are added to a 1 L flask and the system attains equilibrium at T(K) with the number of moles of being 3, 3 and 9 mol respectively (equilibrium moles). Under this condition of equilibrium, 10 mol of Z(g) is added to the flask and the temperature is mainta

Question

$X_{2} (g) + Y_{2} (g) ⇌ 2 Z (g)$

$X_{2} (g)$ and $Y_{2} (g)$ are added to a 1 L flask and it is found that the system attains the above equilibrium at T(K) with the number of moles of $X_{2} (g), Y_{2} (g) and Z (g)$ being 3, 3 and 9 mol respectively (equilibrium moles). Under this condition of equilibrium, 10 mol of Z(g) is added to the flask and the temperature is maintained at T(K). Then, the number of moles of Z(g) in the flask when the new equilibrium is established is ______. (Nearest integer) [2026]

Smart Achievers · Accepted Answer

(15)

$X_{2} (g) + Y_{2} (g) ⇌ 2 Z (g)$

$K_{c} = \frac{{(9)}^{2}}{3 \times 3} = 9$

Now 10 moles of Z are added then reaction will move in backward direction.

$X_{2} (g) + Y_{2} (g) ⇌ 2 Z (g)$

$\begin{matrix} 3 + X & 3 + X & 19 - 2 X \end{matrix}$

$K_{c} = \frac{{(19 - 2 X)}^{2}}{(3 + X) (3 + X)} = 9$

$\frac{19 - 2 X}{3 + X} = 3$

$19 - 2 X = 9 + 3 X$

$10 = 5 X$

$X = 2$

$At equilibrium \Rightarrow moles of Z = 19 - 2 \times 2 = 15 moles$

Solution

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Solution

Ans. (15) X2(g)+Y2(g)⇌2Z(g) Kc=(9)23×3=9 Now 10 moles of Z are added then reaction will move in backward direction. X2(g)+Y2(g)⇌2Z(g) 3+X3+X19-2X Kc=(19-2X)2(3+X)(3+X)=9 19-2X3+X=3 19-2X=9+3X 10=5X X=2 At equilibrium ⇒ moles of Z =19-2×2=15 moles

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