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Chapter 17: Problem 1

A change in reaction conditions increases the rate of a certain forward reaction more than that of the reverse reaction. What is the effect on the equilibrium constant and on the concentrations of reactants and products at equilibrium?

Short Answer

Expert verified
The equilibrium constant, K, increases, resulting in higher concentrations of products and lower concentrations of reactants at equilibrium.

Step by step solution

01

Identify the Reaction Rate Effect

Determine that the change in reaction conditions increases the rate of the forward reaction more than the reverse reaction.
02

Analyze Equilibrium Constant (K)

Recall that the equilibrium constant, K, is given by the ratio of the rate constants of the forward reaction to the reverse reaction, i.e., \( K = \frac{k_{forward}}{k_{reverse}} \). An increase in the rate of the forward reaction more than the reverse suggests an increase in K.
03

Investigate Effect on Concentrations

Recognize that at a new equilibrium with a larger K, the system will have a higher concentration of products and a lower concentration of reactants compared to the original equilibrium.

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Key Concepts

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

Rate of Reaction
The rate of a reaction tells us how quickly reactants are converted into products over time.
Several factors can influence the reaction rate, including temperature, concentration of reactants, and the presence of catalysts.
To measure the rate, we often look at the change in concentration of reactants or products per unit time.
For example, if a change in conditions causes the forward reaction rate to increase more than the reverse:
* This means the forward reaction is converting reactants to products faster.
* The reaction could reach equilibrium more quickly, but this new equilibrium will be different.
Equilibrium Constant
The equilibrium constant, often denoted as K, describes the ratio of concentrations of products to reactants at equilibrium according to the balanced equation.
Mathematically, it is expressed as:
\[K = \frac{[Products]}{[Reactants]} \]
Where the square brackets indicate concentration.
K is determined by the rate constants of the forward and reverse reactions:
\[K = \frac{k_{forward}}{k_{reverse}} \]
When the rate of the forward reaction increases more than the reverse reaction, the value of K increases.
This shift means there will be a greater concentration of products compared to reactants at the new equilibrium.
Concentration of Reactants and Products
At equilibrium, the concentrations of reactants and products remain constant.
However, if K increases due to a higher forward reaction rate:
* The equilibrium position shifts toward producing more products.
* The concentration of products will be higher and the concentration of reactants will be lower compared to the initial equilibrium.
In this way, the system adjusts to maintain equilibrium under the new conditions.
Understanding how concentrations change at equilibrium is crucial for predicting reaction behaviors in different conditions.

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Most popular questions from this chapter

In an experiment to study the formation of \(\mathrm{HI}(g)\) $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) were placed in a sealed container at a certain temperature. At equilibrium, \(\left[\mathrm{H}_{2}\right]=6.50 \times 10^{-5} \mathrm{M},\left[\mathrm{I}_{2}\right]=1.06 \times 10^{-3} \mathrm{M},\) and \([\mathrm{HI}]=1.87 \times 10^{-3} \mathrm{M} .\) Calculate \(K_{\mathrm{c}}\) for the reaction at this temperature.

Gaseous ammonia was introduced into a sealed container and heated to a certain temperature: $$ 2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) $$ At equilibrium, \(\left[\mathrm{NH}_{3}\right]=0.0225 M,\left[\mathrm{~N}_{2}\right]=0.114 M,\) and \(\left[\mathrm{H}_{2}\right]=\) \(0.342 M .\) Calculate \(K_{c}\) for the reaction at this temperature.

White phosphorus, \(P_{4},\) is produced by the reduction of phosphate rock, \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\). If exposed to oxygen, the waxy, white solid smokes, bursts into flames, and releases a large quantity of heat: $$ \mathrm{P}_{4}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{P}_{4} \mathrm{O}_{10}(s)+\text { heat } $$ Does this reaction have a large or small equilibrium constant? Explain.

Compound A decomposes according to the equation $$ \mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)+\mathrm{C}(g) $$ A sealed 1.00-L container initially contains \(1.75 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{A}(g)\) \(1.25 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{B}(g),\) and \(6.50 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{C}(g)\) at \(100^{\circ} \mathrm{C}\) At equilibrium, [A] is \(2.15 \times 10^{-3} \mathrm{M}\). Find [B] and [C].

Using \(\mathrm{CH}_{4}\) and steam as a source of \(\mathrm{H}_{2}\) for \(\mathrm{NH}_{3}\) synthesis requires high temperatures. Rather than burning \(\mathrm{CH}_{4}\) separately to heat the mixture, it is more efficient to inject some \(\mathrm{O}_{2}\) into the reaction mixture. All of the \(\mathrm{H}_{2}\) is thus released for the synthesis, and the heat of reaction for the combustion of \(\mathrm{CH}_{4}\) helps maintain the required temperature. Imagine the reaction occurring in two steps: $$ \begin{array}{r} 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+4 \mathrm{H}_{2}(g) \\ K_{\mathrm{p}}=9.34 \times 10^{28} \mathrm{at} 1000 . \mathrm{K} \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \quad K_{\mathrm{p}}=1.374 \text { at } 1000 . \mathrm{K} \end{array} $$ (a) Write the overall equation for the reaction of methane, steam, and oxygen to form carbon dioxide and hydrogen. (b) What is \(K_{\mathrm{p}}\) for the overall reaction? (c) What is \(K_{c}\) for the overall reaction? (d) A mixture of \(2.0 \mathrm{~mol}\) of \(\mathrm{CH}_{4}, 1.0 \mathrm{~mol}\) of \(\mathrm{O}_{2},\) and \(2.0 \mathrm{~mol}\) of steam with a total pressure of \(30 .\) atm reacts at \(1000 . \mathrm{K}\) at constant volume. Assuming that the reaction is complete and the ideal gas law is a valid approximation, what is the final pressure?
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