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Chemical Equilibrium question

2018 · 15 Apr · Shift 2 · Q18

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Chemical Equilibrium
2018 · 15 Apr · Shift 2 · Q18

Chemical Equilibrium question

2018 · 15 Apr · Shift 2 · Q18

JEE MainChemistryChemical EquilibriumMCQ+4 / −1

At a certain temperature in a

5L

vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,

CO

+

Cl_2

\rightleftharpoons

COCl_2

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

View written solutionFree

Correct answer: B

Write the reaction and initial moles

The reaction is:

$CO + Cl_2 \rightleftharpoons COCl_2$

Initial moles in a $5\,L$ vessel:

$CO = 2$ mol
$Cl_2 = 3$ mol
$COCl_2 = 0$ mol

Set up change in moles

Let $x$ moles of $CO$ react.

Then at equilibrium:

$CO = 2 - x$
$Cl_2 = 3 - x$
$COCl_2 = x$

Given that at equilibrium, $1$ mole of $CO$ is present:

$2 - x = 1$ $x = 1$

So equilibrium moles are:

$CO = 1$
$Cl_2 = 2$
$COCl_2 = 1$

Convert equilibrium moles to concentrations

Volume $= 5\,L$

Therefore,

$[CO] = \frac{1}{5} = 0.2\,M$ $[Cl_2] = \frac{2}{5} = 0.4\,M$ $[COCl_2] = \frac{1}{5} = 0.2\,M$

Write expression for $K_c$

For the reaction,

$K_c = \frac{[COCl_2]}{[CO][Cl_2]}$

Substitute the values:

$K_c = \frac{0.2}{(0.2)(0.4)}$

$K_c = \frac{0.2}{0.08} = 2.5$

Check options

A: $2$ ❌
B: $2.5$ ✅
C: $3$ ❌
D: $4$ ❌

Hence, the correct answer is Option B.

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