Problem 100
Question
In which of the following reactions, equilibrium is independent of pressure? (a) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \Delta \mathrm{H}=+\mathrm{ve}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (c) \(3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=+\mathrm{ve}\)
Step-by-Step Solution
Verified Answer
Equilibrium is independent of pressure in reaction (a).
1Step 1: Understand Equilibrium and Pressure Independence
Equilibrium of a gas phase reaction is independent of pressure when the number of moles of gaseous reactants is equal to the number of moles of gaseous products. In this scenario, a change in pressure doesn't affect the equilibrium position.
2Step 2: Analyze Reaction (a)
In reaction (a), \ \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} \) \ the number of moles on both sides is 2 (1 mole of \(\mathrm{N_2}\) + 1 mole of \(\mathrm{O_2}\) on the reactants side and 2 moles of \(\mathrm{NO}\) on the products side). Since they are equal, equilibrium is independent of pressure.
3Step 3: Analyze Reaction (b)
In reaction (b), \ \( 2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} \) \ on the reactant side, there are 3 moles (2 moles of \(\mathrm{SO_2}\) + 1 mole of \(\mathrm{O_2}\)), but the product side has 2 moles of \(\mathrm{SO_3}\). The moles are not equal, so equilibrium is not independent of pressure.
4Step 4: Analyze Reaction (c)
In reaction (c), \ \( 3 \mathrm{H}_{2} + \mathrm{N}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \) \ there are 4 moles of reactants (3 moles of \(\mathrm{H_2}\) + 1 mole of \(\mathrm{N_2}\)) and 2 moles of products. The moles are not equal, so equilibrium is not independent of pressure.
5Step 5: Analyze Reaction (d)
In reaction (d), \ \( \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \) \ 1 mole of reactant results in 2 moles of products. The moles are not equal, so equilibrium is not independent of pressure.
6Step 6: Conclusion on Pressure Independence
Among the given reactions, only in reaction (a), the number of moles of reactants equals the number of moles of products. This makes its equilibrium position independent of pressure changes.
Key Concepts
Pressure Independence in Chemical EquilibriumMole Balance in Chemical ReactionsGas Phase Reactions and their Equilibrium
Pressure Independence in Chemical Equilibrium
Chemical equilibrium plays a vital role in determining the behavior of reactions, especially in the gas phase. An interesting aspect of chemical equilibrium is pressure independence. This occurs when a gas phase reaction's equilibrium remains unchanged even with pressure variations. The key to identifying pressure independence lies in examining the moles of gaseous substances involved.
- If the total number of moles of gaseous reactants equals the total number of moles of gaseous products, the system's equilibrium state is unaffected by pressure changes.
- This is because pressure changes affect both sides of the reaction equally when the moles are balanced, maintaining the equilibrium constant value.
Mole Balance in Chemical Reactions
Mole balance is a fundamental concept in chemistry that allows us to grasp the stoichiometry of reactions. In gas phase reactions, we often use mole balances to determine whether a reaction's equilibrium is pressure independent.
- Consider the equation: \( a \text{A} + b \text{B} \rightleftharpoons c \text{C} + d \text{D} \), where \( a, b, c, \) and \( d \) represent the coefficients in a balanced chemical equation.
- A mole balance involves comparing the total moles of reactants (\( a + b \)) and products (\( c + d \)).
- If these totals are equal, then the pressure doesn't affect the equilibrium position of the reaction.
Gas Phase Reactions and their Equilibrium
Gas phase reactions are unique due to the role of gases, which are more compressible and sensitive to pressure changes. This sensitivity impacts the equilibrium state, making it a fascinating area of study in chemistry.
- In the context of gas phase reactions, the equilibrium constant is defined based on the partial pressures of the gases involved.
- Changing the total pressure can shift the equilibrium position, unless the number of moles of reactants and products are equal, ensuring pressure independence.
- Another factor to consider is Le Chatelier's Principle, which states that a system at equilibrium will adjust to counteract any imposed change, such as pressure alteration.
Other exercises in this chapter
Problem 98
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For which of the following reaction, \(\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{c}}\) ? (a) \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\m
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In an equilibrium reaction, \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\), the partial pres-
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The equilibrium constant of mutarotation of \(\alpha\)-D-glucose to \(\beta\)-D-glucose is \(1.8\). What per cent of the \(\alpha\)-form remains under equilibri
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