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

Introduction to Le Chatelier's Principle

SO3 g + CO g ...

Question

$S O_{3} (g) + C O (g) ⇌ S O_{2} (g) + C O_{2} (g)$
Which of the following activity can reduce the equilibrium concentration of $C O$ to half?

Increasing pressure twice

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

Increasing the volume of reaction container to two times of its initial value

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

Adding two moles of

C O_{2}

gas

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

Adding two mole of helium gas at constant pressure

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

Open in App

Solution

The correct option is B Increasing the volume of reaction container to two times of its initial value
Increasing the volume of reaction container twice, the concentration of all the species will reduce to half.

Increasing the pressure will not have any effect on the concentration of the reactants and the products, so equilibrium position will not shift.
For the same reason adding helium gas will also does not affect the equilibrium.

Adding $C O_{2}$ gas will shift the reaction towards backward direction according to Le Chatelier's principle.

Suggest Corrections

Similar questions

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

C O

S O_{3} (g), C O (g), S O_{2} (g) a n d C O_{2} (g)

K_{C}

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

2

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

C O_{2}

K_{c}

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

1 / 9

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

S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {S O}_{3} (g)

K_{c}

1.7 \times 10^{12}

300 K

300 K

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

{S O}_{3} (g) ⇌ S O_{2} (g) + \frac{1}{2} O_{2} (g)

Join BYJU'S Learning Program