CameraIcon

SearchIcon

MyQuestionIcon

1

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

Law of Mass Action

At 675 K, ...

Question

At $675 K$ , $H_{2} (g)$ and $C O_{2} (g)$ react to form $C O (g)$ and $H_{2} O (g), K_{p}$ for the reaction is $0.16$ . If a mixture of $0.25$ mole of $H_{2}$ (g) and $0.25$ mole of $C O_{2}$ is heated at $675 K$ , mole percentage of $C O (g)$ in equilibrium mixture is:

A

7.14

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

B

14.28

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

C

28.57

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

D

33.33

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

Open in App

Solution

The correct option is B $14.28$

$H_{2}$
$C O_{2}$
$C O$
$H_{2} O$
Initial moles:
$0.25$
$0.25$
$0$
$0$
Moles at equilibrium: $0.25 - x$ $0.25 - x$
$x$
$x$
The expression for the equilibrium constant is $\frac{x}{0.25 - x} \times \frac{x}{0.25 - x} = 0.16$
Hence, $x = 0.2$
Thus, mole percentage of $C O$ in the equilibrium mixture is $\frac{0.2}{0.5} \times 100 = 14.28 %$

Suggest Corrections

thumbs-up

0

Similar questions

675

H_{2} (g)

C O_{2} (g)

C O (g)

H_{2} O (g)

K_{p}

for the reaction is

0.16

. If a mixture of

0.25

H_{2} (g)

0.25

C O_{2}

675

%

C O (g)

in equilibrium mixture is:

Q. A mixture of one mole of

{C O}_{2}

and one mole of

H_{2}

attains equilibrium at a temperature of

250^{o} C

and a total pressure of 0.1 atm for the change

{C O}_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

K_{P}

if the analysis of final reaction mixture shows 0.16 volume percent of

C O

.

Q. A mixture of

3.8

N O

0.924

C O_{2}

was allowed to react in a

1.0 L

container. The equilibrium mixture contains

0.124

C O_{2}

K_{c}

for the reaction:

N O_{(g)} + C O_{(2 (g)} ⇌ N O_{2 (g)} + C O_{(g)}

700 K

{C O}_{2}

H_{2}

C O

H_{2} O

. For this process

K_{c}

0.11

. If a mixture of

0.45

{C O}_{2}

0.45

H_{2}

700 K

.
(i) Find out the amount of each gas at equilibrium state.
(ii) After equilibrium is reached another

0.34 m o l e

{C O}_{2}

0.34 m o l e

H_{2}

are added to the reaction mixture. Find the composition of the mixture at the new equilibrium state.

Q. A mixture of one mole of

C O_{2}

and one mole of

H_{2}

attains equilibrium at

250^{\circ} C

C O_{2} + H_{2} \to C O + H_{2} O

K_{p}

, if the analysis of final reaction mixture shows 0.16 volume percent of CO :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Laws of Mass Action

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Law of Mass Action

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon