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In which of the following cases does the reaction go farthest to completion: (a) \(\mathrm{K}=1\) (b) \(\mathrm{K}=10\) (c) \(\mathrm{K}=10^{-2}\) (d) \(\mathrm{K}=10^{2}\)

Short Answer

Expert verified
The reaction with \( K = 10^2 \) goes farthest to completion.

Step by step solution

01

Understanding Equilibrium Constant

The equilibrium constant \( K \) indicates the extent to which a chemical reaction proceeds to completion. It is the ratio of the concentrations of products to reactants raised to the power of their coefficients at equilibrium. A larger \( K \) value suggests a reaction that favors the formation of products.
02

Analyzing the Given Options

We are given four options: \( K = 1 \), \( K = 10 \), \( K = 10^{-2} \), and \( K = 10^2 \). The task is to determine which \( K \) value corresponds to the reaction going farthest to completion.
03

Comparing Equilibrium Constants

Compare the four equilibrium constants. The reaction with \( K = 10^2 \) indicates that the products are significantly favored over the reactants at equilibrium, compared to the other \( K \) values.

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Key Concepts

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

Equilibrium Constant (K)
In the context of chemical reactions, the equilibrium constant, represented as \( K \), plays a fundamental role in understanding how reactions proceed to completion. It's a measure that tells us about the balance between products and reactants at equilibrium. To calculate \( K \), we use the concentrations of the products and reactants, each raised to the power of their stoichiometric coefficients. The equation looks like this:
\[ K = \frac{{[product]^m}}{{[reactant]^n}} \]
Here, the variables m and n are determined by the balanced chemical equation. A higher \( K \) value denotes that the reaction, at equilibrium, consists largely of products, meaning the reactants have been largely converted into products. Conversely, a low \( K \) value suggests that reactants are abundant compared to products.
Reaction Completion
Chemical reactions strive towards reaching a state of equilibrium. However, not all reactions go to the same extent of completion. The concept of reaction completion refers to how far a reaction goes until it establishes equilibrium, where no further net changes occur without external influences.
For instance, a condition where \( K = 10^2 \) indicates most of the reactants are turned into products, suggesting near-completion. Accordingly, a reaction with a small \( K \), such as \( K = 10^{-2} \), proceeds only slightly towards products, thus staying much closer to the starting reactants at equilibrium.
Product Formation
Product formation is inherently tied to the value of the equilibrium constant. In any given chemical reaction, as the reaction progresses, reactants form products. This transformation is reflected in the equilibrium constant's magnitude.
  • If \( K \) is large, such as \( K = 10^2 \), it indicates that, at equilibrium, products are predominantly formed over reactants.
  • A moderate \( K \) value, like \( K = 10 \), means the formation of products is significant, but not overwhelmingly dominant.
  • Meanwhile, a small \( K \), like \( K = 10^{-2} \), shows that very little product is formed in comparison to the starting quantity of reactants.
The larger the \( K \), the greater the extent of product formation.
Equilibrium Ratio
The equilibrium ratio is a pivotal concept that describes the proportion of products to reactants at the state of equilibrium in a reaction. It is intrinsically linked to the equilibrium constant \( K \).
When thinking about the equilibrium ratio, it helps to visualize the direction a reaction favors. If \( K > 1 \), the equilibrium ratio implies a higher concentration of products relative to reactants, resulting in a product-favored reaction. Conversely, \( K < 1 \) results in reactant dominance, where only a small fraction of reactants transform into products.
Ultimately, the equilibrium ratio provides insight into the position of the equilibrium and how a reaction behaves under given conditions without further external influence.

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Most popular questions from this chapter

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at 300 is kept in a closed container under one atmosphere. It is heated to 600 when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}\) (g). The resultant pressure is: (a) \(1.2 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.0 \mathrm{~atm}\)

Which of the following reaction will be favoured at low pressure: (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) (c) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (d) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\)

For the reaction, \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) \(\left(\mathrm{K}_{\mathrm{c}}=1.8 \times 10^{-6}\right.\) at \(\left.184^{\circ} \mathrm{C}\right)\) \((\mathrm{R}=0.0831 \mathrm{~kJ} /(\mathrm{mol} \mathrm{K}))\) when \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{\mathrm{c}}\) are compared at \(184^{\circ} \mathrm{C}\) it is found that: (a) \(\mathrm{K}_{\mathrm{p}}\) is greater than \(\mathrm{K}_{\mathrm{c}}\) (b) \(\mathrm{K}_{\mathrm{p}}\) is less than \(\mathrm{K}_{\mathrm{c}}\) (c) \(K_{p}=K_{c}\) (d) Whether \(\mathrm{K}_{\mathrm{p}}\) is greater than, less than or equal to \(\mathrm{K}_{\mathrm{c}}\) depends upon the total gas pressure

In which of the following reactions, equilibrium is independent of pressure: (a) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \Delta \mathrm{H}=+\mathrm{ve}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (c) \(3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=+\mathrm{ve}\)

In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+\) heat, relationship between \(\mathrm{K}_{\mathrm{P}}\) and \(\mathrm{K}_{\mathrm{c}}\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{2}\) (c) \(K_{p}=K_{c}(R T)^{-3}\) (d) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})^{3}\)

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