For the reaction- 2 SO 3 g- 2S O 2 g- O 2 g- the equilibrium constant K c -2-52- 10 -2 at 27oC- Calculate Kp- Given R-0-082 litre atm deg -1 mol -1

The given reaction is : 2 SO _ 3 ( g ) ⇌ 2 SO _ 2 ( g ) + O _ 2 ( g ) We have, #160; K _ P = K _ C ( RT ) #160; ^ n ; T= 27 ^ 0 C=27+273=300 ...

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Derivation of Kp and Kc

For the react...

Question

For the reaction, $2 {S O}_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ , the equilibrium constant $K_{c} = 2.52 \times 10^{- 2}$ at $27^{o} C$ . Calculate $K_{p}$ . (Given $R = 0.082 l i t r e a t m {d e g}^{- 1} {m o l}^{- 1}$ )

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450 K, K_{P} = 2.0 \times 10^{10} b a r

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

K_{c}

K_{p} = 2.0 \times 10^{10}

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

K_{c}

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),

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)

K_{c}

2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}, K_{p} = 3.4 b a r^{- 1}

1000^{o} C

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