/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Consider the statement "The equi... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the statement "The equilibrium constant for a reaction at \(400 \mathrm{~K}\) is 792 . It must be a very fast reaction." What is wrong with the statement?

Short Answer

Expert verified
Answer: The flaw in the statement is the incorrect assumption that a high equilibrium constant indicates a very fast reaction. The equilibrium constant reflects the position of equilibrium and has no direct relation to the reaction rate or speed.

Step by step solution

01

Understanding the Equilibrium Constant

The equilibrium constant (K) is a measure of the relative amounts of reactants and products present at equilibrium. It indicates the position of equilibrium but does not provide any information about the rate (speed) at which the reaction takes place.
02

Distinguishing between Equilibrium Constant and Reaction Rate

It's important to differentiate between equilibrium constant (K) and reaction rate. A high equilibrium constant means that the reaction reaches an equilibrium state where the concentration of products is higher compared to the reactants. On the other hand, reaction rate relates to how fast the reaction reaches the equilibrium.
03

Assessing the Given Statement

The given statement claims that because the equilibrium constant for the reaction is 792 at 400 K, it must be a very fast reaction. But as discussed earlier, the equilibrium constant merely indicates the position of equilibrium and does not provide any information about the rate or speed of the reaction. Therefore, the statement is incorrect in making the assumption that a high equilibrium constant translates into a fast reaction rate.
04

Conclusion

The flaw in the statement is the assumption that a high equilibrium constant (792 in this case) indicates a very fast reaction. The equilibrium constant reflects the position of equilibrium, and it has no direct relation to the reaction rate or speed.

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Key Concepts

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

Reaction Rate
In chemistry, the reaction rate describes how quickly a reaction proceeds. It measures the speed at which reactants are converted into products. Understanding the reaction rate is crucial when analyzing chemical processes. This speed can be influenced by various factors, such as:
  • Concentration of reactants: Higher concentrations typically increase the rate of reaction.
  • Temperature: Raising the temperature generally speeds up a reaction.
  • Presence of a catalyst: Catalysts lower the energy barrier for reaction, increasing the rate without being consumed.
  • Surface area: Finer particles or more surface exposure often lead to faster reactions.
The reaction rate provides important information regarding reaction kinetics, but it's essential to recognize that a fast reaction rate does not necessarily correlate with a high equilibrium constant.
Position of Equilibrium
The position of equilibrium in a chemical reaction refers to the ratio of concentrations of products to reactants when the reaction has reached a state where the rates of the forward and reverse reactions are equal. At this point, the system is in a steady state, meaning there is no net change in concentration over time. The equilibrium constant, often represented as \( K \), is used to express this balance.
A large \( K \) value implies that, at equilibrium, products are favored over reactants, suggesting that the substance mostly consists of products. Conversely, a small \( K \) value means reactants are favored. However, it is necessary to distinguish that the equilibrium constant strictly indicates the position of equilibrium, not the pathway or time needed to achieve this state.
Reaction Kinetics
Reaction kinetics focuses on the factors that affect the rates of chemical reactions and the pathways taken from reactants to products. It deals with understanding how reactions occur on a molecular level. Reaction kinetics involves rate laws that quantitatively define the relationship between reactant concentrations and the reaction rate. These laws are determined through experiments and help predict reaction behaviors under different conditions.
It is crucial to grasp that the concepts of reaction kinetics and equilibrium constant are independent of each other. While kinetics relates to the speed and mechanism of a reaction, the equilibrium constant provides no insights into how long that state will take to achieve. Instead, the equilibrium constant only reflects the ratio of product to reactant concentrations once equilibrium is reached.

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Most popular questions from this chapter

At \(460^{\circ} \mathrm{C}\), the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g)$$ has \(K=84.7\). All gases are at an initial pressure of \(1.25\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

For the decomposition of \(\mathrm{CaCO}_{3}\) at \(900^{\circ} \mathrm{C}, K=1.04\). $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{O}_{2}(g)$$ Find the smallest mass of \(\mathrm{CaCO}_{3}\) needed to reach equilibrium in a 5.00-L vessel at \(900^{\circ} \mathrm{C}\).

WEB Nitrogen dioxide can decompose to nitrogen oxide and oxygen. $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ \(K\) is \(0.87\) at a certain temperature. A 5.0-L flask at equilibrium is determined to have a total pressure of \(1.25\) atm and oxygen to have a partial pressure of \(0.515\) atm. Calculate \(P_{\mathrm{NO}}\) and \(P_{\mathrm{NO}}\), at equilibrium.

A 1.0-L reaction vessel at \(90^{\circ} \mathrm{C}\) contains \(8.00 \mathrm{~g}\) of sulfur, hydrogen, and hydrogen sulfide gases with partial pressures of \(6.0 \mathrm{~atm}\) and \(0.40 \mathrm{~atm}\), respectively, at equilibrium: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ (a) Calculate \(K\) for the reaction at equilibrium. (b) The mass of sulfur is increased to \(10.0\) grams. What are the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{~S}\) when equilibrium is reestablished? (c) The pressure of \(\mathrm{H}_{2} \mathrm{~S}\) is increased to \(1.0 \mathrm{~atm}\). What are the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{~S}\) when equilibrium is reestablished?

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{ccccccc}\hline \text { Time (s) } & 0 & 30 & 45 & 60 & 75 & 90 \\\ P_{\mathrm{A}} \text { (atm) } & 0.500 & 0.390 & 0.360 & 0.340 & 0.325 & 0.325 \\\ P_{\text {B }} \text { (atm) } & 0.000 & 0.220 & 0.280 & 0.320 & 0.350 & 0.350 \\\\\hline\end{array}$$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after 45 s? After 90 s?

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