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What does a value of \(K_{\text {eq }}\) greater than \(10^{3}\) imply? Prove that your answer is correct by using the general expression \(K_{\mathrm{eq}}=[\) Products \(] /[\) Reactants \(]\).

Short Answer

Expert verified
A value of \(K_{eq} > 10^3\) implies that the reaction heavily favors the formation of the products over the reactants at equilibrium. This can be proved using the general expression for the equilibrium constant, \(K_{eq} = \frac{\text{[Products]}}{\text{[Reactants]}}\). If \(K_{eq}\) is much greater than 1, it indicates that the concentration of products is much higher than that of reactants.

Step by step solution

01

Recall the general expression for equilibrium constant

The general expression for the equilibrium constant (\(K_{eq}\)) is given by: \(K_{eq} = \frac{\text{[Products]}}{\text{[Reactants]}}\)
02

Compare the value of \(K_{eq}\) to 1

If the value of \(K_{eq} > 1\), it means there are more products than reactants at equilibrium. On the other hand, if \(K_{eq} < 1\), there are more reactants than products at equilibrium. Therefore, when \(K_{eq} > 10^3\), it implies that the concentration of products is much greater than the concentration of reactants.
03

Interpret the value of \(K_{eq}\)

Given that \(K_{eq} > 10^3\), the reaction favors the formation of products significantly. This means that, at equilibrium, the concentration of products will be much higher compared to the concentration of reactants. In conclusion, a value of \(K_{eq} > 10^3\) implies that the reaction heavily favors the formation of the products over the reactants at equilibrium. This can be proved using the general expression for the equilibrium constant, which states that \(K_{eq} = \frac{\text{[Products]}}{\text{[Reactants]}}\). If \(K_{eq}\) is much greater than 1, it indicates that the concentration of products is much higher than that of reactants.

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Key Concepts

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

Chemical Equilibrium
When a chemical reaction occurs, the reactants transform into products, but under certain conditions, these products can also revert to reactants. This reversible process continues until a state is reached where the rate of the forward reaction equals the rate of the reverse reaction, known as chemical equilibrium. At this point, the concentrations of reactants and products remain constant over time, not because the reactions have stopped, but because they proceed at the same rate in both directions.

Understanding the equilibrium state is crucial as it tells us the extent to which a reaction proceeds before reaching stability. When we say that a reaction favors the formation of products or reactants at equilibrium, we are referring to their relative concentrations at this state. For instance, if the equilibrium constant, denoted by \(K_{eq}\), is higher than 10^3, it suggests that at equilibrium, the reaction comprises predominantly products, insinuating a successful chemical process towards product formation.
Reaction Quotient
For a reaction that has not yet reached equilibrium, the reaction quotient, \(Q\), serves as a predictive tool for determining in which direction the reaction will proceed to achieve equilibrium. It is calculated using the same formula as the equilibrium constant, with the concentration of the reactants and products at any given moment during the reaction.

\(Q = \frac{\text{[Products]}}{\text{[Reactants]}}\)

When \(Q\) is compared to the equilibrium constant \(K_{eq}\), we can infer the status of the reaction. If \(Q < K_{eq}\), the reaction will proceed forward to create more products. Conversely, if \(Q > K_{eq}\), the reaction will shift backward to produce more reactants. This is because the system naturally moves towards a state of equilibrium, where \(Q\) will eventually equal \(K_{eq}\). Thus, understanding the reaction quotient is essential for predicting the movement of a reaction that has not yet reached its point of equilibrium.
Le Chatelier's Principle
Le Chatelier's principle provides insight into how a system at equilibrium responds to external changes such as concentration, temperature, and pressure. According to this principle, if a system at equilibrium is subjected to a change, the equilibrium will shift in the direction that tends to minimize the effect of the change.

For example, increasing the concentration of reactants will shift the equilibrium to produce more products, as the system seeks to reduce the reactant concentration. Similarly, a temperature increase for an exothermic reaction will shift the equilibrium towards the reactants, as the system tries to absorb the added heat. These shifts occur as the system strives to re-establish equilibrium under new conditions.

Le Chatelier's principle is instrumental in chemical manufacturing and various applications where controlling product yields is essential. By understanding how a system at equilibrium responds to disturbances, chemists can manipulate reaction conditions to favor the formation of a desired product.

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Most popular questions from this chapter

Would the solubility of \(\mathrm{PbI}_{2}(s)\) be greater in water or in an aqueous solution of NaI? Explain your answer. (Hint: If \(\left[\mathrm{Pb}^{2+}\right] \times\left[\mathrm{I}^{-}\right]^{2}>K_{\mathrm{sp}}\) ' precipitation will occur.)

Consider the two curves, one showing supply of tapes, the other showing demand for tapes. At what price per quantity is equilibrium established?

The solubility of iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2}\), is \(1.43 \times 10^{-3} \mathrm{~g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) (a) Write a balanced equation for the solubility equilibrium. (b) Write the expression for \(K_{\mathrm{sp}}\) and calculate its value. (c) \(\mathrm{A} 60.0 \mathrm{~mL}\) sample of \(3.00 \times 10^{-3} \mathrm{M} \mathrm{FeSO}_{4}\) solution is added to \(60.0 \mathrm{~mL}\) of \(4.00 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) solution. Does a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{2}\) form? (Hint: Use the concentration values for \(\mathrm{Fe}^{2+}\) and \(\mathrm{OH}^{-}\) and plug them into the \(K_{\mathrm{sp}}\) expression. If the value you get is larger than \(K_{\mathrm{sp}}\) ' precipitation will occur.)

At \(25^{\circ} \mathrm{C}\), the solubility of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), in water is \(6.1 \mathrm{mg} / \mathrm{L}\). (a) What are the equilibrium molar concentrations of \(\mathrm{Ca}^{2+}(a q)\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) ?\) (b) Calculate \(K_{\mathrm{sp}}\) for calcium oxalate.

For the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) \(K_{\text {eq }}=3.59\) at \(900^{\circ} \mathrm{C}\). After the reaction has run for 10 min at \(900^{\circ} \mathrm{C}\), the concentrations are \(\left[\mathrm{CH}_{4}\right]=\) \(1.15 \mathrm{M} ;\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.20 \mathrm{M} ;\left[\mathrm{CS}_{2}\right]=1.51 \mathrm{M} ;\left[\mathrm{H}_{2}\right]=\) \(1.08 \mathrm{M}\). Is this reaction at equilibrium?

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