MyQuestionIcon

1

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

Reaction Quotient

2 moles each ...

Question

2 moles each of $S O_{3} (g), C O (g), S O_{2} (g) a n d C O_{2} (g)$ is taken in a one lit. vessel. If $K_{C}$ for $S O_{3} (g) + C O (g) ⇌ S O_{2} (g) + C O_{2} (g)$ is 1/9 then :

A

total no. of moles at equilibrium are less than 8

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

B

n (S O_{3}) + n (C O_{2}) = 4

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

C

[n (S O_{2}) / n (C O)] < 1

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

D

both (B) and (C)

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

Open in App

Solution

The correct option is D both (B) and (C)

Suggest Corrections

thumbs-up

0

Similar questions

2

S O_{3}, C O, S O_{2}

C O_{2}

is taken in a one lit. vessel. If

K_{c}

S O_{3} (g) + C O (g) ⇌ S O_{2} (g) + C O_{2} (g)

1 / 9

2

S O_{3}, C O, S O_{2}

C O_{2}

is taken in a one litre vessel. If

K_{c}

S O_{3} (g) + C O (g) ⇌ S O_{2} (g) + C O_{2} (g)

\frac{1}{9}

Q. A mixture of

3

{S O}_{2}

4

{N O}_{2}

4

N O

2.0 L

S O_{2} (g) + {N O}_{2} (g) ⇌ {S O}_{3} (g) + N O (g)

.

At equilibrium, the vessel is found to contain

1

S O_{2}

. Calculate the value of

K_{c}

.

Q. The equilirbium constant

K_{c}

for the reaction

{S O}_{2} (g) + {N O}_{2} (g) ⇌ {S O}_{3} (g) + N O (g)

16

1

mole of each of all the four gases is taken in

1 {d m}^{3}

vessel, the equilibrium concentration of

N O

Q. The equilibrium

S O_{2} (g) + 1 / 2 O_{2} (g) ⇌ S O_{3} (g)

is established in a container of

4 L

at a particular temperature. If the number of moles of

S O_{2} (g), O_{2} (g)

S O_{3} (g)

at equilibrium are

2, 1

4

respectively, then find the value of the equilibrium constant.

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