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In which one of the followng gaseous equ...

In which one of the followng gaseous equilibria, `K_(p)` is less than `K_(c)`?

A

`N_(2)O_(4)hArr 2NO_(2)`

B

`2SO_(2)+O_(2)hArr 2SO_(3)`

C

`2HIhArr H_(2)+I_(2)`

D

`N_(2)+O_(2)hArr 2NO`

Text Solution

AI Generated Solution

The correct Answer is:
To determine in which gaseous equilibrium \( K_p \) is less than \( K_c \), we need to understand the relationship between these two equilibrium constants. The relationship is given by the formula: \[ K_p = K_c \cdot (RT)^{\Delta n} \] where: - \( R \) is the ideal gas constant, - \( T \) is the temperature in Kelvin, - \( \Delta n \) is the change in the number of moles of gas, calculated as the number of moles of products minus the number of moles of reactants. ### Step-by-Step Solution: 1. **Identify the Relationship**: We know that \( K_p \) is related to \( K_c \) through the equation \( K_p = K_c \cdot (RT)^{\Delta n} \). 2. **Understand \( \Delta n \)**: - \( \Delta n = n_{\text{products}} - n_{\text{reactants}} \) - If \( \Delta n \) is positive, \( K_p \) will be greater than \( K_c \). - If \( \Delta n \) is zero, \( K_p \) will be equal to \( K_c \). - If \( \Delta n \) is negative, \( K_p \) will be less than \( K_c \). 3. **Determine Conditions for \( K_p < K_c \)**: For \( K_p \) to be less than \( K_c \), we require \( \Delta n < 0 \). This means that there are more moles of gas on the reactant side than on the product side. 4. **Analyze Given Equilibria**: We need to analyze the provided equilibria to find which one has a negative \( \Delta n \). For example: - If we have a reaction like \( A \rightleftharpoons 2B \), then \( \Delta n = 2 - 1 = 1 \) (positive). - If we have a reaction like \( 2A \rightleftharpoons B \), then \( \Delta n = 1 - 2 = -1 \) (negative). 5. **Select the Correct Option**: Based on the analysis of the reactions, we can conclude that the equilibrium where \( \Delta n < 0 \) will be the one where \( K_p < K_c \). ### Conclusion: The correct option is the one where the number of gaseous moles on the reactant side exceeds that on the product side, leading to a negative \( \Delta n \).
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