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Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) A friend asks the following: "I know we have been told that if a mixture of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) is at equilibrium and more of \(\mathrm{A}\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more \(\mathrm{B} ?^{\prime \prime}\) What do you tell your friend?

Short Answer

Expert verified
Adding more reactant A to the system at equilibrium according to Le Châtelier's principle will shift the reaction in the direction that favors the formation of products C and D. As a result, the reaction will consume more A and B to restore equilibrium, leading to higher concentrations of C and D, even without adding more reactant B.

Step by step solution

01

Explain Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction (formation of products) is equal to the rate of the reverse reaction (breakdown of products). In other words, the concentrations of reactants and products no longer change, even though the reaction is still happening at the molecular level, just at equal rates in both directions.
02

Describe Le Châtelier's Principle

Le Châtelier's Principle states that if a system at equilibrium is subjected to a change in conditions, such as concentration, temperature, or pressure, the system will adjust itself to counteract the change and restore a new equilibrium. Essentially, the reaction will shift in the direction that opposes the change.
03

Adding More Reactant A

Now, let's consider the case where we add more of reactant A to the system at equilibrium. According to Le Châtelier's principle, the reaction will shift in a direction to counteract the increase in A's concentration. In this case, the forward reaction (A + B ⇌ C + D) will increase its rate in order to consume the added A and restore equilibrium.
04

No Need to Add More Reactant B

Because only A has been added, the concentration of B initially remains the same. However, as the forward reaction increases its rate and consumes more A and B to form C and D, the concentration of B will start to decrease as well. As a result, the equilibrium will shift to accommodate the decrease in B by further favoring product formation and consume more A to restore the balance. This process will continue until a new equilibrium is reached, with higher concentrations of C and D than before. So, even though we did not add more B to the system, the reaction's shift to restore equilibrium will lead to the formation of more products C and D, as predicted by Le Châtelier's principle.

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Key Concepts

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

Le Châtelier's Principle
Imagine you have a balance scale perfectly leveled. When you add more weight to one side, the scale tips. Le Châtelier's Principle is quite similar. It describes how a chemical reaction at equilibrium responds to changes such as concentration, temperature, or pressure. Consider a reaction at equilibrium, like our example \[\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons\mathrm{C}(g)+\mathrm{D}(g) \]. When this equilibrium is disturbed, say by adding more \( \mathrm{A} \), the system behaves like the scale, trying to rebalance itself.
  • The reaction will "shift" towards the side that counteracts the change. In our case, this means producing more \( \mathrm{C} \) and \( \mathrm{D} \) to consume the excess \( \mathrm{A} \).
  • This shift continues until a new state of balance (equilibrium) is established.
Understanding Le Châtelier's Principle helps predict how a chemical reaction will adapt to different changes in conditions.
Reaction Rates
In any chemical reaction, the rate at which reactants turn into products can be dynamic. Reactants \( \mathrm{A} \) and \( \mathrm{B} \) need energy to collide and react, turning into products \( \mathrm{C} \) and \( \mathrm{D} \). When the system is at equilibrium, these rates are equal, meaning the forward reaction forms products as quickly as the reverse reaction breaks them back down.
When more \( \mathrm{A} \) is added, the forward reaction rate increases because there are more \( \mathrm{A} \) molecules available to collide with \( \mathrm{B} \).
  • This means more products will form, shifting equilibrium towards \( \mathrm{C} \) and \( \mathrm{D} \).
  • Even though \( \mathrm{B} \) isn’t added directly, it will get consumed faster due to the increased forward reaction rate.
Eventually, when the extra \( \mathrm{A} \) is sufficiently reacted, a new equilibrium is reached. Reaction rates are crucial in understanding how quickly and in what direction a system approaches equilibrium.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. Consider different molecules, such as \( \mathrm{A}(g) \) and \( \mathrm{B}(g) \), that come together to create \( \mathrm{C}(g) \) and \( \mathrm{D}(g) \) through a chemical reaction. This process involves breaking and forming bonds, where energy changes play a pivotal role. The initial reaction rate can be altered by changing conditions or concentrations.
  • In our example, adding more \( \mathrm{A} \) doesn't just mean more reactants; it means those extra molecules are likely to find \( \mathrm{B} \) faster to form \( \mathrm{C} \) and \( \mathrm{D} \).
  • This interaction between molecules highlights how changes in a reaction's conditions can influence the progress of the reaction.
Chemical reactions give us insight into how substances change and why some products might be favored over others under different conditions. Understanding these interactions helps predict the outcomes in various scenarios, such as the one described.

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Most popular questions from this chapter

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction$$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$.In an experiment, 1.00 mole of \(\mathrm{H}_{2}, 1.00\) mole of \(\mathrm{I}_{2}\), and 1.00 mole of HI are introduced into a 1.00 -L container. Calculate the concentrations of all species when equilibrium is reached.

Consider the reaction $$\mathbf{P}_{4}(g) \longrightarrow 2 \mathbf{P}_{2}(g)$$, where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{K},\) the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium.

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of \(K\) for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of \(K\) for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{K}\) at an initial pressure of 1.00 atm, where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)$$. At equilibrium, the partial pressure of \(S_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{K}\)

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is \(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \quad K=5.6 \times 10^{-11}\) If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into \(1.00 \mathrm{~L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}{ }^{2-}\) ?

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