MyQuestionIcon

1

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

Law of Mass Action

The equilibri...

Question

The equilibrium constant ( $K_{P}$ ) for the reaction: $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is 0.13 $a t m^{- 1}$ at $830^{\circ}$ C. In one experiment 2.00 mol $S O_{2}$ and 2.00 mol $O_{2}$ were initially present in a flask. If 0.4 mol of $S O_{3}$ is present at equilibrium, find the total pressure.

A

1 atm

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

B

0.5 atm

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

C

2 atm

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

D

4 atm

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

Open in App

Solution

The correct option is A 1 atm
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
2 2 0
1.6 1.8 0.4 at eqb
total moles =3.8 mol
$∴$ , Mole fraction:
$X_{S O_{2}} = \frac{1.6}{3.8} = 0.42$
$X_{O_{2}} = \frac{1.8}{3.8} = 0.474$
$X_{S O_{3}} = \frac{0.4}{3.8} = 0.105$
$P_{S O_{3}} = X_{S O_{3}} P_{T}$
$P_{S O_{2}} = X_{S O_{2}} P_{T}$
$P_{O_{2}} = X_{O_{2}} P_{T}$
Now given,
$K_{P} = 0.13 = \frac{(P_{S O_{3}})^{2}}{P (S O_{2})^{2} . P_{O_{2}}}$ .
$\Rightarrow 0.13 = \frac{(0.105 P_{T})^{2}}{(0.42 \times P_{T})^{2} (0.479 \times P_{T})}$
$\Rightarrow P_{T} = 1 a t m$ .

Suggest Corrections

thumbs-up

0

Similar questions

Q. A mixture of

S O_{3} . S O_{2}

O_{2}

gases is maintained at equilibrium in 10 litre flask at a temperature at which

K_{c}

for the reaction,

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

m o l^{- 1} l i t r e

At equilibrium.
a) If no. of mole of

S O_{3}

S O_{2}

in flask are same, how many mole of

O_{2}

are present?
b) If no. of mole of

S O_{3}

in flask are twice the no.of mole of

S O_{2}

. how many mole of

O_{2}

Q. The value of

K_{p}

for the reaction,

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

5

.
What will be the partial pressure of

O_{2}

at equilibrium when equal moles of

S O_{2}

S O_{2}

are present at equilibrium?

Q. The equilibrium constant

K_{p}

for the reaction

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

1000 K

is 3.5. What would be the partial pressure of

O_{2}

gas with equal moles of

S O_{2}

S O_{3}

at equilibrium?

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}

?

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}

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Law of Mass Action

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon