At T(K) , the equilibrium constant for the reaction H ₂( ~g )+ Br ₂( ~g ) 2 HBr ( g ) is 1.6 10^5 . If 10 bar ' of HBr is introduced into a sealed vessel at T ( K ) , the equilibrium pressure of HBr (in bar) is approximately

Step-by-step solution

Given H ₂( ~g )+ Br ₂( ~g ) 2 HBr ( ~g ) K _ p =1.6 10^5 Reverse reaction, 2 HBr ( ~g ) H ₂( ~g )+ Br ₂( ~g ), K _ p ^ = 1 ~K _ p HBr ( ~g ) 1 2 H ₂( ~g )+ 1 2 Br ( ~g ), K _ p ^ = ( 1 ~K _ p )^ 1 / 2 At t=0 10 bar 00 At eqm, 10- x x 2 x 2 ( 1 ~K _ p )^ 1 2 = p _ H ₂ ^ 1 / 2 p _ Br ^ 1 / 2 P _ HBr ( 1 1.6 10^5 )^ 1 / 2 = ( x 2 )^ 1 / 2 ( x 2 )^ 1 / 2 (10-x) 10 x2.5 10⁻³= x 20 x =0.05 p _ HBr =10-0.05=9.95 bar.