Question:

At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
3A2+B22A3B,K1 3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1
A3B32A2+12B2,K2 A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2
The relation between K1 K_1 and K2 K_2 is:

Updated On: Dec 9, 2024
  • K12=2K2 K_1^2 = 2K_2
  • K2=K12 K_2 = \frac{K_1}{2}
  • K1=1K2 K_1 = \frac{1}{\sqrt{K_2}}
  • K2=1K1 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:

K2=1K1K_{2}=\frac{1}{\sqrt{K_{1}}}

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