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What is the effect of a change in concentration of a reactant or product on a chemical reaction initially at equilibrium?

Short Answer

Expert verified
A change in the concentration of a reactant or product will shift the equilibrium position to counteract the change, according to Le Chatelier's Principle.

Step by step solution

01

Understanding Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This is true for changes in concentration of reactants or products, pressure, volume, or temperature.
02

Applying the Principle to Changes in Concentration

When the concentration of a reactant is increased, the system will respond by consuming more of that reactant to produce additional product, thereby shifting the equilibrium position to the right. Conversely, increasing the concentration of a product will shift the equilibrium to the left, to produce more reactants.
03

Results of Concentration Changes

The effect of a change in concentration of a reactant or product on a chemical reaction initially at equilibrium is to shift the equilibrium position to oppose the change in concentration, either by producing more products or more reactants.

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Key Concepts

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

Chemical Equilibrium
In the delicate balance of a chemical reaction, chemical equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, though this does not imply that the reactants and products are present in equal amounts. Instead, there is a constant exchange of molecules, a dynamic process where the forward and backward reactions continue to occur at the same rate.

Imagine a dance where the number of dancers entering and leaving the floor is the same. The population of dancers on the floor remains steady, but individuals are always in motion. Similarly, in a chemical reaction, molecules transform back and forth, but in a balanced, proportioned manner. This state represents a perfect balance in the microscopic world of molecules, one that can be disrupted by changes in concentration, pressure, or temperature.
Reactant and Product Concentration
The concentration of reactants and products plays a crucial role in the behavior of a reaction at equilibrium. For any chemical reaction, the concentration is a measure of how much of a substance is present in a given volume of solution. It is tantamount to the number of dancers available on the floor; with more available, the dance becomes more vigorous. However, at equilibrium, the system has found an ideal ratio of dancers, or in this case, reactants and products.

In the context of equilibrium, the proportions are maintained in a delicate harmony that can be disrupted if additional reactants or products are introduced. An increase in the concentration of reactants will lead the system to offset this change by converting more reactants into products. Conversely, if more products are added, the system will endeavor to convert some products back into reactants. This adjustment helps to restore the balance, reestablishing the equilibrium state.
Equilibrium Shift
When discussing an equilibrium shift, we often refer to Le Chatelier's Principle, which predicts how a system at equilibrium will respond to external changes. If there is a change in concentration of a reactant or a product, the equilibrium will 'shift' or adjust to minimize this change. It is as if the system is attempting to maintain its original state. This shift can be visualized as the seesaw effect, where the system balances itself out.

For instance, if we add more reactants to a system in equilibrium, it's like adding more dancers to one side of the seesaw. The system responds by converting some of the extra reactants into products, which is equivalent to moving dancers to the other side to level the seesaw again. If, instead, products are added, the system counteracts by reverting some of the products back into reactants. The direction in which the equilibrium shifts — towards more reactants or more products — will depend on which side of the reaction the change has occurred. This self-adjusting mechanism is fundamental for understanding how chemical reactions adapt to changes in their environment and is widely applied in chemical engineering, pharmaceuticals, and various industrial processes.

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Most popular questions from this chapter

For the reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{B}(g),\) a reaction vessel initially containsonly \(\mathrm{A}\) at a pressure of \(P_{\mathrm{A}}=1.32\) atm. At equilibrium, \(P_{\mathrm{A}}=0.25\) atm. Calculate the value of \(K_{\mathrm{P}}\) . (Assume no changes in volume or temperature.)

What is the significance of the equilibrium constant? What does a large equilibrium constant tell us about a reaction? A small one?

Many equilibrium calculations involve finding the equilibrium concen- trations of reactants and products given their initial concentrations and the equilibrium constant. Outline the general procedure used in solving these kinds of problems.

Consider the reaction: $$\begin{array}{c}{2 \mathrm{H}_{2} \mathrm{S}(g) \Longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)} \\ {K_{\mathrm{p}}=2.4 \times 10^{-4} \mathrm{at} 1073 \mathrm{K}}\end{array}$$ A reaction mixture contains 0.112 atm of \(\mathrm{H}_{2}, 0.055\) atm of \(\mathrm{S}_{2},\) and 0.445 atm of \(\mathrm{H}_{2} \mathrm{S} .\) Is the reaction mixture at equilibrium? If not, in what direction will the reaction proceed?

This reaction has an equilibrium constant of \(K_{\mathrm{p}}=2.2 \times 10^{6}\) at 298 \(\mathrm{K}\) \begin{equation}2 \mathrm{COF}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{CF}_{4}(g)\end{equation} Calculate \(K_{\mathrm{p}}\) for each reaction and predict whether reactants or products will be favored at equilibrium. \begin{equation}\begin{array}{l}{\text { a. } \operatorname{COF}_{2}(g) \rightleftharpoons \frac{1}{2} \mathrm{CO}_{2}(g)+\frac{1}{2} \mathrm{CF}_{4}(g)} \\ {\text { b. } 6 \mathrm{COF}_{2}(g) \rightleftharpoons 3 \mathrm{CO}_{2}(g)+3 \mathrm{CF}_{4}(g)} \\ {\text { c. } 2 \mathrm{CO}_{2}(g)+2 \mathrm{CF}_{4}(g) \rightleftharpoons 4 \mathrm{COF}_{2}(g)}\end{array}\end{equation}

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