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?
C. 0.717
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
Related: Chemistry — Chemical Equilibrium · All PYQ Banks