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What is the effect of decreasing the pressure of a reaction mixture at equilibrium if the reactant side has fewer moles of gas particles than the product side?

Short Answer

Expert verified
Decreasing the pressure of a reaction mixture at equilibrium will shift the equilibrium towards the product side if the product side has more moles of gas particles.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's Principle states that if an external change is applied to a system at equilibrium, the system adjusts in such a way as to minimize the effect of that change. In the context of gas reactions, if the pressure is changed, the system will shift towards the side with fewer or more moles of gas particles depending on whether the pressure is increased or decreased.
02

Analyze the Initial Conditions

The initial condition given is that the reaction mixture at equilibrium has fewer moles of gas particles on the reactant side than on the product side.
03

Predict the Shift in Equilibrium

When the pressure is decreased, the system will attempt to increase the pressure. According to Le Chatelier's Principle, the system will shift towards the side with more moles of gas particles to occupy more volume and counteract the decrease in pressure. Since the product side has more moles of gas particles, the equilibrium will shift towards the product side.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is essential when analyzing the behavior of reactions. At equilibrium, the rate of the forward reaction (reactants converting to products) equals the rate of the backward reaction (products converting to reactants). This creates a dynamic balance where the concentrations of reactants and products remain constant over time, even though the reactions continue to occur.

Le Chatelier's Principle is crucial in explaining how a system at equilibrium responds to changes in conditions. When a system in equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will adjust to counteract the imposed change and re-establish a new state of balance. This self-adjusting behavior of chemical systems is fundamental to understanding how reactions adapt to the stresses placed upon them.
Reaction Mixture
In a chemical reaction, the reaction mixture comprises both the reactants and the products at various stages of the reaction. Initially, the mixture may contain only reactants, but as the reaction proceeds, products form and their concentration increases. If the reaction is reversible, the products can also revert to reactants, resulting in an equilibrium state where the concentrations of reactants and products reach a constant ratio.

At equilibrium, although the proportions of reactants and products remain steady, the actual mixture is incredibly dynamic. Particles are always reacting, colliding, and converting from reactants to products and vice versa. The particular composition of a reaction mixture, in terms of the number of moles of gaseous reactants and products, has a profound impact on how the equilibrium will respond to changes in physical conditions like pressure.
Pressure Effects on Equilibrium
The pressure effect on equilibrium is an application of Le Chatelier's Principle concerning gaseous equilibria. If a system at equilibrium in a closed container experiences a change in pressure, the equilibrium position will shift in order to counteract this change. For reactions involving gases, the number of moles of gas on each side of the reaction plays a determining role.

When pressure is decreased on a reaction mixture at equilibrium, as in the exercise, the system compensates by shifting the equilibrium to increase pressure. This is achieved by favoring the production of more gas particles. Therefore, if the product side of the reaction contains more moles of gas than the reactant side, the equilibrium shifts towards the products in response to the pressure decrease. This shift allows for more gas molecules to occupy the increased volume, which in turn, attempts to restore the lost pressure.

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Most popular questions from this chapter

Consider the reaction. $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ A solution is made containing initial \(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=0.020 \mathrm{M}\). At equilibrium, \(\left[\mathrm{Cl}_{2}\right]=1.2 \times 10^{-2} \mathrm{M}\). Calculate the value of the equilibrium constant. Hint: Use the chemical reaction stoichiometry to calculate the equilibrium concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\).

For each equilibrium constant, indicate if you would expect an equilibrium reaction mixture to be dominated by reactants or by products, or to contain significant amounts of both. (a) \(K_{\mathrm{eq}}=5.2 \times 10^{17}\) (b) \(K_{\mathrm{eq}}=1.24\) (c) \(K_{\mathrm{eq}}=3.22 \times 10^{-21}\) (d) \(K_{\mathrm{eq}}=0.47\)

Consider the reaction. $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ An equilibrium mixture of this reaction at a certain temperature has \(\left[\mathrm{COCl}_{2}\right]=0.225 \mathrm{M},[\mathrm{CO}]=0.105 \mathrm{M}\), and \(\left[\mathrm{Cl}_{2}\right]=0.0844 \mathrm{M}\). What is the value of the equilibrium constant at this temperature?

Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products. (a) \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g)\) (b) \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) (c) \(\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q)\) (d) \(\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

The solubility of \(\mathrm{CaCrO}_{4}\) at \(25^{\circ} \mathrm{C}\) is \(4.15 \mathrm{~g} / \mathrm{L}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{CaCrO}_{4}\).

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