You visited us 1 times! Enjoying our articles? Unlock Full Access!

Law of Mass Action

NO and Br2 at...

Question

$N O$ and $B r_{2}$ 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 ( $K p$ ) 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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

164 a t m^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

234 a t m^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

264 a t m^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

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

Suggest Corrections

Similar questions

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.

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

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) .

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

. At equilibrium if nitrosyl bromide (

N O B r

) is

40

% dissociated at certain temp. and a total pressure of

0.30 a t m

K_{p}

for the reaction

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

is:

Q. For the reaction equilibrium,

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

.
If

P_{B_{1}} = \frac{P}{9}

at equilibrium and

P

is total pressure, the ratio of

\frac{K_{P}}{P}

is equal to:

Q. For the reaction equilibrium

2 N O B r_{(g)} ⇋ 2 N O_{(g)} + B r_{2 (g)}

, if

P_{B r_{2}} = \frac{P}{9}

at equilibrium and P is the initial total pressure, then the ratio

\frac{K_{P}}{P}

is equal to :

Join BYJU'S Learning Program

Explore more

Law of Mass Action

Standard XIII Chemistry

Join BYJU'S Learning Program

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).