The equilibrium constant for the reaction- SO3 g - SO2 g -1-2O2 g is Kc- 4-9-10-2- The value of Kc for the reaction 2SO2 g - O2 g - 2SO3 g will be -

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

Law of Mass Action

The equilibri...

Question

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 $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
will be :

9.8 \times 10^{- 2}

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

4.9 \times 10^{- 2}

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

416

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

2.40 \times 10^{- 3}

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

Open in App

Solution

Suggest Corrections

Similar questions

K_{c} = 4.9 \times 10^{- 2}

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

K_{c}

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)

K_{c} = 4.9 \times 10^{- 2}

K_{c}

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

K_{c}

S O_{3 (g)} ⇌

+ \frac{1}{2} O_{2} (g)

4.9 \times 10^{- 2}

K_{c}

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); K_{c} = 4.9 \times 10^{- 2} m o l^{1 / 2} l i t r e^{- 1 / 2}

K_{c}

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); K_{c} = 5 \times 10^{- 2}

K_{c}

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

Join BYJU'S Learning Program