At a certain temperature in a 5 L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,CO + Cl2 COCl2 At equilibrium, if one mole of CO is present then equilibrium constant (Kc) for the reaction is :

Question

At a certain temperature in a 5 L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,CO + Cl2 <=> COCl2 At equilibrium, if one mole of CO is present then equilibrium constant (Kc) for the reaction is : (A) 2 (B) 2.5 (C) 3 (D) 4

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We have <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/5d9e3390662990e4e61c1f1f027a8ae3-.png" /> Equilibrium constant can be calculated as K c =([ COCl 2]/[ CO ][ Cl 2]) K c =((1 mol / 5 L )/(1 mol / 5 L ) ×(2 mol / 5 L ))=2.5

Q. At a certain temperature in a 5 L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction, $CO + C l_{2} <=> COC l_{2}$
At equilibrium, if one mole of CO is present then equilibrium constant $(K_{c})$ for the reaction is :

2

2.5

3

4

We have

Equilibrium constant can be calculated as $K_{c} = \frac{[ COC l _{2} ]}{[ CO ] [ C l _{2} ]}$

$K_{c} = \frac{( 1 m o l /5 L )}{( 1 m o l /5 L ) \times ( 2 m o l /5 L )} = 2.5$

Q. At a certain temperature in a 5 L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,CO+Cl2​<=>COCl2​ At equilibrium, if one mole of CO is present then equilibrium constant (Kc​) for the reaction is :

2

2.5

3

4

We have Equilibrium constant can be calculated as Kc​=[CO][Cl2​][COCl2​]​ Kc​=(1mol/5L)×(2mol/5L)(1mol/5L)​=2.5

Q. At a certain temperature in a 5 L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction, $CO + C l_{2} <=> COC l_{2}$
At equilibrium, if one mole of CO is present then equilibrium constant $(K_{c})$ for the reaction is :

We have

Equilibrium constant can be calculated as $K_{c} = \frac{[ COC l _{2} ]}{[ CO ] [ C l _{2} ]}$

$K_{c} = \frac{( 1 m o l /5 L )}{( 1 m o l /5 L ) \times ( 2 m o l /5 L )} = 2.5$