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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2019 · 10 April · Evening Shift · Chemistry · Chemical Equilibrium

For the reaction, 2S O _ 2 (g) + O _ 2 (g) ⇌ 2S O _ 3 (g), ∆ H = -...

For the reaction,

2SO2(g)+O2(g)⇌2SO3(g),2S{O}_{2}\left(g\right) + {O}_{2}\left(g\right) ⇌ 2S{O}_{3}\left(g\right),

∆H=−57.2kJmol−1andKc=1.7×1016.∆ H = - 57.2kJmo{l}^{- 1}and{K}_{c} = 1.7 \times {10}^{16} .

Which of the following statements is incorrect?

Options

  1. AThe equilibrium constant is large suggestive of reaction going to completion and so, no catalyst is required.
  2. BThe equilibrium will shift in forward direction as the pressure increases.
  3. CThe equilibrium constant decreases as the temperature increases.
  4. DThe addition of inert gas at constant volume will not affect the equilibrium constant.

Correct answer

A. The equilibrium constant is large suggestive of reaction going to completion and so, no catalyst is required.

Step-by-step solution

In option \left(B\right) - Δ{n}_{g} is −-ve therefore, increase in pressure will bring reaction in forward direction.

In option (C)−\left(C\right) - as the reaction is exothermic therefore, increase in temperature will decrease the equilibrium constant.

In option (D)−\left(D\right) - Equilibrium constant changes only with temperature.

Hence, option \left(B\right), \left(C\right) and \left(D\right) are correct. Therefore, option \left(A\right) is incorrect choice.
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