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Suppose a reaction has the equilibrium constant \(K=1.3 \times 10^{8} .\) What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?

Short Answer

Expert verified
The equilibrium constant (K) for the reaction is 1.3 × 10^8, which is significantly greater than 1. As a result, the reaction will proceed towards the products and favor the formation of products over reactants at equilibrium. Since the reaction is highly favorable for the formation of products, it is likely to be a good source of the products.

Step by step solution

01

Understanding the Equilibrium Constant

The equilibrium constant, K, is a ratio that compares the concentrations of products and reactants at equilibrium. It is defined as: \( K = \frac{[products]}{[reactants]} \) where [products] and [reactants] represent the concentrations of products and reactants at equilibrium. The value of K indicates the extent to which a reaction will proceed: - If K > 1, the reaction will proceed towards the products (favoring products). - If K = 1, the reaction is balanced between both products and reactants at equilibrium. - If K < 1, the reaction will proceed towards the reactants (favoring reactants).
02

Analyzing the Magnitude of K

For this problem, we are given the equilibrium constant, K = 1.3 × 10^8: - Since K = 1.3 × 10^8, we can identify that K > 1. Therefore, the reaction will proceed towards the products and will favor the formation of products over reactants at equilibrium.
03

Determining the Suitability of the Reaction as a Product Source

Since the equilibrium constant (K = 1.3 × 10^8) is significantly greater than 1, this tells us that the reaction is highly favorable for the formation of products, with a high concentration of products compared to reactants once equilibrium is reached. Based on this information, we can say that the reaction is likely to be a good source of the products.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs when the forward and reverse reactions happen at the same rate in a chemical reaction, resulting in no net change in the concentration of reactants and products. At this point, the reaction is said to be in a state of balance or dynamic equilibrium. It's important to understand that even though the concentrations remain constant, the reactions themselves are still occurring — just at equal rates.
Here are key aspects of chemical equilibrium:
  • The concentrations of reactants and products remain constant over time.
  • This state can be achieved from either the complete reaction of reactants or partial reaction, depending on the reaction conditions and the equilibrium constant.
  • The system must be closed, meaning no products or reactants are added or removed while equilibrium is occurring.
Thus, chemical equilibrium provides insight into the final concentrations of substances once a reaction has reached its completion under a given set of conditions, as described by the equilibrium constant.
Reaction Quotient
The reaction quotient, denoted as Q, is similar to the equilibrium constant but pertains to the current concentrations of products and reactants at any point before equilibrium is reached. It is calculated using the same formula as the equilibrium constant:
\[ Q = \frac{[products]}{[reactants]} \]
The main role of Q is to predict the direction in which a reaction will proceed in order to reach equilibrium:
  • If Q < K, the reaction will move in the forward direction, converting reactants into products to reach equilibrium.
  • If Q = K, the system is already at equilibrium, and no shift in reaction direction is necessary.
  • If Q > K, the reaction will proceed in the reverse direction, converting products back into reactants.
By comparing Q and K, chemists can determine how far a reaction is from equilibrium and in which direction it needs to shift to achieve equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that explains how a system at equilibrium responds to external changes or stresses. When a chemical system at equilibrium is disturbed by changes in concentration, temperature, or pressure, the system will adjust itself to counteract the effect of the disturbance and restore a new equilibrium.
Some important ways in which Le Chatelier's Principle can be applied include:
  • Concentration Changes: Adding or removing products or reactants will cause the system to shift in a direction that opposes the change. For example, adding more reactants will shift the equilibrium towards more products.
  • Temperature Changes: For exothermic reactions, increasing the temperature shifts the equilibrium toward the reactants, while for endothermic reactions, it shifts towards the products.
  • Pressure Changes: For gaseous reactions, increasing pressure by decreasing volume will shift the equilibrium toward the side with fewer gas moles.
Understanding Le Chatelier's Principle helps in predicting the outcome of changing reaction conditions and can be crucial for industrial applications involving chemical reactions.

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Most popular questions from this chapter

Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\) $$\begin{array}{ll}\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\\\\mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\\\\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\\\\mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14}\end{array}$$,determine the values for the equilibrium constants for the following reactions: a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

In which direction will the position of the equilibrium.$$2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g).$$,be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\mathrm{HI}(g)\) is removed. d. In a rigid reaction container, some \(\operatorname{Ar}(g)\) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\) \(\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} \mathrm{M},\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=\) \(2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\).Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\). b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\). d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\).

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