JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
For a reaction \(X + Y \rightleftharpoons 2\, Z , 1.0 mol\) of \(X , 1.5\) mol of \(Y\) and \(0.5\, mol\) of \(Z\) were taken in a \(1\) \(L\) vessel and allowed to react. At equilibrium, the concentration of \(Z\) was \(1.0\, mol\, L ^{-1}\). The equilibrium constant of the reaction is \(-\frac{x}{15} .\) The value of \(x\) is........
- A \(18\)
- B \(20\)
- C \(16\)
- D \(22\)
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
\(\quad\quad \quad x\quad +\quad Y\quad =\quad 2 Z\) \(t=0 \quad 1 \quad 1.5 \quad 0.5\) \(At\,eq. \quad 0.75\quad 1.25 \quad 1\) \(K_{e q .}=\frac{1^{2}}{\frac{3}{4} \times \frac{5}{4}}=\frac{16}{15}\)
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