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QJEEM22Ch13f5: JEE Main 2022 Chemistry - Class 11 JEEM22Ch13f5 Solution: 2NOCl(g) 2NO(g) + Cl 2 (g) In an experiment, 2.0 moles of NOCl…

2NOCl(g) ⇌\rightleftharpoons 2NO(g) + Cl2(g)

In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30 ∘^\circ C is ______________ ×\times 10 −- 4.

Step-by-Step Verified Solution

JEE Main 2022 (Online) 27th June Morning Shift Chemistry - Chemical Equilibrium Question 38 English Explanation

Given that at equilibrium, concentration of NO = 0.4 mol/L

∴\therefore 2x = 0.4

⇒\Rightarrow x = 0.2

∴\therefore Concentration of NOCl at equilibrium,

[NOCl]eq = 2 −- 2 ×\times 0.2 = 1.6

and [NO]eq = 0.4

and [Cl2]eq = 0.2

We know,

KC=[NO]2[Cl2][NOCl]2{K_C} = {{{{[NO]}^2}[C{l_2}]} \over {{{[NOCl]}^2}}}

=[0.4]2[0.2][1.6]2= {{{{[0.4]}^2}[0.2]} \over {{{[1.6]}^2}}}

⇒KC=12.5×10−3\Rightarrow {K_C} = 12.5 \times {10^{ - 3}}

⇒KC=125×10−4\Rightarrow {K_C} = 125 \times {10^{ - 4}}

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