CameraIcon

SearchIcon

MyQuestionIcon

1

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

Reaction Quotient

A 100 dm3 f...

Question

A $100 d m^{3}$ flask contains $10$ moles each of $N_{2}$ and $H_{2}$ at $777 K$ . After equilibrium was reached, the partial pressure of $H_{2}$ was $1 a t m$ . At this point, $5 l i t r e s$ of $H_{2} O_{(I)}$ was injected and the gas mixture was cooled to $298 K$ . Find out the pressure of the gaseous mixture left.

Open in App

Solution

$N_{2} + 3 H_{2} \to 2 N H_{3}$
At $t = 0;$ $1010$
At $t = t_{e q}$ $(10 - x) (10 - 3 x) 2 x$
Given,
$P_{H_{2}} = 1 a t m$

$P V = n R T$
$n_{H_{2}} = \frac{P V}{R T} = \frac{1 \times 100}{0.0821 \times 777} = 1.74 m o l e s$

$10 - 3 x = 1.74$

$x = 2.75 m o l e s$

$n_{N_{2}} = (10 - x) = 7.25 m o l e s$

$N H_{3} + H_{2} O \to N H_{4} O H$ (completely dissolved)

Volume remaining= $100 - 5 = 95 l i t r e s$

Total number of moles= $n_{H_{2}} + n_{N_{2}} = 1.74 + 7.25 = 8.99 m o l e s$
$P V = n R T$

Substituting the values we get $P = 2.26 a t m$
$P = 2 a t m$

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}

)

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}

)

Q. A gaseous mixture was prepared by taking equal moles of

N_{2}

O_{2}

at room temperature. If the total pressure of the gaseous mixture was found to be 1 atm, then the partial pressure of nitrogen

(N_{2})

in the gaseous mixture is:

Q. A mixture of 2 moles of

C H_{4}

H_{2} S

was placed in an evacuated container, which was then heated to & maintained at

727^{\circ} C

. When equilibrium was established in the gaseous reaction

C H_{4} + 2 H_{2} S ⇌ C S_{2} + 4 H_{2}

; the total pressure in the container was 0.92 atm & the partial pressure of hydrogen was 0.2 atm. What was the volume of the container?

Q. A gaseous mixture was prepared by taking equal moles of

C O

N_{2}

. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen

(N_{2})

in the mixture is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Equilibrium Constants

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Reaction Quotient

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon