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

Short Answer

Expert verified
Decreasing the pressure of a reaction mixture at equilibrium will cause the equilibrium to shift towards the side with more moles of gas, which in this case is the reactants side.

Step by step solution

01

Understand the effect of pressure change on equilibrium

According to Le Chatelier's principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. Decreasing the pressure of a reaction mixture at equilibrium will cause the equilibrium to shift towards the side with more moles of gas, which will result in an increase in pressure.
02

Identify the number of moles of gas on each side of the equilibrium

In this scenario, the product side has fewer moles of gas particles than the reactant side.
03

Predict the direction of the equilibrium shift

Since decreasing pressure favors the side with more moles of gas, the equilibrium will shift towards the reactants side (more moles of gas) to increase pressure and counteract the change in condition.

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Key Concepts

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

Equilibrium Shift
When we talk about an equilibrium shift, we're referring to the change in concentrations of reactants and products in a chemical reaction that's at equilibrium. This shift occurs in response to an external change, such as alterations in pressure, temperature, volume, or the addition of a catalyst.

According to Le Chatelier's principle, the system will try to counteract that change. Imagine it like a seesaw: if you add weight to one side, the other side lifts up. In the context of a chemical reaction, if you decrease the pressure of the system, the equilibrium will shift towards the side with more moles of gas to balance it out. This helps maintain a stable state within the reaction mixture.

For students working through textbook exercises, it's important to visualize the reaction as a dynamic balance, where reactants and products are constantly interconverting. When a change is made, this balance is disturbed, and the system adjusts to a new position of equilibrium in response. Remember, the shifting doesn't create or destroy matter—it simply redistributes it to relieve the stress caused by the change.
Reaction Mixture Pressure Change
The pressure within a reaction mixture can significantly influence chemical equilibrium. This is most apparent in reactions involving gases. According to the gas laws, pressure and volume are inversely related for a gas at a constant temperature, which is to say, if the volume increases, pressure decreases, and vice versa, assuming the amount of gas stays the same.

In a sealed container, reducing the pressure often means increasing the volume available for the gases. This leads to dilution of the gas molecules; they have more space to move around and are less likely to collide and react. Le Chatelier's principle tells us the system will adjust by favoring the direction that produces more particles—that is, more gas molecules—to increase the pressure again.

When dealing with complicated exercises on this topic, remember to keep in mind the direct relationship between the number of moles of gas and pressure. Fewer moles mean less pressure, so to compensate for a decrease in pressure, the equilibrium will favor the production of more moles of gas.
Moles of Gas in Chemical Equilibrium
Understanding the role of moles of gas in chemical equilibrium is essential when predicting how a reaction will respond to changes. A mole is basically a count of particles—specifically, Avogadro's number (6.022 x 10^{23}) of them. In a gaseous reaction at equilibrium, these moles can tell us a lot about the reaction's behavior.

The number of moles is directly related to volume, temperature, and pressure by the ideal gas law: PV = nRTwhere P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

So, when you're faced with a situation where the pressure is changed, you need to think about how the moles of gas will also respond. If the product side of a reaction at equilibrium has fewer moles of gas, decreasing the pressure will cause the equilibrium to shift towards the reactants' side where there are more moles, thereby producing more gas to increase pressure. This is a crucial concept for students to understand, as it forms the basis for predicting the behavior of gases in all sorts of chemical reactions.

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Most popular questions from this chapter

Coal, which is primarily carbon, can be converted to natural gas, primarily \(\mathrm{CH}_{4}\), by this exothermic reaction. $$ \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g) $$ If this reaction mixture is at equilibrium, predict the effect (shift right, shift left, or no effect) of these changes. (a) adding more \(C\) to the reaction mixture (b) adding more \(\mathrm{H}_{2}\) to the reaction mixture (c) raising the temperature of the reaction mixture (d) lowering the volume of the reaction mixture (e) adding a catalyst to the reaction mixture

The body temperature of cold-blooded animals varies with the ambient temperature. From the point of view of reaction rates, explain why cold- blooded animals are more sluggish at cold temperatures.

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) \(\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Consider the effect of a volume change on this reaction at equilibrium. $$ \mathrm{I}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{ICl}(g) $$ Predict the effect (shift right, shift left, or no effect) of these changes. (a) increasing the reaction volume (b) decreasing the reaction volume

Write an equilibrium expression for each chemical equation. (a) \(2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)\) (b) \(2 \mathrm{BrNO}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) (c) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\)

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