Question:
At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
\[ 3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1 \]
\[ A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2 \]
The relation between \( K_1 \) and \( K_2 \) is:
At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
\[ 3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1 \]
\[ A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2 \]
The relation between \( K_1 \) and \( K_2 \) is:
\[ 3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1 \]
\[ A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2 \]
The relation between \( K_1 \) and \( K_2 \) is:
Updated On: Dec 9, 2024
- \( K_1^2 = 2K_2 \)
- \( K_2 = \frac{K_1}{2} \)
- \( K_1 = \frac{1}{\sqrt{K_2}} \)
- \( K_2 = \frac{1}{\sqrt{K_1}} \)
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The Correct Option is D
Solution and Explanation
The second equilibrium reaction is the reverse of the first reaction divided by 2.
Therefore, the relationship between the equilibrium constants is:
\[K_{2}=\frac{1}{\sqrt{K_{1}}}\]
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