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Reaction Quotient

NO and Br 2 a...

Question

NO and $B r_{2}$ at initial partial pressure of 98.4 and 41.3 torr, respectively were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of KP for the following reaction at 300K.

$2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g)$

12.4

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13.4

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15.4

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17.4

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Solution

The correct option is D
17.4

The givan reaction is,

$\begin{matrix} 2 N O (g) + B r_{2} (g) & ⇋ 2 N O B r (g) I n t i a l n o . o f m o l e s & 98.4 41.3 & 0 N o . o f m o l e s a t E q b & (98.4 - x) (41.3 - \frac{x}{2}) & x \end{matrix}$

Total pressure at equilibrium = 110.5 torr.

$(98.4 - x) + (41.3 \frac{x}{2}) + x = 110.5$

$x = 58.4 t o r r (760 t o r r = 1 a t m)$

$P (N O B r) = 58.4 t o r r = 7.68 \times 10^{- 2} a t m$

$P (N O) = 98.4 - 58.4 = 40 t o r r = 5.26 \times 10^{- 2} a t m$

$P (B r_{2}) = 41.3 - \frac{58.4}{2} = 12.1 t o r r$
$= 1.59 \times 10^{- 2} a t m$

$K_{P} = \frac{P^{2} (N O B r)}{P^{2} (N O) \times P_{B r_{2}}}$

$= \frac{(7.68 \times 10^{- 2})^{2}}{(5.26 \times 10^{- 2})^{2} (1.59 \times 10^{- 2})} = 17.47$

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NO and Br2 at initial partial pressure of 98.4 and 41.3 torr, respectively were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of KP for the following reaction at 300K. 2NO(g)+Br2(g)⇌2NOBr(g)

NO and $B r_{2}$ at initial partial pressure of 98.4 and 41.3 torr, respectively were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of KP for the following reaction at 300K.

$2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g)$