At 700K- K p for reaction 2S O 3 - 2S O 2 - O 2 K p -1-8- 10 -3 kPaWhat is the numerical value of K c-

K_p=1.8×10^ 3kPa=1.8 Pa #10; #10; T=700 K #10; #10;We know that #10; #10; K_p=K_c( RT )^Δ n #10; #10; K_c=K_p( RT )^Δ #1 ...

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Characteristics of Equilibrium Constant

At 700K, K ...

Question

At 700K, $K_{p}$ for reaction
$2 S O_{3} ⇌ 2 S O_{2} + O_{2}$
$K_{p} = 1.8 \times 10^{- 3} k P a$
What is the numerical value of $K_{c}$ ?

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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}

K_{p}

700 K

1.3 \times 10^{- 3} {a t m}^{- 1}

K_{C}

2 {S O}_{2} + O_{2} ⇌ 2 {S O}_{3}

K_{p},

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

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

K_{C}

R = 0.0821 L a t m m o l^{- 1} K^{- 1}

K_{p},

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

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

K_{C}

R = 0.0821 L a t m m o l^{- 1} K^{- 1}

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