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N(2(g)) + 3H(2(g)) harr 2NH(3(g)) is a g...

`N_(2(g)) + 3H_(2(g)) harr 2NH_(3(g))` is a gaseous phase equilibrium reaction taking place at 400K in a 5 L flask. For this

A

`K_(c) = K_(x)`

B

`K_(c) = 25K_(x)`

C

`K_(x) = 25K_(c)`

D

`K_(c) = 5 K_(x)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem regarding the equilibrium reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) taking place at 400 K in a 5 L flask, we need to establish the relationship between the equilibrium constants \( K_c \) and \( K_x \). ### Step-by-Step Solution: 1. **Understanding the Reaction**: The reaction is given as: \[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \] This is a gaseous phase equilibrium reaction. 2. **Equilibrium Constants**: - The equilibrium constant in terms of concentration \( K_c \) is defined as: \[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \] - The equilibrium constant in terms of mole fraction \( K_x \) is defined similarly, but using mole fractions instead of concentrations. 3. **Relating \( K_c \) and \( K_x \)**: The relationship between \( K_c \) and \( K_x \) can be expressed as: \[ K_c = K_x \cdot \left( \frac{RT}{V} \right)^{\Delta n} \] where \( \Delta n \) is the change in the number of moles of gas during the reaction, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume of the flask. 4. **Calculating \( \Delta n \)**: - For the given reaction: - Reactants: 1 mole of \( N_2 \) + 3 moles of \( H_2 \) = 4 moles - Products: 2 moles of \( NH_3 \) - Thus, \( \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 4 = -2 \). 5. **Substituting Values**: - The volume \( V \) is given as 5 L. - Substituting \( \Delta n \) into the equation: \[ K_c = K_x \cdot \left( \frac{RT}{5} \right)^{-2} \] - This simplifies to: \[ K_c = K_x \cdot \frac{25}{(RT)^2} \] 6. **Final Relation**: - Rearranging gives: \[ K_c = \frac{25 K_x}{(RT)^2} \] - Since \( R \) and \( T \) are constants, we can express \( K_c \) in terms of \( K_x \) and the volume. ### Conclusion: Thus, the relationship between \( K_c \) and \( K_x \) in this equilibrium reaction at 400 K in a 5 L flask is: \[ K_c = 5 K_x \] This indicates that the correct option is option 4.
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