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Standard XII

Chemistry

Characteristics of Equilibrium Constant

In a reaction...

Question

In a reaction between hydrogen and iodine, $6.34$ moles of hydrogen and $4.02 m o l e s$ of iodine are found to be present in equilibrium with $42.85 m o l e s$ of hydrogen iodine at $35^{o} C$ . Calculate the equilibrium constant.

72.04

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65.38

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52.08

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45.20

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Solution

The correct option is B $72.04$
$H_{2} 6.34 \frac{6.34}{V}$ $+$ $I_{2} 4.02 \frac{4.02}{V}$ $⇌$ $2 H I 42.85 \frac{42.85}{V}$ Let $v o l u m e = V$
$K_{e q} = \frac{(4 I)^{2}}{H_{2} \times I_{2}} = \frac{(\frac{42.85}{V})^{2}}{(\frac{4.02}{V}) \times (\frac{6.34}{V})} = 72.04$

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In a reaction between hydrogen and iodine, 6.34 moles of hydrogen and 4.02 moles of iodine are found to be present in equilibrium with 42.85 moles of hydrogen iodine at 35oC. Calculate the equilibrium constant.

The correct option is B 72.04H26.346.34V + I24.024.02V ⇌ 2HI42.8542.85V Let volume=VKeq=(4I)2H2×I2=(42.85V)2(4.02V)×(6.34V)=72.04

In a reaction between hydrogen and iodine, $6.34$ moles of hydrogen and $4.02 m o l e s$ of iodine are found to be present in equilibrium with $42.85 m o l e s$ of hydrogen iodine at $35^{o} C$ . Calculate the equilibrium constant.

The correct option is B $72.04$
$H_{2} 6.34 \frac{6.34}{V}$ $+$ $I_{2} 4.02 \frac{4.02}{V}$ $⇌$ $2 H I 42.85 \frac{42.85}{V}$ Let $v o l u m e = V$
$K_{e q} = \frac{(4 I)^{2}}{H_{2} \times I_{2}} = \frac{(\frac{42.85}{V})^{2}}{(\frac{4.02}{V}) \times (\frac{6.34}{V})} = 72.04$