Question

\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) in the above reaction \(K_{p}\) and \(K_{c}\) are related as: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-1}\) (c) \(K_{c}=K_{p} \times(R T)^{2}\) (d) \(K_{p}=K_{c} \times(R T)^{-2}\)

Short answer

The correct answer is (b) \(K_p = K_c \times (RT)^{-1}\).

Step-by-step solution

Write the Reaction Equation

The given reaction is \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\). This is a gaseous reaction reaching equilibrium.

Understand Kp and Kc Relation

For a general reaction involving gases: \(aA + bB \rightleftharpoons cC + dD\), the relation between \(K_p\) (equilibrium constant in terms of pressure) and \(K_c\) (equilibrium constant in terms of concentration) is \(K_p = K_c (RT)^{\Delta n}\), where \(\Delta n\) is the change in moles of gas phase reactants and products.

Determine Δn for the Reaction

Calculate \(\Delta n = (moles\ of\ gaseous\ products) - (moles\ of\ gaseous\ reactants)\). For the reaction \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\), \(\Delta n = 2 - (2 + 1) = 2 - 3 = -1\).

Apply the Formula to Find Kp-Kc Relation

Substitute \(\Delta n = -1\) into the relation \(K_p = K_c (RT)^{\Delta n}\). This gives \(K_p = K_c (RT)^{-1}\).

Select the Correct Option

Compare the obtained expression \(K_p = K_c (RT)^{-1}\) with the given options. The correct option is (b) \(K_p = K_c \times (RT)^{-1}\).

Key concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant is a crucial value that helps predict the extent of a reaction. It tells us the ratio of the concentration of products to reactants at equilibrium.
For a reaction represented as \( aA + bB \rightleftharpoons cC + dD \), where \( A, B, C, \) and \( D \) are substances and \( a, b, c, \) and \( d \) are their coefficients in the balanced equation, the equilibrium constant \( K_c \) is calculated using concentrations:

\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]

This formula uses molar concentrations, meaning it depends on the amounts of substances, in moles per liter, that are present at equilibrium. The equilibrium constant, whether in terms of concentration \( K_c \) or pressure \( K_p \), is specific to a particular reaction at a given temperature.

Gaseous Reactions

Gaseous reactions are those where reactants and products are in the gaseous state. These reactions are often studied at equilibrium, where the forward and reverse reactions occur at the same rate.
The behavior of gases under equilibrium conditions is expressed in terms of partial pressures, which is why we often use \( K_p \) for these reactions. The partial pressure is simply the pressure a gas would exert if it alone occupied the entire volume of the reaction vessel.

• The ideal gas law \( PV = nRT \) helps in relating pressure \( P \), volume \( V \), and temperature \( T \) for gases.
• Equilibrium in gaseous reactions is dynamic and can be shifted by changing conditions, such as temperature and pressure, according to Le Chatelier’s Principle.

Understanding these principles is key to predicting how a gaseous reaction will behave under different circumstances.

Kp and Kc Relation

The relationship between \( K_p \) and \( K_c \) is fundamental when dealing with gaseous reactions. It allows conversion of the equilibrium constants based on partial pressures \( K_p \) to those based on concentration \( K_c \), and vice versa.

• This conversion uses the equation: \( K_p = K_c (RT)^{\Delta n} \); where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.
• \( \Delta n \) represents the change in moles of gas from reactants to products.

Since \( \Delta n \) can be zero, positive, or negative, the impact on the relationship varies. For example, when \( \Delta n = 0 \), \( K_p = K_c \), meaning there's no effect from changes in pressure. However, if \( \Delta n \) is positive or negative, the equation adjusts the exponential term accordingly, modifying how pressure and concentration relate at equilibrium.

Delta n in Reactions

In the context of gaseous equilibria, \( \Delta n \) plays a significant role in the relationship between \( K_p \) and \( K_c \). It signifies the difference in moles of gaseous products and reactants, calculated as:

\[ \Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) \]

For the reaction \( 2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) \), the calculation becomes:

• Products: 2 moles of \( \mathrm{SO}_{3} \)
• Reactants: 2 moles of \( \mathrm{SO}_{2} \) + 1 mole of \( \mathrm{O}_{2} \)
• Therefore, \( \Delta n = 2 - 3 = -1 \).

A negative \( \Delta n \) indicates fewer moles of gaseous products than reactants, thus affecting how pressure impacts the equilibrium state. This has direct implications on the \( K_p \) and \( K_c \) relation, as seen in the equation \( K_p = K_c (RT)^{-1} \). Understanding \( \Delta n \) helps in predicting reaction behavior with varying pressures and concentrations.