Consider the following reaction in a sealed vessel at equilibrium with concentrations of N ₂=3.0 10⁻³ M , O ₂=4.2 10⁻³ M and NO =2.8 10⁻³ M . 2 NO _ ( g ) N _ 2( ~g ) + O _ 2( ~g ) If 0.1 mol L ⁻¹ of NO _ ( g ) is taken in a closed vessel, what will be degree of dissociation ( ) of NO _ ( g ) at equilibrium?

Step-by-step solution

aligned & 2 NO _ ( g ) N _ 2( ~g ) + O _ 2( ~g ) & K _ c = [ N ₂ ] [ O ₂ ] [ NO ]^2 &= 3 10⁻³ 4.2 10⁻³ 2.8 10⁻³ 2.8 10⁻³ &=1.607 & t =0 2 NO _ ( g ) N _ 2( ~g ) + O _ 2( ~g ) & 0.1-0.1 0.05 0.05 & K _ c = 0.05 0.05 (0.1-0.1 )^2 & ~K _ c = 0.05 0.05 0.01(1- )^2 & 1.607= (0.05)^2 ^2 0.01(1- )^2 & ^2 (1- )^2 = 1.607 (0.1)^2 (0.05)^2 aligned gathered 1- = 1.27 0.1 0.05 1- =2.54 =2.54-2.54 3.54 =2.54 = 2.54 3.54 =0.717 gathered