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Law of Mass Action

NO and Br 2 a...

Question

$N O$ and $B r_{2}$ at initial partial pressures of 98.4 and 41.3 $t o r r$ , respectively, were allowed to react at $300 K$ . At equilibrium the total pressure was $110.5 t o r r$ . Calculate the value of the equilibrium constant ( $K p$ ) at $300 K$ , for the reaction $2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g) .$

134 a t m^{- 1}

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164 a t m^{- 1}

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234 a t m^{- 1}

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264 a t m^{- 1}

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Solution

The correct option is A $134 a t m^{- 1}$
$\begin{matrix} 2 N O (g) & + & B r_{2} (g) & ⇌ & 2 N O B r (g) I n i t i a l p r e s s u r e & 98.4 & 41.3 & 0 A t e q u i l i b r i u m & 98.4 - x & 41.3 - \frac{x}{2} & x \end{matrix}$
Total pressure at equilibrium is 110.5 torr
$∴ 98.4 - x + 41.3 - \frac{x}{2} + x = 110.5$
$x = 58.4 t o r r$
Now, $1 a t m = 760 t o r r, ∴ x = 7.68 \times 10^{- 2} a t m$
$P_{N O B r} = 7.68 \times 10^{- 2} a t m; P_{N O} = 98.4 - x = 40 t o r r = 5.26 \times 10^{- 2} a t m$
$P_{B r_{2}} = 41.3 - \frac{x}{2} = 12.1 t o r r = 1.59 \times 10^{- 2} a t m$

$K_{P} = \frac{[P_{N O B r}]^{2}}{[P_{N O}]^{2} [P_{B r_{2}}]}$
$= \frac{(7.68 \times 10^{- 2})^{2}}{(5.26 \times 10^{- 2})^{2} (1.59 \times 10^{- 2})} = 134 a t m^{- 1}$

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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.

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Law of Mass Action

Standard XII Chemistry

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