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Ozone, \(\mathrm{O}_{3}\), decomposes to molecular oxygen in the stratosphere according to the reaction \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)\). Would an increase in pressure favor the formation of ozone or of oxygen?

Short Answer

Expert verified
An increase in pressure will favor the formation of ozone.

Step by step solution

01

Understand the Chemical Reaction

The decomposition of ozone into molecular oxygen is given by the reaction \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)\). This indicates that two moles of ozone decompose to produce three moles of oxygen gas.
02

Apply Le Chatelier's Principle

According to Le Chatelier's Principle, if an external pressure is applied to a system at equilibrium, the equilibrium will shift to counteract the pressure change. This means the reaction will favor the side with fewer moles of gas.
03

Count Moles of Gas on Both Sides

Count the total number of moles of gas on each side of the equation: \(2 \mathrm{O}_{3}(g)\) means there are 2 moles of gas reactants, and \(3 \mathrm{O}_{2}(g)\) means there are 3 moles of gas products.
04

Determine Direction of Equilibrium Shift

With an increase in pressure, the equilibrium will shift towards the side with fewer gas moles to reduce pressure. In this reaction, the reactant side (\(\mathrm{O}_{3}\)) has fewer moles (2) than the product side (\(\mathrm{O}_{2}\)), which has 3 moles.
05

Conclusion on the Effect of Pressure

An increase in pressure will favor the formation of ozone (\(\mathrm{O}_{3}\)) because the side with fewer moles of gas is the reactant side, which helps lower the pressure.

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Key Concepts

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

Equilibrium
Equilibrium in chemistry refers to a state where the forward and reverse reactions occur at the same rate. This means that the concentrations of reactants and products remain constant over time. It's important to understand that equilibrium does not mean the amounts are equal, but rather that their rates of production and consumption are balanced.
The concept of equilibrium is crucial in predicting how a chemical system will respond to changes. Whether involving gases, liquids, or solids, each system seeks a state of equilibrium where any disturbance will be counteracted. This is where Le Chatelier's Principle comes into play, helping predict shifts in equilibrium when external conditions such as pressure, temperature, or concentration change.
  • Static equilibrium: No net change occurs over time, but microscopic changes can still happen.
  • Dynamic equilibrium: Both reactions continue to happen, but at an equal rate, leading to no overall change.
Chemical Reaction
A chemical reaction is a process that leads to the transformation of one set of chemical substances to another. It involves breaking and forming bonds between atoms. During reactions, reactants are converted to products. The reaction between ozone ( O_3) and oxygen ( O_2) is a classic example of decomposition.
In the decomposition reaction:
  • The reactants are 2 moles of ozone (O3).
  • The products are 3 moles of molecular oxygen (O2).
This reaction highlights how a change in the number of moles on each side of an equation can be significant, especially under varying conditions like pressure.
Understanding the nature of chemical reactions helps predict how chemical systems will behave. When discussing reactions at equilibrium, it is essential to count moles for both sides, as they often determine how a system responds to external changes.
Pressure Effect on Equilibrium
Pressure can significantly affect a gaseous equilibrium. Le Chatelier's Principle helps predict the direction in which an equilibrium will shift in response to an external pressure change.
When pressure increases, systems will shift the equilibrium position to reduce that pressure. This usually means favoring the side of the reaction with fewer moles of gas. In the ozone decomposition reaction:
  • Left side (reactants): 2 moles of gas (O3)
  • Right side (products): 3 moles of gas (O2)
Hence, if pressure is increased, the system will shift towards the side with fewer moles, which in this case is the reactant side (O3).
This change helps minimize the effect of added pressure and brings the system back towards equilibrium. Thus, understanding pressure effects helps in controlling and optimizing reactions in industrial and laboratory settings.

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Most popular questions from this chapter

Assume that the equilibrium constant for the dissociation of molecular bromine, \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\), at 800 K is \(K_{c}=5.4 \times 10^{-3}\). (a) Which species predominates at equilibrium, \(\mathrm{Br}_{2}\) or Br, assuming that the concentration of \(\mathrm{Br}_{2}\) is larger than \(5.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?(\mathbf{b})\) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger numeric value of the rate constant, the forward or the reverse reaction?

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$ 2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\operatorname{Br}_{2}(g) $$ An equilibrium mixture in a 5.00-L vessel at \(100^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of \(\mathrm{NOBr}, 3.08 \mathrm{~g}\) of \(\mathrm{NO},\) and \(4.19 \mathrm{~g}\) of \(\mathrm{Br}_{2}\). (a) Calculate \(K_{c}\). (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of \(\mathrm{NOBr}\) ?

Which of the following statements are true and which are false? (a) For the reaction \(2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{~B}(g) K_{c}\) and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

Consider the equilibrium $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{array}{lr} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c}=2.0 \\ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c}=2.1 \times 10^{30} \end{array} $$

For the equilibrium $$ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) $$ \(K_{p}=5.27\) at \(60^{\circ} \mathrm{C} .(\mathbf{a})\) Calculate \(K_{c} .(\mathbf{b})\) After \(3.00 \mathrm{~g}\) of solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed 1.500-L vessel at \(60^{\circ} \mathrm{C}\), the vessel is charged with \(0.0500 \mathrm{~g}\) of \(\mathrm{BCl}_{3}(g)\). What is the equilibrium concentration of \(\mathrm{PH}_{3}\) ?

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