Question

For given reaction $X_2(g) + Y_2(g) \rightleftharpoons 2Z (g)$. Moles of $X_2, Y_2 & Z$ are at equilibrium are $3 \text{ mole}, 3 \text{ mole} & 9 \text{ mole}$ respectively. If $10 \text{ moles}$ of $Z$ are added at constant $T$ then find moles of $Z$ at re-established equilibrium.

• A. 15
• B. 18
• C. 19
• D. 13

Hint:

Le Chatelier's Principle dictates that adding product ($Z$) shifts equilibrium backward. When the number of moles of gaseous reactants equals gaseous products ($\Delta n_g = 0$), $K_c$ calculated using moles is equivalent to $K_p$ calculated using partial pressures.

Answer 1

The Correct Option is A

To solve this equilibrium problem, we must understand the shifts in equilibrium according to Le Châtelier's principle. We start by evaluating the initial equilibrium condition and the changes caused by the addition of moles of \( Z \). 

1. Initial equilibrium:
   • Given reaction: \( X_2(g) + Y_2(g) \rightleftharpoons 2Z(g) \)
   • Equilibrium moles: \( X_2 = 3 \, \text{moles}, \, Y_2 = 3 \, \text{moles}, \, Z = 9 \, \text{moles} \)
2. Calculating the equilibrium constant \( K_c \):
   • For the reaction given, the expression for the equilibrium constant \( K_c \) is: \(K_c = \frac{{[Z]^2}}{{[X_2][Y_2]}}\)
   • At equilibrium:
     • \([Z] = 9 \, \text{moles}\)
     • \([X_2] = 3 \, \text{moles}\)
     • \([Y_2] = 3 \, \text{moles}\)
   • Substituting values: \(K_c = \frac{9^2}{3 \times 3} = \frac{81}{9} = 9\)
3. Impact of adding 10 moles of \( Z \):
   • New moles of \( Z \): \( 9 + 10 = 19 \, \text{moles} \)
   • According to Le Châtelier’s Principle, the system will shift to counteract the increase in \( Z \), favoring the reverse reaction until a new equilibrium is reached.
4. Let the change in moles of \( X_2 \) used up be \( x \). Then:
   • Moles of \( X_2 \) at new equilibrium = \((3 - x)\)
   • Moles of \( Y_2 \) at new equilibrium = \((3 - x)\)
   • Moles of \( Z \) at new equilibrium = \((19 - 2x)\)
   • At new equilibrium: \(K_c = \frac{{(19 - 2x)^2}}{{(3-x)(3-x)}} = 9\)
   • Expanding and equating: \((19 - 2x)^2 = 9(3-x)^2\)
   • Solving, we find \( x = 2 \).
5. Thus, the moles of \( Z \) at re-established equilibrium is:
   • \(19 - 2 \times 2 = 15 \, \text{moles}\)

Therefore, the correct answer is 15 moles of \( Z \).