CameraIcon

SearchIcon

MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Common Ion Effect

2 mol of SO 2...

Question

2 mol of $S O_{2}$ and I mol of $O_{2}$ are heated in a closed vessel to reach the equilibrium:
$2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}$ . The equilibrium mixture exerted a pressure of 5 atm and required $\frac{1}{3}$ mol of $K_{2} C r_{2} O_{7}$ in acidic medium. $K_{P}$ for the reaction is

A

2.0

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

B

0.5

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

C

1.0

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

D

none of these

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

Open in App

Solution

The correct option is C
1.0

$S O_{2} \to S O_{3}$ , Eq.mass=molar mass/2
$C r_{2} O_{7}^{2} \to 2 C r^{3 +}$ , Eq. mass = molar mass/6
$\frac{1}{3}$ mol of $K_{2} C r_{2} O_{7} = \frac{6}{3} = 2$ equiv.of $K_{2} C r_{2} O_{7}$
= 2 equiv. of $S O_{2} = 1$ mol of $S O_{2}$
$\begin{matrix} 2 S O_{2} & + & O_{2 (g)} & ⇌ & 2 S O_{3 (g)} 2 m o l & 1 m o l & 0 m o l 1 m o l & 0.5 m o l & 1 m o l \end{matrix}$

$P_{S O_{2}} = \frac{1}{2.5} \times 5 = 2 a t m, P_{O_{2}} = \frac{0.5 \times 5}{2.5} = 1 a t m, P_{S O_{3}} = \frac{1}{2} .5 \times 5 = 2 a t m K_{p} = \frac{P_{S O_{3}^{2}}}{P_{S O_{3}^{2}} \times P_{o_{2}}} = \frac{2^{2}}{2^{2} \times 1} = 1.0$

Suggest Corrections

thumbs-up

2

Similar questions

Q. In the equilibrium,

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g),

the equilibrium partial pressure of

S O_{2}, O_{2}

S O_{3}

0.6, 0.1

0.3

atm respectively. what would be the

K_{p}

?

Q. The equilibrium constant (

K_{P}

) for the reaction:

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

a t m^{- 1}

830^{\circ}

C. In one experiment 2.00 mol

S O_{2}

O_{2}

were initially present in a flask. If 0.4 mol of

S O_{3}

is present at equilibrium, find the total pressure.

K_{p}

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

5.0 a t m^{- 1}

.What is the equilibrium partial pressure of

O_{2}

if the equilibrium pressure of

S O_{2}

S O_{3}

K_{p}

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

5.0 a t m^{- 1}

. What is the equilibrium partial pressure of

O_{2}

if the equilibrium pressures of

S O_{2}

S O_{3}

Q. The equilibrium mixture for

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

present in 1L vessel at

600^{p} C

contains 0.50, 0.12, and 5.0 mole of

S O_{2}, O_{2} a n d S O_{3}

respectively.
Also calculate

K_{p}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Degree of Dissociation

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Common Ion Effect

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon