the following statements is true about the solution? (A) A spectrophotometer set to the wavelength of red light would read a high absorbance. (B) If the solution is diluted, the amount of light reflected by the solution will decrease. (C) All light with a frequency that is lower than that of red light will be absorbed by it. (D) Electronic transmissions within the solution match the wavelength of red light.

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.

\(2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) During the reaction above, the product yield can be increased by increasing the temperature of the reaction. Why is this effective? (A) The reaction is endothermic; therefore adding heat will shift it to the right. (B) Increasing the temperature increases the speed of the molecules, meaning there will be more collisions between them. (C) The reactants are less massive than the products, and an increase in temperature will cause their kinetic energy to increase more than that of the products. (D) The increase in temperature allows for a higher percentage of molecular collisions to occur with the proper orientation to create the product.

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