MyQuestionIcon

1

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

Ideal Gas Law

A mixture in ...

Question

A mixture in which the mole ratio of $H_{2}$ and $O_{2}$ is $2 : 1$ is used to prepare water by the reaction:
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} (g) O$

The total pressure in the container is $0.8$ atm at 20 $^{o} C$ before the reaction. The final pressure (in atm) at $120$ $^{o} C$ after the reaction, assuming $80 %$ yield of water is (as nearest integer) :

Open in App

Solution

The reaction is as follows:
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} (g) O$
Initial moles: $2 a$ $a$ $0$
Final moles: $(2 a - 2 x)$ $(a - x)$ $2 x$
Given, $2 x = 2 \times \frac{80}{100} \times 1.6 a$
$∴ x = 0.8 a$
Thus, after the reaction
left $H_{2} = 2 a - 1.6 a = 0.4 a$ mol
left $O_{2}$ = 0.2a mol
formed $H_{2} O = 1.6 a$ mol
Total moles after the reaction at $120^{o} C$ in gaseous phase $= 0.4 a + 0.2 a + 1.6 a = 2.2 a$
At initial conditions, P=0.8 atm , T= 293K
Total moles = $3 a$
from $P V = n R T$ , we get
$0.8 \times V = 3 a \times R \times 293$
$V = \frac{3 a \times R \times 293}{0.8}$
Now using PV=nRT after the reaction,volume remains constant
$\frac{P \times 3 a \times R \times 293}{0.8} = 2.2 a \times R \times 393$
$P = 0.787$ atm
So, the answer is $1$ atm.

Suggest Corrections

thumbs-up

0

Similar questions

Q. A mixture in which the mole ration of

H_{2}

O_{2}

is 2:1 is used to prepare water by the reaction

{2 H}_{2} (g) + O_{2} (g) ⟶ {2 H}_{2} O (g)

The total pressure of the container is 0.8 atm at 20 before the reaction. Determine the final pressure at 120 after reaction assuming 80% yield of water.

Q. A mixture in which the mole ratio of

H_{2}

O_{2}

is :1 is used to prepare water by the reaction,

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)

The total pressure in the container is 0.8 atm at

20^{o} C

before the reaction. Determine the final pressure at

120^{o} C

after reaction assuming 80% yield of water.

Q. A reaction is carried out by 1 mole of

N_{2}

H_{2}

shows at equilibrium, the mole fraction of

N H_{3}

is 0.012 at 500

^{o} C

and 10 atm pressure.

K_{p}

is the equilibrium constant. The pressure (in atm) at which mole % of

N H_{3}

in equilibrium mixture increased to 10.4 is_________.
[Give your answer in closest integer]

20^{o} C

and a pressure of 1 atm each gas, 1 litre of water dissolves 0.043 g of pure

O_{2}

or 0.019 g of pure

N_{2}

. Assuming that dry air is composed of 20%

O_{2}

N_{2}

(by volume) determine the masses of

O_{2}

N_{2}

dissolved by 1 litre of water at

20^{o} C

exposed to air at a total pressure of 1 atm. Assume V.P. of

H_{2}

20^{o} C

Sulphur dioxide and oxygen react according to the equation :- $2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$ .If 50.0 g of $S O_{2}$ (g) is added to a flask resulting in a pressure of 0.750 atm.What will be the total pressure in the flask when a stochiometric amount of oxygen is added?What is the pressure after the completion of reaction and before the beginning of the reaction (with $O_{2}$ just added)?

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon