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Chapter 14: Problem 46

If \(k_{\mathrm{f}}>k_{\mathrm{r}}\), will \(K_{\mathrm{eq}}\) be less than 1 or greater than 1? Explain your answer.

Short Answer

Expert verified
When \(k_{f} > k_{r}\), the equilibrium constant, \(K_{eq}\), is greater than 1. This is because \(K_{eq} = \frac{k_{f}}{k_{r}}\) and with a larger numerator, the fraction becomes greater than 1. As a result, at equilibrium, the concentration of the products will be greater than the concentration of the reactants.

Step by step solution

01

Understand the meaning of rate constants and equilibrium constant.

Rate constants (kf and kr) are used to describe the rates of forward and reverse reactions in a chemical equilibrium system. The forward rate constant (kf) indicates how fast the reactants are converted into products, while the reverse rate constant (kr) indicates the rate at which products are converted back into reactants. The equilibrium constant (Keq) is a measure of the ratio of concentrations of products to reactants when the reaction has reached a state of equilibrium. In other words, it represents the point at which the rates of the forward and reverse reactions are equal and there is no net change in the concentrations of the reactants and products.
02

Relate rate constants to the equilibrium constant.

The relationship between the rate constants (kf and kr) and the equilibrium constant (Keq) can be represented by the following equation: \[ K_{eq} = \frac{k_{f}}{k_{r}} \] This equation shows that the equilibrium constant is equal to the ratio of the forward rate constant to the reverse rate constant.
03

Determine the relationship between Keq and the given condition.

Now, we are given the condition that the forward rate constant (kf) is greater than the reverse rate constant (kr). \[ k_{f} > k_{r} \] According to the equation relating rate constants and the equilibrium constant: \[ K_{eq} = \frac{k_{f}}{k_{r}} \] If kf > kr, it means the numerator (kf) is greater than the denominator (kr), so the fraction kf/kr will be greater than 1.
04

Conclusion.

Based on the given condition, kf > kr, and the relationship between rate constants and the equilibrium constant, we can conclude that when kf > kr, the equilibrium constant (Keq) will be greater than 1. This means that at equilibrium, the concentration of the products will be greater than the concentration of the reactants.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is foundational to grasping more complex concepts in chemistry. A chemical reaction is said to have reached equilibrium when the rate at which the reactants form products equals the rate at which the products decompose back into reactants. At this point, the concentration of reactants and products does not change with time. It's important to note that chemical equilibrium refers to a dynamic balance where reactions continue to occur, but in such a way that there is no net change in the concentrations of substances involved.

An easy way to think about this is by imagining a room with two doors, where people are entering at the same rate as they are leaving. The number of people in the room (analogous to the concentration of reactants and products) stays the same, even though there is movement (reaction) going on.
Rate Constants
Rate constants are crucial in the study of chemical kinetics. Each reaction has a forward rate constant (\(k_f\)) and a reverse rate constant (\(k_r\)), which are measures of the speed at which a chemical reaction proceeds in both the forward and reverse directions. Essentially, \(k_f\) indicates how rapidly reactants are transformed into products, while \(k_r\) indicates the speed at which products revert to reactants.

These constants are dependent on various factors, including temperature and the presence of catalysts, but are independent of the concentrations of reactants and products. They can greatly affect the position of equilibrium. For example, a large forward rate constant compared to the reverse one (\( k_f > k_r \)) suggests that the reaction favors the formation of products, impacting the equilibrium constant as we will explore.
Reaction Rates
Speaking of reaction rates, they are a measure of how quickly concentrations change over time in a chemical reaction. A higher rate means that the concentration of a substance decreases or increases more quickly. In the context of equilibrium, we often look at the rate of the forward reaction (the rate at which reactants turn into products) compared to the rate of the reverse reaction (the rate at which products revert to reactants).

When a reaction first starts, the rate of the forward reaction is typically higher because the concentration of reactants is at its maximum. As the reaction proceeds, reactant concentrations decrease, thus decreasing the forward reaction rate, while product concentrations increase, increasing the reverse reaction rate. Equilibrium is achieved when these two rates are equal. Understanding how these rates interact and how they are defined by rate constants is integral to predicting the behavior of chemical systems at equilibrium.

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Most popular questions from this chapter

The solubility of \(\mathrm{PbI}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is \(1.52 \times 10^{-3}\) M. How many grams of \(\mathrm{PbI}_{2}\) will dissolve in \(2.50 \times 10^{6}\) gallons of water at \(25^{\circ} \mathrm{C}\) ?

(a) Write the equilibrium constant expression for the reaction $$ \mathrm{PbI}_{2}(s) \leftrightarrows \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$ (b) How would the equilibrium be affected if \(\mathrm{PbI}_{2}(s)\) were added? (c) How would the equilibrium be affected if \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s)\) were added? (Hint: Don't forget that \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is a water-soluble salt.)

What do we mean by the position of a reaction's equilibrium, and what practical consequence can it have?

Cooling an exothermic reaction for which \(K_{\text {eq }}\) is very low shifts the reaction to the right, so that more product is formed, but there is a trade-off. What is the downside of cooling such a reaction, as far as forming product is concerned?

Diamond and graphite are two forms of elemental carbon. Under the appropriate conditions they will be in equilibrium with each other: \(C_{\text {diamond }} \rightleftarrows C_{\text {graphite }}\) If graphite is subjected to very high pressure and temperature, it will convert into the diamond form. (a) Is the above equilibrium reaction exothermic or endothermic? Explain how you know. (b) Which form, graphite or diamond, has the higher density? (Hint: Think about what increasing the pressure of a gas does to its density. It works the same for the solid and liquid phases as well.)
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