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Standard XII

Chemistry

Relation between Kp and Kc

I 2 gets 5 % ...

Question

$I_{2}$ gets $5 %$ dissociated into atoms when1.0 mole of $I_{2} (g)$
is introduced into an evacuated flask of 1 L capacity at 150K.The $K_{p}$ for the reaction $I_{2} (g) ⇌ 2 I (g)$ at 150K is:

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Solution

The correct option is A
$I_{2} (g) ⇌ 2 I (g)$ $α = \frac{5}{100} = 0.05$
Initial conc. 1 0
Equilibrium conc. $1 - α$ $2 α$
$K_{c} = \frac{[I]^{2}}{[I_{2}]} = \frac{[2 α]^{2}}{1 - α} = [2 \times 0.05] = 1.05 \times 10^{- 2}$

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When 3.06 g of solid $N H_{4}$ HS is introduced into a two-litre evacuated flask at 27 $^{o} C$ , 30 $%$ of the solid decomposes into gaseous ammonia and hydrogen sulphide. Choose the most appropriate one regarding $K_{c}$ , $K_{p}$ and degree of dissociation $α$ :

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I2 gets 5% dissociated into atoms when1.0 mole of I2(g) is introduced into an evacuated flask of 1 L capacity at 150K.The Kp for the reaction I2(g)⇌2I(g) at 150K is:

The correct option is A I2(g)⇌2I(g) α=5100=0.05 Initial conc. 1 0 Equilibrium conc. 1−α 2α Kc=[I]2[I2]=[2α]21−α=[2×0.05]=1.05×10−2

$I_{2}$ gets $5 %$ dissociated into atoms when1.0 mole of $I_{2} (g)$
is introduced into an evacuated flask of 1 L capacity at 150K.The $K_{p}$ for the reaction $I_{2} (g) ⇌ 2 I (g)$ at 150K is:

The correct option is A
$I_{2} (g) ⇌ 2 I (g)$ $α = \frac{5}{100} = 0.05$
Initial conc. 1 0
Equilibrium conc. $1 - α$ $2 α$
$K_{c} = \frac{[I]^{2}}{[I_{2}]} = \frac{[2 α]^{2}}{1 - α} = [2 \times 0.05] = 1.05 \times 10^{- 2}$