Problem 187 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(700 \mathrm{~K}\), the equilibrium constant \(\mathrm{K}\) for the reaction \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) is \(1.80 \times 10^{-3} .\) What is the numerical value in mole per litre of equilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for this reaction at the same temperature? (a) \(8.1 \times 10^{-8}\) (b) \(9.1 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(3.1 \times 10^{-7}\) (d) \(6.1 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\)

Step-by-Step Solution

Answer

\(K_c\) calculated is \(3.13 \times 10^{-5}\); contrast with options reflecting analysis might require further accuracy or depiction review.

Step 1: Understand the Given Information

The equilibrium constant K is provided for the reaction \(2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g)\) at \(700\, \text{K}\). The value of \(K\) is \(1.80 \times 10^{-3}\). We are to find the numerical value of the equilibrium constant \(K_c\) for this gaseous reaction at the same temperature.

Step 2: Use the Relation Between \(K\) and \(K_c\)

For gaseous reactions, the relationship between \(K_p\) (equilibrium constant in terms of partial pressures) and \(K_c\) (equilibrium constant in terms of concentrations) is given by the formula: \[ K_p = K_c (RT)^{\Delta n} \]where \( R \) is the ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\)), \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas (moles of gaseous products - moles of gaseous reactants).

Step 3: Calculate \(\Delta n\)

For the given reaction, the change in moles \( \Delta n = (2 + 1) - 2 = 1 \). This means \( \Delta n = 1 \) for this reaction.

Step 4: Substitute and Solve for \(K_c\)

Using the relationship from Step 2, we can rearrange to solve for \(K_c\):\[ K_c = \frac{K_p}{(RT)^{\Delta n}} \]Substituting the known values:\[ K_c = \frac{1.80 \times 10^{-3}}{(0.0821 \times 700)^1} \]Calculating the denominator:\[ 0.0821 \times 700 = 57.47 \]Thus,\[ K_c = \frac{1.80 \times 10^{-3}}{57.47} = 3.13 \times 10^{-5} \]

Step 5: Compare Calculated \(K_c\) with the Options

The quoted value of \( K_c = 3.13 \times 10^{-5} \) does not directly match any given option. Upon reassessment, there may have been confusion in applying or interpreting the question with respect to units typically. The options are interpreted potentially with a referenced system of units or misinterpretation, which should guide final verification. Understanding respective multiplication or theoretical derivation typically aligns as, but reevaluation brings mindful checking.

Key Concepts

Gaseous Equilibria

In the realm of chemistry, understanding gaseous equilibria is crucial as it involves reactions where both reactants and products are gases. These reactions reach a state of dynamic balance where the rate of the forward reaction equals the rate of the reverse reaction. However, this doesn't mean the concentrations remain static; instead, they continue to change but at a rate that balances out overall.

For these reactions, the equilibrium constant can be expressed in terms of partial pressures, denoted as \(K_p\), or in terms of concentrations, indicated as \(K_c\). When a reaction involves gases, the relationship between \(K_p\) and \(K_c\) becomes essential, as seen in the original exercise. The formula used is \(K_p = K_c (RT)^{\Delta n}\), where \(\Delta n\) represents the change in moles of gas, \(R\) is the ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\)), and \(T\) is the temperature in Kelvin.

In our specific example, the reaction \(2\text{SO}_3(g) \rightleftharpoons 2\text{SO}_2(g) + \text{O}_2(g)\) had a \(\Delta n\) of 1, which implies a change due simply to having one more mole of gaseous products than reactants. This change significantly influences the relationship between \(K_p\) and \(K_c\). Understanding this balance helps chemists manipulate conditions to favor the formation of desired products.

Equilibrium in Chemistry

Equilibrium in chemistry refers to the stage in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time, although not necessarily equal to each other. This is a dynamic equilibrium, meaning that reactions are ongoing, but their effects cancel each other out.

Equilibrium constants provide a numerical representation of this balance. The constant is specific to a given temperature and its value can help predict the position of equilibrium and how changes in temperature, pressure, or concentration can shift it. A low equilibrium constant (\(K\)) indicates a reaction where reactants are favored, while a high \(K\) suggests products are more prevalent at equilibrium.

In the context of the exercise, calculating the equilibrium constant \(K_c\) from \(K_p\) involves understanding these principles. Given that changes in molecular amounts (\(\Delta n\)) affect the equilibrium expressions, converting between \(K_p\) and \(K_c\) requires careful manipulation of the gas constant and temperature components. The intricacy of these calculations illustrates the interconnected nature of chemical equilibria variables, aiding chemists in predicting and controlling reaction outcomes.

Thermodynamics in Chemistry

Thermodynamics in chemistry is a branch that explores energy changes, particularly those associated with chemical reactions and phase changes. It helps clarify why certain reactions occur spontaneously, how energy is transferred, and how systems reach equilibrium.

A fundamental aspect involves understanding how temperature and energy influence equilibrium constants. For gaseous systems, these constants tie into the larger framework of energy distribution among reactants and products. The relationship \(K_p = K_c (RT)^{\Delta n}\) demonstrates how the equilibrium position can shift with temperature since \(R\), the gas constant, is inherently tied to temperature (\(T\)).

Important concepts like Gibbs free energy (\(\Delta G\)) also relate closely to equilibrium. The condition \(\Delta G = 0\) reflects a system at equilibrium, where no net change occurs. When \(\Delta G < 0\), reactions are spontaneous, suggesting a favorable energy distribution towards products, whereas positive \(\Delta G\) indicates non-spontaneity.

Thus, thermodynamics provides a comprehensive backdrop for interpreting the behavior of reactions at equilibrium, ensuring a deeper grasp of the forces driving chemical processes and aiding in the manipulation of conditions to yield desired reactions.