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

Consider the following gaseous equilibrium in a closed container of volume "V" at T(K):
\(P_{2}(g)+Q_{2}(g)\rightleftharpoons2PQ(g)\).
2 moles each of \(P_{2}(g)\), \(Q_{2}(g)\) and PQ (g) are present at equilibrium. Now one mole each of '\(P_{2}\)' and '\(Q_{2}\)' are added to the equilibrium keeping the temperature at T(K). The number of moles of \(P_{2}\), \(Q_{2}\) and PQ at the new equilibrium, respectively, are -

  1. A 2.67, 2.67, 2.67
  2. B 1.21, 2.24, 1.56
  3. C 1.66, 1.66, 1.66
  4. D 2.56, 1.62, 2.24
Verified Solution

Answer & Solution

Correct Answer

(A) 2.67, 2.67, 2.67

Step-by-step Solution

Detailed explanation

\(\underset {t = t_{eq}}{} ~\underset {2 ~mole}{P_2(g)} +~\underset {2 ~mole}{Q_2(g)} \rightleftharpoons ~\underset {2 ~mole}{2PQ(g)}\) \(K _{ eq }=\frac{2^2}{2.2}=1\) Now 1 mole of each \(P_2\) and \(Q_2\) is added So reaction will move in forward direction…
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