At 450 K - K p -2-0 - 1010 - bar for the given reaction at equilibrium- 2 SO 2g- O 2g-n- T 2 SO 3gWhat is Kc at this temperature-

For the given reaction, δn = 2 – 3 = – 1 T = 450 K R = 0.0831 bar L bar K–1 mol–1 = 2.0 × 1010 bar –1 We know that, ...

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Abundance and Isotopes

At 450 K , K ...

Question

At 450 K, K_p= 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

What is K_cat this temperature?

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

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

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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}

350

K_{p}

3.0 t i m e s 10^{10} b a r^{- 1}

K_{c}

2 N_{2 (g)} + O_{2 (g)} ⇌ 2 N_{2} O_{(} g)

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

K_{c} = 2.52 \times 10^{- 2}

27^{o} C

K_{p}

R = 0.082 l i t r e a t m {d e g}^{- 1} {m o l}^{- 1}

700 K

K_{p}

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

1.8 \times 10^{- 3} k P a

d m^{3}

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