JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
Consider the following reactions in which all the reactants and products are present in gaseous state
\(2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5\times 10^5\)
\(xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5\times 10^{-3}\)
The value of \(K_3\) for the equilibrium \(\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz\) is:
- A \(2.5\times 10^{-3}\)
- B \(2.5\times 10^{3}\)
- C \(1.0\times 10^{-5}\)
- D \(5\times 10^{-3}\)
Answer & Solution
Correct Answer
(C) \(1.0\times 10^{-5}\)
Step-by-step Solution
Detailed explanation
The given reactions are: \(2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5 \times 10^5\) \(xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5 \times 10^{-3}\) We need to find the equilibrium constant \(K_3\) for the reaction:…
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