You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Reaction Quotient

A mixture of ...

Question

A mixture of $N_{2}$ and $H_{2}$ in the volume ratio $1 : 3$ is allowed to attain equilibrium : $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ at $670 K$ and 1000 atm, 0.2491 mole fraction of $N H_{3}$ is formed. $K_{p}$ is $X \times 10^{- 6} {a t m}^{- 2}$ . $100 X$ is__________.

Open in App

Solution

$N_{2}$ $H_{2}$ $N H_{3}$
Initial moles
1
3
0
Moles at equilibrium
$1 - x$ $3 - 3 x$ $2 x$
Mole fraction
$\frac{1 - x}{4 - 2 x}$ $\frac{3 - 3 x}{4 - 2 x}$ $\frac{2 x}{4 - 2 x}$
$\frac{2 x}{4 - 2 x} = 0.2491$
$x = \frac{4}{10.029} = 0.4$

$N_{2}$ $H_{2}$ $N H_{3}$
Mole fraction
0.1875
0.5625
0.25

Partial pressures
at equilibrium
187.5
562.5
250
$K_{p} = \frac{(250)^{2}}{(187.5) (562.5)^{3}} = 1.87 \times 10^{- 6} = X \times 10^{- 6}$
$100 X = 187$

Suggest Corrections

Similar questions

Q. A mixture of

N_{2}

and

H_{2}

is the molar ratio

1 : 3

at 50 atm and

650

is allowed to react till equilibrium is obtained. The

N H_{3}

present at equilibrium is 25% by weight,

K_{p}

for,

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

X \times 10^{- 4} {a t m}^{- 2}

(1000 X)

is_________.

Q. In a mixture of

N_{2}

and

H_{2}

in the ratio 1:3 at 30 atm and

300^{0} C

, the % of

{N H}_{3}

at equilibrium is 17.8.
Calculate

K_{P}

for

{N_{2} + {3 H}_{2} ⇌ 2 N H}_{3}

Q. For

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, one mole of

N_{2}

and three moles of

H_{2}

are at pressure of

4

atm. Equilibrium pressure is found to be

3

atm. Hence,

K_{p}

is:

Q. For

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, one mole of

N_{2}

and three mole of

H_{2}

are at pressure of 4 atm. Equilibrium pressure is found to be 3 atm. Hence,

K_{p}

is :

Q. A reaction mixture containing

H_{2}, N_{2}

and

N H_{3}

has partial pressures 2 atm,1 atm, and 3 atm respectively at 725K.

If the value of

K_{p}

for the reaction,

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

4.28 \times 10^{- 5} {a t m}^{- 2}

at 725K,in which direction the net reaction will go :

Join BYJU'S Learning Program

	$N_{2}$	$H_{2}$	$N H_{3}$
Initial moles	1	3	0
Moles at equilibrium	$1 - x$	$3 - 3 x$	$2 x$
Mole fraction	$\frac{1 - x}{4 - 2 x}$	$\frac{3 - 3 x}{4 - 2 x}$	$\frac{2 x}{4 - 2 x}$

A mixture of N2 and H2 in the volume ratio 1:3 is allowed to attain equilibrium : N2+3H2⇌2NH3 at 670K and 1000 atm, 0.2491 mole fraction of NH3 is formed. Kp is X×10−6atm−2. 100X is__________.

N2H2NH3Initial moles130Moles at equilibrium1−x3−3x2xMole fraction1−x4−2x3−3x4−2x2x4−2x2x4−2x=0.2491x=410.029=0.4N2H2NH3Mole fraction0.18750.56250.25Partial pressures at equilibrium187.5562.5250Kp=(250)2(187.5)(562.5)3=1.87×10−6=X×10−6100X=187

A mixture of $N_{2}$ and $H_{2}$ in the volume ratio $1 : 3$ is allowed to attain equilibrium : $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ at $670 K$ and 1000 atm, 0.2491 mole fraction of $N H_{3}$ is formed. $K_{p}$ is $X \times 10^{- 6} {a t m}^{- 2}$ . $100 X$ is__________.