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JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
In reaction \(A + 2B\) \( \rightleftharpoons \) \(2C + D\), initial concentration of \(B\) was \(1.5\, times\) of \([A]\) , but at equilibrium the concentrations of \(A\) and \(B\) became equal. The equilibrium constant for the reaction is
- A \(8\)
- B \(4\)
- C \(12\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(\mathop {\mathop A\limits_a }\limits_{(a - x)} + \mathop {\mathop {2B}\limits_{1.5a} }\limits_{(1.5a - 2x)} \leftrightarrow \mathop {\mathop {2C}\limits_0 }\limits_{2x} + \mathop {\mathop D\limits_0 }\limits_x \) Hence…
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