🛠️ JEE🧪 Chemistry

The equilibrium constant for the reaction \(S{O}_{3}(g)⇌S{O}_{2}(g)+\frac{1}{2}{O}_{2}(g)\) is \({K}_{c}=4.9\times {10}^…

Q1

The equilibrium constant for the reaction
\(S{O}_{3}(g)⇌S{O}_{2}(g)+\frac{1}{2}{O}_{2}(g)\)
is \({K}_{c}=4.9\times {10}^{-2}\). The value of \({K}_{c}\) for the reaction given below is \(2S{O}_{2}(g)+{O}_{2}(g)⇌2S{O}_{3}(g)\) is :

a

4.9

b

49

c

41.6

d

416

🔒
Answer & explanation — PYQ Pass

Options are free to see. Unlock the correct answer and full explanation with Pass.

Unlock · ₹149

Practice more JEE Chemistry PYQs

See every question on Equilibrium, or browse the full JEE question bank.

See all questions on Equilibrium →