The equilibrium constant for the reversible reaction
2A(g) \( \rightleftharpoons \) 2B(g) + C(g) is K1
\( {3 \over 2} \)A(g) \( \rightleftharpoons \) \( {3 \over 2} \)B(g) + \( {3 \over 4} \)C(g) is K2.
K1 and K2 are related as :
The equilibrium constant for the reversible reaction
2A(g) \( \rightleftharpoons \) 2B(g) + C(g) is K1
\( {3 \over 2} \)A(g) \( \rightleftharpoons \) \( {3 \over 2} \)B(g) + \( {3 \over 4} \)C(g) is K2.
K1 and K2 are related as :
Show Hint
- \( {K_1} = \sqrt {{K_2}} \)
- \( {K_2} = \sqrt {{K_1}} \)
- \( {K_2} = K_1^{3/4} \)
- \( {K_1} = K_2^{3/4} \)
The Correct Option is C
Solution and Explanation
Step 2: Multiplying an equation by n gives K \(\to\) Kn, so K2 = (K1)3/4.
Answer: (C) K2 = K13/4
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