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JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2018 · 15 April · Evening Shift · Online · Chemistry · Chemical Equilibrium

At a certain temperature in a 5L vessel, 2 moles of carbon monoxide...

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+Cl2⇌COCl2\mathrm{CO}+\mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2. At equilibrium, if one mole of CO\mathrm{CO} is present then equilibrium constant (Kc)\left(K_{\mathrm{c}}\right) for the reaction is:

Options

  1. A2.52.5
  2. B4
  3. C2
  4. D3

Correct answer

A. 2.52.5

Step-by-step solution

Initially 2 moles of CO\mathrm{CO} are present. At equilibrium, 1 mole of CO\mathrm{CO} is present Hence, 2−1=12-1=1 moles of CO\mathrm{CO} has reacted. 1 mole of CO will react with 1 mole of Cl2\mathrm{Cl}_2 to form 1 mole of COCl2\mathrm{COCl}_2. 3−1=23-1=2 moles of Cl2\mathrm{Cl}_2 remains at equilibrium The equilibrium constant Kc=[COCl2][CO][Cl2]=1 mol5 L1 mol5 L×2 mol5 L=K_{\mathrm{c}}=\frac{\left[\mathrm{COCl}_2\right]}{[\mathrm{CO}]\left[\mathrm{Cl}_2\right]}=\frac{\frac{1 \mathrm{~mol}}{5 \mathrm{~L}}}{\frac{1 \mathrm{~mol}}{5 \mathrm{~L}} \times \frac{2 \mathrm{~mol}}{5 \mathrm{~L}}}=2.5
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