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Standard XII

Chemistry

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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Solution

For the given reaction,

Δn = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K^–1mol^–1

= 2.0 × 10¹⁰ bar ^–1

We know that,

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Similar questions

At 450 K, $K_{p} = 2.0 \times 10^{10} / b a r$ for the given reaction at equilibrium.
$2 S O_{2 (g)} + O_{2 (g)} \leftrightarrow 2 S O_{3 (g)}$
What is $K_{c}$ at this temperature?

Q. At

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

for the given reaction at equilibrium.

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

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K_{p}

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Q. 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}

27^{o} C

. Calculate

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

)

Q. For the reaction,

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

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

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

What is K_cat this temperature?

For the given reaction,

Δn = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K^–1mol^–1

= 2.0 × 10¹⁰ bar ^–1

We know that,