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

Characteristics of Chemical Equilibrium

At 450K, K ...

Question

At 450K, $K_{p} = 2.0 \times 10^{10}$ /bar for the given reaction at equilibrium.

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

What is $K_{c}$ at this temperature?

Open in App

Solution

$Δ n = n_{p r o d u c t s} - n_{r e a c t a n t s} = 2 - (2 + 1) = - 1$

we know that $K_{p} = K_{c} (R T)^{Δ n}$

$∴ K_{c} = K_{p} (R T)^{- Δ n} = (2.0 \times 10^{10}) \times (0.0831 \times 450)^{1} = 7.48 \times 10^{11} M / m o l$

Suggest Corrections

Similar questions

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

What is K_cat this temperature?

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}

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}

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)

Join BYJU'S Learning Program