Problem 186
Question
One of the following equilibria is not affected by change in volume of the flask: (a) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g}) \mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) (d) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\)
Step-by-Step Solution
Verified Answer
Equilibrium (c) is not affected by a change in volume as it has \( \Delta n = 0 \).
1Step 1: Understand the Problem
We need to identify which equilibrium is not affected by a change in the volume of the flask. For a change in volume to affect the equilibrium, the number of moles of gas on each side of the reaction must be different.
2Step 2: Calculate Change in Moles for Each Reaction
For each reaction, calculate the change in the number of moles of gas, \( \Delta n \), between the reactants and products.(a) \( \mathrm{PCl}_{5} \rightarrow \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \); \( \Delta n = 2 - 1 = 1 \)(b) \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \); \( \Delta n = 2 - 4 = -2 \)(c) \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightarrow 2 \mathrm{NO} \); \( \Delta n = 2 - 2 = 0 \)(d) \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{SO}_{2} + \mathrm{Cl}_{2} \); \( \Delta n = 2 - 1 = 1 \).
3Step 3: Identify Equilibrium Not Affected by Volume Change
A change in volume affects equilibria where the number of moles of gas changes from reactants to products. From our calculations, reaction (c) has \( \Delta n = 0 \), meaning the number of gas moles on each side is equal. Therefore, the equilibrium of this reaction will not shift with a change in volume.
Key Concepts
Le Chatelier's PrincipleGas Laws in ChemistryReaction Stoichiometry
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemical equilibrium that helps predict how a system at equilibrium responds to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts itself to counteract the change and restore a new equilibrium.When applied to changes in volume, the principle suggests that a shift will occur to decrease or increase the pressure depending on the change in volume:- If volume decreases, pressure increases, shifting the reaction toward the side with fewer moles of gas.- Conversely, if volume increases, pressure decreases, and the equilibrium shifts toward the side with more moles of gas.However, as illustrated in the original exercise, when the number of moles of gases on both sides of the reaction is equal (as seen in reaction \( c: \ \mathrm{N}_{2} + \mathrm{O}_{2} \rightarrow 2 \mathrm{NO} \)), there is no change in pressure, and thus the equilibrium is unaffected by volume changes.
Gas Laws in Chemistry
Gas laws in chemistry describe the behavior of gases in response to changes in pressure, volume, temperature, and moles. The ideal gas law is commonly used, expressed as \( PV = nRT \), where:
- \( P \) is pressure.
- \( V \) is volume.
- \( n \) is the number of moles.
- \( R \) is the gas constant.
- \( T \) is temperature.
Reaction Stoichiometry
Reaction stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. Understanding stoichiometry allows us to predict the amounts of substances consumed and produced in reactions, ensuring that mass and energy are conserved.For chemical equilibria, knowing the stoichiometric coefficients of each substance is essential to determine changes in mole numbers (9 n) as the reaction progresses. This concept ties back to volume changes, as reactions where 9 n is zero indicate no change in the number of moles of gases between reactants and products:- A reaction like \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightarrow 2 \mathrm{NO} \), with 9 n = 0, implies that changes in volume will not affect equilibrium, as both sides have an equal number of gas moles.- In contrast, reactions with non-zero 9 n will experience shifts in equilibrium upon volume changes due to unequal gas mole numbers.Accurate stoichiometry calculations are vital for proper application of principles like Le Chatelier's Principle and understanding gas behavior under varying conditions.
Other exercises in this chapter
Problem 184
For the equilibrium \(\mathrm{AB}(\mathrm{g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})\), at a given temperature \(\frac{1}{3}\) rd of \
View solution Problem 185
For the following reaction in gaseous phase \(\mathrm{CO}+1 / 2 \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\) \(\mathrm{K}_{c} / \mathrm{K}_{\mathrm{p}}\)
View solution Problem 187
At \(700 \mathrm{~K}\), the equilibrium constant \(\mathrm{K}_{\text {s }}\) for the reaction \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_
View solution Problem 188
Consider the reaction equilibrium, \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \math
View solution