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Standard XII

Chemistry

Derivation of Kp and Kc

Equimolar con...

Question

Equimolar concentrations of $H_{2} a n d I_{2}$ are heated to equilibrium in a 2litre flask. At equilibrium, the forward and backward rate constants are found to be equal. What percentage of initial concentration of $H_{2}$ has reacted at equilibrium?

33.33%

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75%

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50%

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15%

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Solution

The correct option is C 50%
Solution:Correct option is c).

$\begin{matrix} H_{2} (g) + & l_{2} (g) ⇋ & 2 H I (g) I n i t i a l m o l e s & 1 & 1 & 0 A t e q u i l i b r i u m & 1 - x & 1 - x & 2 x M o l a r c o n c e n t r a t i o n (m o l / L) & (1 - x) / 2 & (1 - x) / 2 & 2 x / 2 = x \end{matrix}$

$K = x^{2} / \frac{(1 - x) \times (1 - x)}{2 \times 2} = 4 x^{2} / (1 - x)^{2} \approx 4 x^{2}$
$K = K_{f} / K_{b} = 1$

$4 x^{2} = 1 x = \frac{1}{2}, i . e ., 50 %$

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Similar questions

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Equimolar concentrations of H2 and I2 are heated to equilibrium in a 2litre flask. At equilibrium, the forward and backward rate constants are found to be equal. What percentage of initial concentration of H2has reacted at equilibrium?

The correct option is C 50%Solution:Correct option is c). H2(g) +l2(g) ⇋2HI(g)Initial moles110At equilibrium1−x1−x2x Molarconcentration(mol/L)(1−x)/2(1−x)/22x/2=x K=x2/ (1−x) ×(1−x)2×2=4x2/(1−x)2 ≈ 4x2 K=Kf/Kb=1 4x2=1 x=12,i.e.,50%

Equimolar concentrations of $H_{2} a n d I_{2}$ are heated to equilibrium in a 2litre flask. At equilibrium, the forward and backward rate constants are found to be equal. What percentage of initial concentration of $H_{2}$ has reacted at equilibrium?