MyQuestionIcon

1

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

Ideal Gas Equation

11 moles N2 a...

Question

$11 m o l e s$ $N_{2}$ and $12 m o l e s$ of $H_{2}$ mixture reacted in $20 L$ vessel at $800 K$ . After equilibrium was reached, $6 m o l e$ of $H_{2}$ was present. $3.58 L$ of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to $300 K$ . What is the final pressure of gaseous mixture/? Neglect vapour pressure of liquid solution. Assume (i) all $N H_{3}$ dissolved in water (ii) no change in volume of liquid (iii) no reaction of $N_{2}$ and $H_{2}$ at $300 K$

(Given $R = 0.821 L a t m m o l^{- 1} K^{- 1}$ )

A

18.47 a t m

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

B

60 a t m

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

C

22.5 a t m

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

D

45 a t m

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

Open in App

Solution

The correct option is C $22.5 a t m$
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
$I n i t i a l m o l e s 11120$
$a t e q u i l i b r i u m 964$
$Moles of N_{2} a n d H_{2} present at equilibrium = 9 + 6 = 15$
After addition of water
$N H_{3} (g) + H_{2} O (l) \to N H_{4} O H (l)$
$Volume of vessel available for gaseous mixture of N_{2} a n d H_{2} = 20 - 3.58 \Rightarrow 16.42 L$
$Pressure exerted by gaseous mixture at 300 K By using the relation, P V = n R T \Rightarrow P = \frac{n R T}{V} = \frac{15 \times 0.0821 \times 300}{16.42} \Rightarrow 22.5 a t m$

Suggest Corrections

thumbs-up

0

Similar questions

11 m o l e s

N_{2}

12 m o l e s

H_{2}

mixture reacted in

20 L

800 K

. After equilibrium was reached,

6 m o l e

H_{2}

3.58 L

of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to

300 K

. What is the final pressure of gaseous mixture/? Neglect vapour pressure of liquid solution. Assume (i) all

N H_{3}

dissolved in water (ii) no change in volume of liquid (iii) no reaction of

N_{2}

H_{2}

300 K

R = 0.821 L a t m m o l^{- 1} K^{- 1}

)

100 d m^{3}

10

N_{2}

H_{2}

777 K

. After equilibrium was reached, the partial pressure of

H_{2}

1 a t m

. At this point,

5 l i t r e s

H_{2} O_{(I)}

was injected and the gas mixture was cooled to

298 K

. Find out the pressure of the gaseous mixture left.

821 mL N_{2} (g)

was collected over liquid water at

300 K

and 1 atm. If the vapour pressure of

H_{2} O

is 30 torr then number of moles of

N_{2} (g)

in the moist gas mixture is:

Q. A gaseous catalyst X is added in a equilibrium mixture of

N_{2}, H_{2}

N H_{3}

gases at 298 K keeping pressure constant What will be observed at equilibrium reaction:

N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}

Q. A mixture of

2

N_{2}

8

H_{2}

are heated in a

2

lit vessel, till equilibrium is established. At equilibrium

0.4

N_{2}

was present. What will be the equilibrium concentration of

H_{2}

?

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

The Ideal Gas Equation

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Ideal Gas Equation

Standard XI Chemistry

Join BYJU'S Learning Program

CrossIcon