Consider the following gaseous equilibrium in a closed container of volume 'V' at T(K). 2 moles each of , and are present at equilibrium. Now one mole each of and are added to the equilibrium keeping the temperature at T(K). The number of moles of

Question

Consider the following gaseous equilibrium in a closed container of volume 'V' at T(K).

$P_{2} (g) + Q_{2} (g) ⇌ 2 PQ (g)$

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 [2026]

Smart Achievers · Accepted Answer

(1)

$\begin{matrix} P_{2} (g) & + & Q_{2} (g) & ⇌ & 2 PQ (g) \\ t = t_{e q} & 2 mole & 2 mole & 2 mole \end{matrix}$

$K_{e q} = \frac{2^{2}}{2.2} = 1$

$Now 1 mole of each P_{2} and Q_{2} is added$

$So reaction will move in forward direction$

$\begin{matrix} P_{2} (g) & + & Q_{2} (g) & ⇌ & 2 PQ (g) \\ t = t'_{e q} & 3 - x & 3 - x & 2 + 2 x \end{matrix}$

$K_{c} = 1 = \frac{{(2 + 2 x)}^{2}}{(3 - x) (3 - x)}$

$\frac{2 + 2 x}{3 - x} = 1$

$2 + 2 x = 3 - x$

$x = \frac{1}{3}$

$At new equilibrium:$

$Moles of P_{2} = \frac{8}{3} = 2.67$

$Moles of Q_{2} = \frac{8}{3} = 2.67$

$Moles of P Q = \frac{8}{3} = 2.67$

Solution

1 2.67, 2.67, 2.67

2 1.21, 2.24, 1.56

3 1.66, 1.66, 1.66

4 2.56, 1.62, 2.24

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