JEE Main 2026 Jan 23 Shift 2, Chemistry Q21: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2026Jan 23, Shift 2Chemistry

$X_{2} (g) + Y_{2} (g) ⇌ 2 Z (g)$

$X_{2} (g)$ 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 conditions 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).

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Solution

From the first equilibrium in a 1 L flask:

$K_{c} = \frac{[ Z ] ^{2}}{[ X _{2} ] [ Y _{2} ]} = \frac{9 ^{2}}{3 \times 3} = 9$

After adding 10 mol of Z, the new initial moles are: $X_{2} = 3$ , $Y_{2} = 3$ , $Z = 19$ .

The reaction shifts backwards. Let $X$ mol of Z dissociate to form $X /2$ mol each of $X_{2}$ and $Y_{2}$ . Wait, rewrite using stoichiometry: backward shift consumes $2 X$ mol of Z to form $X$ mol of $X_{2}$ and $X$ mol of $Y_{2}$ .

New moles: $X_{2} = 3 + X$ , $Y_{2} = 3 + X$ , $Z = 19 - 2 X$ .

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

$\frac{19 - 2 X}{3 + X} = 3 \Rightarrow 19 - 2 X = 9 + 3 X \Rightarrow 5 X = 10 \Rightarrow X = 2$ .

Moles of Z at new equilibrium $= 19 - 2 (2) = 15$ .

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JEE Main 2026 Jan 23 Shift 2, Chemistry Q21: Law of Chemical Equilibrium And Equilibrium Constant

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