AP EAMCET · Chemistry · Chemical Equilibrium
When the reaction \(A+2 B \rightleftharpoons 2 C+D\) was studied, it was observed that the initial concentration of \(B\) was 1.5 times that of \(A\), and the equilibrium concentrations of \(A\) and \(C\) were equal. Then \(K_C\) for the given equilibrium is equal to
- A \(4.1\)
- B \(0.3\)
- C \(2.5\)
- D \(1.8\)
Answer & Solution
Correct Answer
(B) \(0.3\)
Step-by-step Solution
Detailed explanation
Given, \([A]=[C]\) \(a-x=2 x\) \(\therefore \quad x=a / 3\) Now, \(K_C=\frac{[C]^2[D]}{[A][B]^2}=\frac{\left(\frac{2 a}{3}\right)^2 \times \frac{a}{3}}{\left(a-\frac{a}{3}\right)\left(\frac{3 a}{2}-\frac{2 a}{3}\right)^2}\) \(\therefore \quad K_C=0.3\)
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