For the reaction S O 2 - 1 2 O 2 - S O 3- if we write K p - K c RT x- then x becomes-

The correct option is C 1 2 In the relation, K _ p = K _ c ( RT ) nbsp;^ x , nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; n ...

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

Derivation of Kp and Kc

For the react...

Question

For the reaction $S O_{2} + \frac{1}{2} O_{2} ⇌ S O_{3}$ , if we write $K_{p} = K_{c} {(R T)}^{x}$ , then $x$ becomes:

- 1

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

- \frac{1}{2}

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

\frac{1}{2}

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

1

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

Open in App

Solution

Suggest Corrections

Similar questions

S O_{2} + \frac{1}{2} O_{2} ⇌ S O_{2}

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

K_{1}

K_{2}

K_{1}

K_{2}

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

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

K_{1}

K_{2}

K_{1}

K_{2}

$S (s) + O_{2} (g) ⇌ S O_{2} (g); K_{1} = 10^{52}$

$2 S (s) + 3 O_{2} (g) ⇌ 2 S O_{3} (g); K_{2} = 10^{129}$
Find the equilibrium constant for
$2 S O_{2} + O_{2} (g) ⇌ 2 S O_{3} (g); K_{3}$

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)

N_{2} + O_{2} ⇋ 2 N O, K_{1}; \frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇋ N O, K_{2};

2 N O ⇌ N_{2} + O_{2}, K_{3}; N O ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}, K_{4}

K_{1}, K_{2}, K_{3}

K_{4}

Join BYJU'S Learning Program