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JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)

The reaction \(A(g) \rightleftharpoons B(g) + C(g)\) was initiated with the amount '\(a\)' of \(A(g)\). At equilibrium it is found that the amount of \(A(g)\) remaining is \((a - x)\) at a total pressure of \(p\).
The equilibrium constant \(K_p\) of the reaction can be calculated from the expression :

  1. A \(\dfrac{x^2}{a^2 + x^2} \times p\)
  2. B \(\dfrac{x^2}{a^2 - x^2} \times p\)
  3. C \(\dfrac{a + x^2}{x^2} \times p\)
  4. D \(\dfrac{a^2 - x^2}{x^2} \times p\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\dfrac{x^2}{a^2 - x^2} \times p\)

Step-by-step Solution

Detailed explanation

The reaction is \(A(g) \rightleftharpoons B(g) + C(g)\) Initial moles: \(a\) for \(A\), \(0\) for \(B\), \(0\) for \(C\) Moles at equilibrium: \((a - x)\) for \(A\), \(x\) for \(B\), \(x\) for \(C\) Total moles at equilibrium \(= (a - x) + x + x = a + x\) Partial pressures at…
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